ab-angle->ABCF B

Percentage Accurate: 53.8% → 57.2%

Time: 35.0s

Alternatives: 16

Speedup: 27.9×

• 32.0% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \pi \cdot \frac{angle}{180}\\ \left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin t\_0\right) \cdot \cos t\_0 \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* PI (/ angle 180.0))))
   (* (* (* 2.0 (- (pow b 2.0) (pow a 2.0))) (sin t_0)) (cos t_0))))

C

double code(double a, double b, double angle) {
	double t_0 = ((double) M_PI) * (angle / 180.0);
	return ((2.0 * (pow(b, 2.0) - pow(a, 2.0))) * sin(t_0)) * cos(t_0);
}

Java

public static double code(double a, double b, double angle) {
	double t_0 = Math.PI * (angle / 180.0);
	return ((2.0 * (Math.pow(b, 2.0) - Math.pow(a, 2.0))) * Math.sin(t_0)) * Math.cos(t_0);
}

Python

def code(a, b, angle):
	t_0 = math.pi * (angle / 180.0)
	return ((2.0 * (math.pow(b, 2.0) - math.pow(a, 2.0))) * math.sin(t_0)) * math.cos(t_0)

Julia

function code(a, b, angle)
	t_0 = Float64(pi * Float64(angle / 180.0))
	return Float64(Float64(Float64(2.0 * Float64((b ^ 2.0) - (a ^ 2.0))) * sin(t_0)) * cos(t_0))
end

MATLAB

function tmp = code(a, b, angle)
	t_0 = pi * (angle / 180.0);
	tmp = ((2.0 * ((b ^ 2.0) - (a ^ 2.0))) * sin(t_0)) * cos(t_0);
end

Wolfram

code[a_, b_, angle_] := Block[{t$95$0 = N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]}, N[(N[(N[(2.0 * N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision]]

Initial Program: 53.8% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \pi \cdot \frac{angle}{180}\\ \left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin t\_0\right) \cdot \cos t\_0 \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* PI (/ angle 180.0))))
   (* (* (* 2.0 (- (pow b 2.0) (pow a 2.0))) (sin t_0)) (cos t_0))))

C

double code(double a, double b, double angle) {
	double t_0 = ((double) M_PI) * (angle / 180.0);
	return ((2.0 * (pow(b, 2.0) - pow(a, 2.0))) * sin(t_0)) * cos(t_0);
}

Java

public static double code(double a, double b, double angle) {
	double t_0 = Math.PI * (angle / 180.0);
	return ((2.0 * (Math.pow(b, 2.0) - Math.pow(a, 2.0))) * Math.sin(t_0)) * Math.cos(t_0);
}

Python

def code(a, b, angle):
	t_0 = math.pi * (angle / 180.0)
	return ((2.0 * (math.pow(b, 2.0) - math.pow(a, 2.0))) * math.sin(t_0)) * math.cos(t_0)

Julia

function code(a, b, angle)
	t_0 = Float64(pi * Float64(angle / 180.0))
	return Float64(Float64(Float64(2.0 * Float64((b ^ 2.0) - (a ^ 2.0))) * sin(t_0)) * cos(t_0))
end

MATLAB

function tmp = code(a, b, angle)
	t_0 = pi * (angle / 180.0);
	tmp = ((2.0 * ((b ^ 2.0) - (a ^ 2.0))) * sin(t_0)) * cos(t_0);
end

Wolfram

code[a_, b_, angle_] := Block[{t$95$0 = N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]}, N[(N[(N[(2.0 * N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision]]

Alternative 1: 57.2% accurate, 0.5× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 2 \cdot 10^{-165}:\\ \;\;\;\;\left(t\_0 \cdot \sin \left(\frac{angle\_m \cdot \pi}{180}\right)\right) \cdot \cos \left(\frac{1}{\frac{180}{angle\_m \cdot \pi}}\right)\\ \mathbf{elif}\;{a}^{2} \leq 5 \cdot 10^{+269}:\\ \;\;\;\;\cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle\_m \cdot 0.005555555555555556\right)\right)\right)\right) \cdot \left(t\_0 \cdot \sin \left(\frac{angle\_m}{180} \cdot {\left(\sqrt{\pi}\right)}^{2}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(t\_0 \cdot \sin \left(e^{\log \left(angle\_m \cdot \left(0.005555555555555556 \cdot \pi\right)\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle\_m}{180}\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* 2.0 (* (+ b a) (- b a)))))
   (*
    angle_s
    (if (<= (pow a 2.0) 2e-165)
      (*
       (* t_0 (sin (/ (* angle_m PI) 180.0)))
       (cos (/ 1.0 (/ 180.0 (* angle_m PI)))))
      (if (<= (pow a 2.0) 5e+269)
        (*
         (cos (expm1 (log1p (* PI (* angle_m 0.005555555555555556)))))
         (* t_0 (sin (* (/ angle_m 180.0) (pow (sqrt PI) 2.0)))))
        (*
         (* t_0 (sin (exp (log (* angle_m (* 0.005555555555555556 PI))))))
         (cos (* PI (/ angle_m 180.0)))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if (pow(a, 2.0) <= 2e-165) {
		tmp = (t_0 * sin(((angle_m * ((double) M_PI)) / 180.0))) * cos((1.0 / (180.0 / (angle_m * ((double) M_PI)))));
	} else if (pow(a, 2.0) <= 5e+269) {
		tmp = cos(expm1(log1p((((double) M_PI) * (angle_m * 0.005555555555555556))))) * (t_0 * sin(((angle_m / 180.0) * pow(sqrt(((double) M_PI)), 2.0))));
	} else {
		tmp = (t_0 * sin(exp(log((angle_m * (0.005555555555555556 * ((double) M_PI))))))) * cos((((double) M_PI) * (angle_m / 180.0)));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if (Math.pow(a, 2.0) <= 2e-165) {
		tmp = (t_0 * Math.sin(((angle_m * Math.PI) / 180.0))) * Math.cos((1.0 / (180.0 / (angle_m * Math.PI))));
	} else if (Math.pow(a, 2.0) <= 5e+269) {
		tmp = Math.cos(Math.expm1(Math.log1p((Math.PI * (angle_m * 0.005555555555555556))))) * (t_0 * Math.sin(((angle_m / 180.0) * Math.pow(Math.sqrt(Math.PI), 2.0))));
	} else {
		tmp = (t_0 * Math.sin(Math.exp(Math.log((angle_m * (0.005555555555555556 * Math.PI)))))) * Math.cos((Math.PI * (angle_m / 180.0)));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = 2.0 * ((b + a) * (b - a))
	tmp = 0
	if math.pow(a, 2.0) <= 2e-165:
		tmp = (t_0 * math.sin(((angle_m * math.pi) / 180.0))) * math.cos((1.0 / (180.0 / (angle_m * math.pi))))
	elif math.pow(a, 2.0) <= 5e+269:
		tmp = math.cos(math.expm1(math.log1p((math.pi * (angle_m * 0.005555555555555556))))) * (t_0 * math.sin(((angle_m / 180.0) * math.pow(math.sqrt(math.pi), 2.0))))
	else:
		tmp = (t_0 * math.sin(math.exp(math.log((angle_m * (0.005555555555555556 * math.pi)))))) * math.cos((math.pi * (angle_m / 180.0)))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(2.0 * Float64(Float64(b + a) * Float64(b - a)))
	tmp = 0.0
	if ((a ^ 2.0) <= 2e-165)
		tmp = Float64(Float64(t_0 * sin(Float64(Float64(angle_m * pi) / 180.0))) * cos(Float64(1.0 / Float64(180.0 / Float64(angle_m * pi)))));
	elseif ((a ^ 2.0) <= 5e+269)
		tmp = Float64(cos(expm1(log1p(Float64(pi * Float64(angle_m * 0.005555555555555556))))) * Float64(t_0 * sin(Float64(Float64(angle_m / 180.0) * (sqrt(pi) ^ 2.0)))));
	else
		tmp = Float64(Float64(t_0 * sin(exp(log(Float64(angle_m * Float64(0.005555555555555556 * pi)))))) * cos(Float64(pi * Float64(angle_m / 180.0))));
	end
	return Float64(angle_s * tmp)
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 2e-165], N[(N[(t$95$0 * N[Sin[N[(N[(angle$95$m * Pi), $MachinePrecision] / 180.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(1.0 / N[(180.0 / N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 5e+269], N[(N[Cos[N[(Exp[N[Log[1 + N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision] * N[(t$95$0 * N[Sin[N[(N[(angle$95$m / 180.0), $MachinePrecision] * N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(t$95$0 * N[Sin[N[Exp[N[Log[N[(angle$95$m * N[(0.005555555555555556 * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]

Derivation

1. Split input into 3 regimes

2. if (pow.f64 a 2) < 2e-165

   1. Initial program 56.6%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 56.6%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 56.6%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 56.6%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 56.6%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   5. Step-by-step derivation

      1. associate-*r/ 57.8%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   6. Applied egg-rr 57.8%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   7. Step-by-step derivation

      1. associate-*r/ 56.7%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)} \]

      2. *-commutative 56.7%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \left(\frac{\color{blue}{angle \cdot \pi}}{180}\right) \]

      3. clear-num 60.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{1}{\frac{180}{angle \cdot \pi}}\right)} \]

      4. *-commutative 60.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \left(\frac{1}{\frac{180}{\color{blue}{\pi \cdot angle}}}\right) \]

   8. Applied egg-rr 60.0%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{1}{\frac{180}{\pi \cdot angle}}\right)} \]

   if 2e-165 < (pow.f64 a 2) < 5.0000000000000002e269

   1. Initial program 52.1%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 52.1%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 52.1%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 52.1%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 52.1%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   5. Step-by-step derivation

      1. add-sqr-sqrt 55.9%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\color{blue}{\left(\sqrt{\pi} \cdot \sqrt{\pi}\right)} \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. pow2 55.9%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\color{blue}{{\left(\sqrt{\pi}\right)}^{2}} \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   6. Applied egg-rr 55.9%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\color{blue}{{\left(\sqrt{\pi}\right)}^{2}} \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   7. Step-by-step derivation

      1. div-inv 54.5%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \color{blue}{\left(angle \cdot \frac{1}{180}\right)}\right) \]

      2. metadata-eval 54.5%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot \color{blue}{0.005555555555555556}\right)\right) \]

      3. expm1-log1p-u 45.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right)} \]

   8. Applied egg-rr 45.0%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right)} \]

   if 5.0000000000000002e269 < (pow.f64 a 2)

   1. Initial program 43.3%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 43.3%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 43.3%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 52.7%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 52.7%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   5. Step-by-step derivation

      1. associate-*r/ 51.2%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   6. Applied egg-rr 51.2%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   7. Step-by-step derivation

      1. add-cube-cbrt 48.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\color{blue}{\left(\sqrt[3]{\pi \cdot angle} \cdot \sqrt[3]{\pi \cdot angle}\right) \cdot \sqrt[3]{\pi \cdot angle}}}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 48.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\color{blue}{{\left(\sqrt[3]{\pi \cdot angle}\right)}^{2}} \cdot \sqrt[3]{\pi \cdot angle}}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. associate-/l* 51.8%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{{\left(\sqrt[3]{\pi \cdot angle}\right)}^{2}}{\frac{180}{\sqrt[3]{\pi \cdot angle}}}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      4. add-exp-log 32.4%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(e^{\log \left(\frac{{\left(\sqrt[3]{\pi \cdot angle}\right)}^{2}}{\frac{180}{\sqrt[3]{\pi \cdot angle}}}\right)}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      5. associate-/l* 32.4%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \color{blue}{\left(\frac{{\left(\sqrt[3]{\pi \cdot angle}\right)}^{2} \cdot \sqrt[3]{\pi \cdot angle}}{180}\right)}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      6. unpow2 32.4%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(\frac{\color{blue}{\left(\sqrt[3]{\pi \cdot angle} \cdot \sqrt[3]{\pi \cdot angle}\right)} \cdot \sqrt[3]{\pi \cdot angle}}{180}\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      7. add-cube-cbrt 32.4%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(\frac{\color{blue}{\pi \cdot angle}}{180}\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      8. associate-*r/ 33.7%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \color{blue}{\left(\pi \cdot \frac{angle}{180}\right)}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      9. *-commutative 33.7%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \color{blue}{\left(\frac{angle}{180} \cdot \pi\right)}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      10. div-inv 33.7%

          \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(\color{blue}{\left(angle \cdot \frac{1}{180}\right)} \cdot \pi\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      11. metadata-eval 33.7%

          \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(\left(angle \cdot \color{blue}{0.005555555555555556}\right) \cdot \pi\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      12. associate-*l* 33.7%

          \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \color{blue}{\left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   8. Applied egg-rr 33.7%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(e^{\log \left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
3. Recombined 3 regimes into one program.

4. Final simplification 47.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 2 \cdot 10^{-165}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{angle \cdot \pi}{180}\right)\right) \cdot \cos \left(\frac{1}{\frac{180}{angle \cdot \pi}}\right)\\ \mathbf{elif}\;{a}^{2} \leq 5 \cdot 10^{+269}:\\ \;\;\;\;\cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right) \cdot \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{angle}{180} \cdot {\left(\sqrt{\pi}\right)}^{2}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 57.1% accurate, 0.5× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq -1 \cdot 10^{+273}:\\ \;\;\;\;\left(t\_0 \cdot \sin \left(e^{\log \left(angle\_m \cdot \left(0.005555555555555556 \cdot \pi\right)\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle\_m}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(t\_0 \cdot \sin \left(\frac{\sqrt{\pi}}{\frac{\frac{180}{angle\_m}}{\sqrt{\pi}}}\right)\right) \cdot \cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle\_m \cdot 0.005555555555555556\right)\right)\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* 2.0 (* (+ b a) (- b a)))))
   (*
    angle_s
    (if (<= (- (pow b 2.0) (pow a 2.0)) -1e+273)
      (*
       (* t_0 (sin (exp (log (* angle_m (* 0.005555555555555556 PI))))))
       (cos (* PI (/ angle_m 180.0))))
      (*
       (* t_0 (sin (/ (sqrt PI) (/ (/ 180.0 angle_m) (sqrt PI)))))
       (cos (expm1 (log1p (* PI (* angle_m 0.005555555555555556))))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if ((pow(b, 2.0) - pow(a, 2.0)) <= -1e+273) {
		tmp = (t_0 * sin(exp(log((angle_m * (0.005555555555555556 * ((double) M_PI))))))) * cos((((double) M_PI) * (angle_m / 180.0)));
	} else {
		tmp = (t_0 * sin((sqrt(((double) M_PI)) / ((180.0 / angle_m) / sqrt(((double) M_PI)))))) * cos(expm1(log1p((((double) M_PI) * (angle_m * 0.005555555555555556)))));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if ((Math.pow(b, 2.0) - Math.pow(a, 2.0)) <= -1e+273) {
		tmp = (t_0 * Math.sin(Math.exp(Math.log((angle_m * (0.005555555555555556 * Math.PI)))))) * Math.cos((Math.PI * (angle_m / 180.0)));
	} else {
		tmp = (t_0 * Math.sin((Math.sqrt(Math.PI) / ((180.0 / angle_m) / Math.sqrt(Math.PI))))) * Math.cos(Math.expm1(Math.log1p((Math.PI * (angle_m * 0.005555555555555556)))));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = 2.0 * ((b + a) * (b - a))
	tmp = 0
	if (math.pow(b, 2.0) - math.pow(a, 2.0)) <= -1e+273:
		tmp = (t_0 * math.sin(math.exp(math.log((angle_m * (0.005555555555555556 * math.pi)))))) * math.cos((math.pi * (angle_m / 180.0)))
	else:
		tmp = (t_0 * math.sin((math.sqrt(math.pi) / ((180.0 / angle_m) / math.sqrt(math.pi))))) * math.cos(math.expm1(math.log1p((math.pi * (angle_m * 0.005555555555555556)))))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(2.0 * Float64(Float64(b + a) * Float64(b - a)))
	tmp = 0.0
	if (Float64((b ^ 2.0) - (a ^ 2.0)) <= -1e+273)
		tmp = Float64(Float64(t_0 * sin(exp(log(Float64(angle_m * Float64(0.005555555555555556 * pi)))))) * cos(Float64(pi * Float64(angle_m / 180.0))));
	else
		tmp = Float64(Float64(t_0 * sin(Float64(sqrt(pi) / Float64(Float64(180.0 / angle_m) / sqrt(pi))))) * cos(expm1(log1p(Float64(pi * Float64(angle_m * 0.005555555555555556))))));
	end
	return Float64(angle_s * tmp)
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision], -1e+273], N[(N[(t$95$0 * N[Sin[N[Exp[N[Log[N[(angle$95$m * N[(0.005555555555555556 * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[(t$95$0 * N[Sin[N[(N[Sqrt[Pi], $MachinePrecision] / N[(N[(180.0 / angle$95$m), $MachinePrecision] / N[Sqrt[Pi], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(Exp[N[Log[1 + N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (-.f64 (pow.f64 b 2) (pow.f64 a 2)) < -9.99999999999999945e272

   1. Initial program 51.3%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 51.3%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 51.3%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 51.3%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 51.3%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   5. Step-by-step derivation

      1. associate-*r/ 49.5%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   6. Applied egg-rr 49.5%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   7. Step-by-step derivation

      1. add-cube-cbrt 42.6%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\color{blue}{\left(\sqrt[3]{\pi \cdot angle} \cdot \sqrt[3]{\pi \cdot angle}\right) \cdot \sqrt[3]{\pi \cdot angle}}}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 42.6%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\color{blue}{{\left(\sqrt[3]{\pi \cdot angle}\right)}^{2}} \cdot \sqrt[3]{\pi \cdot angle}}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. associate-/l* 45.7%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{{\left(\sqrt[3]{\pi \cdot angle}\right)}^{2}}{\frac{180}{\sqrt[3]{\pi \cdot angle}}}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      4. add-exp-log 25.9%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(e^{\log \left(\frac{{\left(\sqrt[3]{\pi \cdot angle}\right)}^{2}}{\frac{180}{\sqrt[3]{\pi \cdot angle}}}\right)}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      5. associate-/l* 25.9%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \color{blue}{\left(\frac{{\left(\sqrt[3]{\pi \cdot angle}\right)}^{2} \cdot \sqrt[3]{\pi \cdot angle}}{180}\right)}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      6. unpow2 25.9%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(\frac{\color{blue}{\left(\sqrt[3]{\pi \cdot angle} \cdot \sqrt[3]{\pi \cdot angle}\right)} \cdot \sqrt[3]{\pi \cdot angle}}{180}\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      7. add-cube-cbrt 25.9%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(\frac{\color{blue}{\pi \cdot angle}}{180}\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      8. associate-*r/ 27.5%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \color{blue}{\left(\pi \cdot \frac{angle}{180}\right)}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      9. *-commutative 27.5%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \color{blue}{\left(\frac{angle}{180} \cdot \pi\right)}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      10. div-inv 27.5%

          \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(\color{blue}{\left(angle \cdot \frac{1}{180}\right)} \cdot \pi\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      11. metadata-eval 27.5%

          \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(\left(angle \cdot \color{blue}{0.005555555555555556}\right) \cdot \pi\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      12. associate-*l* 27.5%

          \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \color{blue}{\left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   8. Applied egg-rr 27.5%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(e^{\log \left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   if -9.99999999999999945e272 < (-.f64 (pow.f64 b 2) (pow.f64 a 2))

   1. Initial program 51.1%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 51.1%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 51.1%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 54.8%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 54.8%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   5. Step-by-step derivation

      1. div-inv 55.8%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \color{blue}{\left(angle \cdot \frac{1}{180}\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. metadata-eval 55.8%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot \color{blue}{0.005555555555555556}\right)\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. add-cube-cbrt 55.9%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\left(\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)} \cdot \sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right) \cdot \sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      4. pow3 55.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left({\left(\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right)}^{3}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   6. Applied egg-rr 55.0%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left({\left(\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right)}^{3}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   7. Step-by-step derivation

      1. div-inv 54.5%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \color{blue}{\left(angle \cdot \frac{1}{180}\right)}\right) \]

      2. metadata-eval 54.5%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot \color{blue}{0.005555555555555556}\right)\right) \]

      3. expm1-log1p-u 45.7%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right)} \]

   8. Applied egg-rr 44.7%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left({\left(\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right)}^{3}\right)\right) \cdot \cos \color{blue}{\left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right)} \]
   9. Step-by-step derivation

      1. rem-cube-cbrt 45.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}\right) \cdot \cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right) \]

      2. metadata-eval 45.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot \color{blue}{\frac{1}{180}}\right)\right)\right) \cdot \cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right) \]

      3. div-inv 45.3%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \color{blue}{\frac{angle}{180}}\right)\right) \cdot \cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right) \]

      4. associate-*r/ 44.5%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right) \]

      5. associate-/l* 45.6%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi}{\frac{180}{angle}}\right)}\right) \cdot \cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right) \]

      6. add-sqr-sqrt 46.2%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\color{blue}{\sqrt{\pi} \cdot \sqrt{\pi}}}{\frac{180}{angle}}\right)\right) \cdot \cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right) \]

      7. associate-/l* 46.4%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\sqrt{\pi}}{\frac{\frac{180}{angle}}{\sqrt{\pi}}}\right)}\right) \cdot \cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right) \]

   10. Applied egg-rr 46.4%

       \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\sqrt{\pi}}{\frac{\frac{180}{angle}}{\sqrt{\pi}}}\right)}\right) \cdot \cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 41.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq -1 \cdot 10^{+273}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\sqrt{\pi}}{\frac{\frac{180}{angle}}{\sqrt{\pi}}}\right)\right) \cdot \cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 57.6% accurate, 0.7× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\\ t_1 := \pi \cdot \frac{angle\_m}{180}\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq -2 \cdot 10^{+302}:\\ \;\;\;\;\left(t\_0 \cdot \sin \left(e^{\log \left(angle\_m \cdot \left(0.005555555555555556 \cdot \pi\right)\right)}\right)\right) \cdot \cos t\_1\\ \mathbf{else}:\\ \;\;\;\;\left(t\_0 \cdot \sin t\_1\right) \cdot \cos \left(\frac{angle\_m}{180} \cdot {\left(\sqrt{\pi}\right)}^{2}\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* 2.0 (* (+ b a) (- b a)))) (t_1 (* PI (/ angle_m 180.0))))
   (*
    angle_s
    (if (<= (- (pow b 2.0) (pow a 2.0)) -2e+302)
      (*
       (* t_0 (sin (exp (log (* angle_m (* 0.005555555555555556 PI))))))
       (cos t_1))
      (* (* t_0 (sin t_1)) (cos (* (/ angle_m 180.0) (pow (sqrt PI) 2.0))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b + a) * (b - a));
	double t_1 = ((double) M_PI) * (angle_m / 180.0);
	double tmp;
	if ((pow(b, 2.0) - pow(a, 2.0)) <= -2e+302) {
		tmp = (t_0 * sin(exp(log((angle_m * (0.005555555555555556 * ((double) M_PI))))))) * cos(t_1);
	} else {
		tmp = (t_0 * sin(t_1)) * cos(((angle_m / 180.0) * pow(sqrt(((double) M_PI)), 2.0)));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b + a) * (b - a));
	double t_1 = Math.PI * (angle_m / 180.0);
	double tmp;
	if ((Math.pow(b, 2.0) - Math.pow(a, 2.0)) <= -2e+302) {
		tmp = (t_0 * Math.sin(Math.exp(Math.log((angle_m * (0.005555555555555556 * Math.PI)))))) * Math.cos(t_1);
	} else {
		tmp = (t_0 * Math.sin(t_1)) * Math.cos(((angle_m / 180.0) * Math.pow(Math.sqrt(Math.PI), 2.0)));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = 2.0 * ((b + a) * (b - a))
	t_1 = math.pi * (angle_m / 180.0)
	tmp = 0
	if (math.pow(b, 2.0) - math.pow(a, 2.0)) <= -2e+302:
		tmp = (t_0 * math.sin(math.exp(math.log((angle_m * (0.005555555555555556 * math.pi)))))) * math.cos(t_1)
	else:
		tmp = (t_0 * math.sin(t_1)) * math.cos(((angle_m / 180.0) * math.pow(math.sqrt(math.pi), 2.0)))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(2.0 * Float64(Float64(b + a) * Float64(b - a)))
	t_1 = Float64(pi * Float64(angle_m / 180.0))
	tmp = 0.0
	if (Float64((b ^ 2.0) - (a ^ 2.0)) <= -2e+302)
		tmp = Float64(Float64(t_0 * sin(exp(log(Float64(angle_m * Float64(0.005555555555555556 * pi)))))) * cos(t_1));
	else
		tmp = Float64(Float64(t_0 * sin(t_1)) * cos(Float64(Float64(angle_m / 180.0) * (sqrt(pi) ^ 2.0))));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = 2.0 * ((b + a) * (b - a));
	t_1 = pi * (angle_m / 180.0);
	tmp = 0.0;
	if (((b ^ 2.0) - (a ^ 2.0)) <= -2e+302)
		tmp = (t_0 * sin(exp(log((angle_m * (0.005555555555555556 * pi)))))) * cos(t_1);
	else
		tmp = (t_0 * sin(t_1)) * cos(((angle_m / 180.0) * (sqrt(pi) ^ 2.0)));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision], -2e+302], N[(N[(t$95$0 * N[Sin[N[Exp[N[Log[N[(angle$95$m * N[(0.005555555555555556 * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$1], $MachinePrecision]), $MachinePrecision], N[(N[(t$95$0 * N[Sin[t$95$1], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(N[(angle$95$m / 180.0), $MachinePrecision] * N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (-.f64 (pow.f64 b 2) (pow.f64 a 2)) < -2.0000000000000002e302

   1. Initial program 52.3%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 52.3%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 52.3%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 52.3%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 52.3%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   5. Step-by-step derivation

      1. associate-*r/ 50.4%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   6. Applied egg-rr 50.4%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   7. Step-by-step derivation

      1. add-cube-cbrt 42.3%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\color{blue}{\left(\sqrt[3]{\pi \cdot angle} \cdot \sqrt[3]{\pi \cdot angle}\right) \cdot \sqrt[3]{\pi \cdot angle}}}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 42.3%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\color{blue}{{\left(\sqrt[3]{\pi \cdot angle}\right)}^{2}} \cdot \sqrt[3]{\pi \cdot angle}}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. associate-/l* 45.3%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{{\left(\sqrt[3]{\pi \cdot angle}\right)}^{2}}{\frac{180}{\sqrt[3]{\pi \cdot angle}}}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      4. add-exp-log 26.2%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(e^{\log \left(\frac{{\left(\sqrt[3]{\pi \cdot angle}\right)}^{2}}{\frac{180}{\sqrt[3]{\pi \cdot angle}}}\right)}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      5. associate-/l* 26.2%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \color{blue}{\left(\frac{{\left(\sqrt[3]{\pi \cdot angle}\right)}^{2} \cdot \sqrt[3]{\pi \cdot angle}}{180}\right)}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      6. unpow2 26.2%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(\frac{\color{blue}{\left(\sqrt[3]{\pi \cdot angle} \cdot \sqrt[3]{\pi \cdot angle}\right)} \cdot \sqrt[3]{\pi \cdot angle}}{180}\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      7. add-cube-cbrt 26.2%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(\frac{\color{blue}{\pi \cdot angle}}{180}\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      8. associate-*r/ 27.9%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \color{blue}{\left(\pi \cdot \frac{angle}{180}\right)}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      9. *-commutative 27.9%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \color{blue}{\left(\frac{angle}{180} \cdot \pi\right)}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      10. div-inv 27.9%

          \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(\color{blue}{\left(angle \cdot \frac{1}{180}\right)} \cdot \pi\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      11. metadata-eval 27.9%

          \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(\left(angle \cdot \color{blue}{0.005555555555555556}\right) \cdot \pi\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      12. associate-*l* 27.9%

          \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \color{blue}{\left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   8. Applied egg-rr 27.9%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(e^{\log \left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   if -2.0000000000000002e302 < (-.f64 (pow.f64 b 2) (pow.f64 a 2))

   1. Initial program 50.8%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 50.8%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 50.8%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 54.5%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 54.5%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   5. Step-by-step derivation

      1. add-sqr-sqrt 56.3%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\color{blue}{\left(\sqrt{\pi} \cdot \sqrt{\pi}\right)} \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. pow2 56.3%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\color{blue}{{\left(\sqrt{\pi}\right)}^{2}} \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   6. Applied egg-rr 57.1%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\color{blue}{{\left(\sqrt{\pi}\right)}^{2}} \cdot \frac{angle}{180}\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 50.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq -2 \cdot 10^{+302}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\frac{angle}{180} \cdot {\left(\sqrt{\pi}\right)}^{2}\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 57.4% accurate, 0.8× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \sqrt{angle\_m \cdot \pi}\\ t_1 := 2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{+269}:\\ \;\;\;\;\left(t\_1 \cdot \sin \left(\frac{angle\_m \cdot \pi}{180}\right)\right) \cdot \cos \left(t\_0 \cdot \left(0.005555555555555556 \cdot t\_0\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(t\_1 \cdot \sin \left(e^{\log \left(angle\_m \cdot \left(0.005555555555555556 \cdot \pi\right)\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle\_m}{180}\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (sqrt (* angle_m PI))) (t_1 (* 2.0 (* (+ b a) (- b a)))))
   (*
    angle_s
    (if (<= (pow a 2.0) 5e+269)
      (*
       (* t_1 (sin (/ (* angle_m PI) 180.0)))
       (cos (* t_0 (* 0.005555555555555556 t_0))))
      (*
       (* t_1 (sin (exp (log (* angle_m (* 0.005555555555555556 PI))))))
       (cos (* PI (/ angle_m 180.0))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = sqrt((angle_m * ((double) M_PI)));
	double t_1 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if (pow(a, 2.0) <= 5e+269) {
		tmp = (t_1 * sin(((angle_m * ((double) M_PI)) / 180.0))) * cos((t_0 * (0.005555555555555556 * t_0)));
	} else {
		tmp = (t_1 * sin(exp(log((angle_m * (0.005555555555555556 * ((double) M_PI))))))) * cos((((double) M_PI) * (angle_m / 180.0)));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = Math.sqrt((angle_m * Math.PI));
	double t_1 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if (Math.pow(a, 2.0) <= 5e+269) {
		tmp = (t_1 * Math.sin(((angle_m * Math.PI) / 180.0))) * Math.cos((t_0 * (0.005555555555555556 * t_0)));
	} else {
		tmp = (t_1 * Math.sin(Math.exp(Math.log((angle_m * (0.005555555555555556 * Math.PI)))))) * Math.cos((Math.PI * (angle_m / 180.0)));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = math.sqrt((angle_m * math.pi))
	t_1 = 2.0 * ((b + a) * (b - a))
	tmp = 0
	if math.pow(a, 2.0) <= 5e+269:
		tmp = (t_1 * math.sin(((angle_m * math.pi) / 180.0))) * math.cos((t_0 * (0.005555555555555556 * t_0)))
	else:
		tmp = (t_1 * math.sin(math.exp(math.log((angle_m * (0.005555555555555556 * math.pi)))))) * math.cos((math.pi * (angle_m / 180.0)))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = sqrt(Float64(angle_m * pi))
	t_1 = Float64(2.0 * Float64(Float64(b + a) * Float64(b - a)))
	tmp = 0.0
	if ((a ^ 2.0) <= 5e+269)
		tmp = Float64(Float64(t_1 * sin(Float64(Float64(angle_m * pi) / 180.0))) * cos(Float64(t_0 * Float64(0.005555555555555556 * t_0))));
	else
		tmp = Float64(Float64(t_1 * sin(exp(log(Float64(angle_m * Float64(0.005555555555555556 * pi)))))) * cos(Float64(pi * Float64(angle_m / 180.0))));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = sqrt((angle_m * pi));
	t_1 = 2.0 * ((b + a) * (b - a));
	tmp = 0.0;
	if ((a ^ 2.0) <= 5e+269)
		tmp = (t_1 * sin(((angle_m * pi) / 180.0))) * cos((t_0 * (0.005555555555555556 * t_0)));
	else
		tmp = (t_1 * sin(exp(log((angle_m * (0.005555555555555556 * pi)))))) * cos((pi * (angle_m / 180.0)));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[Sqrt[N[(angle$95$m * Pi), $MachinePrecision]], $MachinePrecision]}, Block[{t$95$1 = N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 5e+269], N[(N[(t$95$1 * N[Sin[N[(N[(angle$95$m * Pi), $MachinePrecision] / 180.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(t$95$0 * N[(0.005555555555555556 * t$95$0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[(t$95$1 * N[Sin[N[Exp[N[Log[N[(angle$95$m * N[(0.005555555555555556 * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a 2) < 5.0000000000000002e269

   1. Initial program 54.5%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 54.5%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 54.5%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 54.5%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 54.5%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   5. Step-by-step derivation

      1. associate-*r/ 55.5%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   6. Applied egg-rr 55.5%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   7. Step-by-step derivation

      1. associate-*r/ 55.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)} \]

      2. *-commutative 55.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \left(\frac{\color{blue}{angle \cdot \pi}}{180}\right) \]

      3. clear-num 56.2%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{1}{\frac{180}{angle \cdot \pi}}\right)} \]

      4. *-commutative 56.2%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \left(\frac{1}{\frac{180}{\color{blue}{\pi \cdot angle}}}\right) \]

   8. Applied egg-rr 56.2%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{1}{\frac{180}{\pi \cdot angle}}\right)} \]
   9. Step-by-step derivation

      1. associate-/r/ 55.2%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{1}{180} \cdot \left(\pi \cdot angle\right)\right)} \]

      2. metadata-eval 55.2%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \left(\color{blue}{0.005555555555555556} \cdot \left(\pi \cdot angle\right)\right) \]

      3. add-sqr-sqrt 30.1%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \left(0.005555555555555556 \cdot \color{blue}{\left(\sqrt{\pi \cdot angle} \cdot \sqrt{\pi \cdot angle}\right)}\right) \]

      4. associate-*r* 30.2%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\left(0.005555555555555556 \cdot \sqrt{\pi \cdot angle}\right) \cdot \sqrt{\pi \cdot angle}\right)} \]

   10. Applied egg-rr 30.2%

       \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\left(0.005555555555555556 \cdot \sqrt{\pi \cdot angle}\right) \cdot \sqrt{\pi \cdot angle}\right)} \]

   if 5.0000000000000002e269 < (pow.f64 a 2)

   1. Initial program 43.3%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 43.3%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 43.3%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 52.7%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 52.7%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   5. Step-by-step derivation

      1. associate-*r/ 51.2%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   6. Applied egg-rr 51.2%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   7. Step-by-step derivation

      1. add-cube-cbrt 48.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\color{blue}{\left(\sqrt[3]{\pi \cdot angle} \cdot \sqrt[3]{\pi \cdot angle}\right) \cdot \sqrt[3]{\pi \cdot angle}}}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 48.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\color{blue}{{\left(\sqrt[3]{\pi \cdot angle}\right)}^{2}} \cdot \sqrt[3]{\pi \cdot angle}}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. associate-/l* 51.8%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{{\left(\sqrt[3]{\pi \cdot angle}\right)}^{2}}{\frac{180}{\sqrt[3]{\pi \cdot angle}}}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      4. add-exp-log 32.4%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(e^{\log \left(\frac{{\left(\sqrt[3]{\pi \cdot angle}\right)}^{2}}{\frac{180}{\sqrt[3]{\pi \cdot angle}}}\right)}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      5. associate-/l* 32.4%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \color{blue}{\left(\frac{{\left(\sqrt[3]{\pi \cdot angle}\right)}^{2} \cdot \sqrt[3]{\pi \cdot angle}}{180}\right)}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      6. unpow2 32.4%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(\frac{\color{blue}{\left(\sqrt[3]{\pi \cdot angle} \cdot \sqrt[3]{\pi \cdot angle}\right)} \cdot \sqrt[3]{\pi \cdot angle}}{180}\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      7. add-cube-cbrt 32.4%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(\frac{\color{blue}{\pi \cdot angle}}{180}\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      8. associate-*r/ 33.7%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \color{blue}{\left(\pi \cdot \frac{angle}{180}\right)}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      9. *-commutative 33.7%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \color{blue}{\left(\frac{angle}{180} \cdot \pi\right)}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      10. div-inv 33.7%

          \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(\color{blue}{\left(angle \cdot \frac{1}{180}\right)} \cdot \pi\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      11. metadata-eval 33.7%

          \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(\left(angle \cdot \color{blue}{0.005555555555555556}\right) \cdot \pi\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      12. associate-*l* 33.7%

          \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \color{blue}{\left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   8. Applied egg-rr 33.7%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(e^{\log \left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 31.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{+269}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{angle \cdot \pi}{180}\right)\right) \cdot \cos \left(\sqrt{angle \cdot \pi} \cdot \left(0.005555555555555556 \cdot \sqrt{angle \cdot \pi}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 57.1% accurate, 0.8× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq 10^{+183}:\\ \;\;\;\;\cos \left(\frac{1}{\frac{180}{angle\_m \cdot \pi}}\right) \cdot \left(t\_0 \cdot \sin \left(\pi \cdot \frac{angle\_m}{180}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;t\_0 \cdot \sin \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle\_m \cdot 0.005555555555555556\right)\right)\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* 2.0 (* (+ b a) (- b a)))))
   (*
    angle_s
    (if (<= (- (pow b 2.0) (pow a 2.0)) 1e+183)
      (*
       (cos (/ 1.0 (/ 180.0 (* angle_m PI))))
       (* t_0 (sin (* PI (/ angle_m 180.0)))))
      (* t_0 (sin (expm1 (log1p (* PI (* angle_m 0.005555555555555556))))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if ((pow(b, 2.0) - pow(a, 2.0)) <= 1e+183) {
		tmp = cos((1.0 / (180.0 / (angle_m * ((double) M_PI))))) * (t_0 * sin((((double) M_PI) * (angle_m / 180.0))));
	} else {
		tmp = t_0 * sin(expm1(log1p((((double) M_PI) * (angle_m * 0.005555555555555556)))));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if ((Math.pow(b, 2.0) - Math.pow(a, 2.0)) <= 1e+183) {
		tmp = Math.cos((1.0 / (180.0 / (angle_m * Math.PI)))) * (t_0 * Math.sin((Math.PI * (angle_m / 180.0))));
	} else {
		tmp = t_0 * Math.sin(Math.expm1(Math.log1p((Math.PI * (angle_m * 0.005555555555555556)))));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = 2.0 * ((b + a) * (b - a))
	tmp = 0
	if (math.pow(b, 2.0) - math.pow(a, 2.0)) <= 1e+183:
		tmp = math.cos((1.0 / (180.0 / (angle_m * math.pi)))) * (t_0 * math.sin((math.pi * (angle_m / 180.0))))
	else:
		tmp = t_0 * math.sin(math.expm1(math.log1p((math.pi * (angle_m * 0.005555555555555556)))))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(2.0 * Float64(Float64(b + a) * Float64(b - a)))
	tmp = 0.0
	if (Float64((b ^ 2.0) - (a ^ 2.0)) <= 1e+183)
		tmp = Float64(cos(Float64(1.0 / Float64(180.0 / Float64(angle_m * pi)))) * Float64(t_0 * sin(Float64(pi * Float64(angle_m / 180.0)))));
	else
		tmp = Float64(t_0 * sin(expm1(log1p(Float64(pi * Float64(angle_m * 0.005555555555555556))))));
	end
	return Float64(angle_s * tmp)
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision], 1e+183], N[(N[Cos[N[(1.0 / N[(180.0 / N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(t$95$0 * N[Sin[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(t$95$0 * N[Sin[N[(Exp[N[Log[1 + N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (-.f64 (pow.f64 b 2) (pow.f64 a 2)) < 9.99999999999999947e182

   1. Initial program 55.8%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 55.8%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 55.8%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 55.8%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 55.8%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   5. Step-by-step derivation

      1. associate-*r/ 55.9%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)} \]

      2. *-commutative 55.9%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \left(\frac{\color{blue}{angle \cdot \pi}}{180}\right) \]

      3. clear-num 56.7%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{1}{\frac{180}{angle \cdot \pi}}\right)} \]

      4. *-commutative 56.7%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \left(\frac{1}{\frac{180}{\color{blue}{\pi \cdot angle}}}\right) \]

   6. Applied egg-rr 57.3%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{1}{\frac{180}{\pi \cdot angle}}\right)} \]

   if 9.99999999999999947e182 < (-.f64 (pow.f64 b 2) (pow.f64 a 2))

   1. Initial program 38.7%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 38.7%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 38.7%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 49.1%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 49.1%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   5. Taylor expanded in angle around 0 47.2%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \color{blue}{1} \]
   6. Step-by-step derivation

      1. div-inv 48.8%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \color{blue}{\left(angle \cdot \frac{1}{180}\right)}\right) \]

      2. metadata-eval 48.8%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot \color{blue}{0.005555555555555556}\right)\right) \]

      3. expm1-log1p-u 43.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right)} \]

   7. Applied egg-rr 42.4%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right)}\right) \cdot 1 \]
3. Recombined 2 regimes into one program.

4. Final simplification 53.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq 10^{+183}:\\ \;\;\;\;\cos \left(\frac{1}{\frac{180}{angle \cdot \pi}}\right) \cdot \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 57.2% accurate, 1.0× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \pi \cdot \frac{angle\_m}{180}\\ t_1 := 2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;a \leq 8.2 \cdot 10^{-91}:\\ \;\;\;\;\left(t\_1 \cdot \sin \left(\frac{angle\_m \cdot \pi}{180}\right)\right) \cdot \cos \left(\frac{1}{\frac{180}{angle\_m \cdot \pi}}\right)\\ \mathbf{elif}\;a \leq 7.2 \cdot 10^{+160}:\\ \;\;\;\;\left(t\_1 \cdot \sin t\_0\right) \cdot \cos \left(\frac{angle\_m}{180} \cdot \sqrt[3]{{\pi}^{3}}\right)\\ \mathbf{else}:\\ \;\;\;\;\cos t\_0 \cdot \left(t\_1 \cdot \sin \left(e^{\log \left(\pi \cdot \left(angle\_m \cdot 0.005555555555555556\right)\right)}\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* PI (/ angle_m 180.0))) (t_1 (* 2.0 (* (+ b a) (- b a)))))
   (*
    angle_s
    (if (<= a 8.2e-91)
      (*
       (* t_1 (sin (/ (* angle_m PI) 180.0)))
       (cos (/ 1.0 (/ 180.0 (* angle_m PI)))))
      (if (<= a 7.2e+160)
        (* (* t_1 (sin t_0)) (cos (* (/ angle_m 180.0) (cbrt (pow PI 3.0)))))
        (*
         (cos t_0)
         (*
          t_1
          (sin (exp (log (* PI (* angle_m 0.005555555555555556))))))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = ((double) M_PI) * (angle_m / 180.0);
	double t_1 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if (a <= 8.2e-91) {
		tmp = (t_1 * sin(((angle_m * ((double) M_PI)) / 180.0))) * cos((1.0 / (180.0 / (angle_m * ((double) M_PI)))));
	} else if (a <= 7.2e+160) {
		tmp = (t_1 * sin(t_0)) * cos(((angle_m / 180.0) * cbrt(pow(((double) M_PI), 3.0))));
	} else {
		tmp = cos(t_0) * (t_1 * sin(exp(log((((double) M_PI) * (angle_m * 0.005555555555555556))))));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = Math.PI * (angle_m / 180.0);
	double t_1 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if (a <= 8.2e-91) {
		tmp = (t_1 * Math.sin(((angle_m * Math.PI) / 180.0))) * Math.cos((1.0 / (180.0 / (angle_m * Math.PI))));
	} else if (a <= 7.2e+160) {
		tmp = (t_1 * Math.sin(t_0)) * Math.cos(((angle_m / 180.0) * Math.cbrt(Math.pow(Math.PI, 3.0))));
	} else {
		tmp = Math.cos(t_0) * (t_1 * Math.sin(Math.exp(Math.log((Math.PI * (angle_m * 0.005555555555555556))))));
	}
	return angle_s * tmp;
}

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(pi * Float64(angle_m / 180.0))
	t_1 = Float64(2.0 * Float64(Float64(b + a) * Float64(b - a)))
	tmp = 0.0
	if (a <= 8.2e-91)
		tmp = Float64(Float64(t_1 * sin(Float64(Float64(angle_m * pi) / 180.0))) * cos(Float64(1.0 / Float64(180.0 / Float64(angle_m * pi)))));
	elseif (a <= 7.2e+160)
		tmp = Float64(Float64(t_1 * sin(t_0)) * cos(Float64(Float64(angle_m / 180.0) * cbrt((pi ^ 3.0)))));
	else
		tmp = Float64(cos(t_0) * Float64(t_1 * sin(exp(log(Float64(pi * Float64(angle_m * 0.005555555555555556)))))));
	end
	return Float64(angle_s * tmp)
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[a, 8.2e-91], N[(N[(t$95$1 * N[Sin[N[(N[(angle$95$m * Pi), $MachinePrecision] / 180.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(1.0 / N[(180.0 / N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[a, 7.2e+160], N[(N[(t$95$1 * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(N[(angle$95$m / 180.0), $MachinePrecision] * N[Power[N[Power[Pi, 3.0], $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[Cos[t$95$0], $MachinePrecision] * N[(t$95$1 * N[Sin[N[Exp[N[Log[N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if a < 8.20000000000000048e-91

   1. Initial program 48.4%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 48.4%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 48.4%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 51.3%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 51.3%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   5. Step-by-step derivation

      1. associate-*r/ 53.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   6. Applied egg-rr 53.0%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   7. Step-by-step derivation

      1. associate-*r/ 53.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)} \]

      2. *-commutative 53.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \left(\frac{\color{blue}{angle \cdot \pi}}{180}\right) \]

      3. clear-num 54.9%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{1}{\frac{180}{angle \cdot \pi}}\right)} \]

      4. *-commutative 54.9%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \left(\frac{1}{\frac{180}{\color{blue}{\pi \cdot angle}}}\right) \]

   8. Applied egg-rr 54.9%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{1}{\frac{180}{\pi \cdot angle}}\right)} \]

   if 8.20000000000000048e-91 < a < 7.20000000000000042e160

   1. Initial program 49.7%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 49.7%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 49.7%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 49.7%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 49.7%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   5. Step-by-step derivation

      1. add-cbrt-cube 54.3%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\color{blue}{\sqrt[3]{\left(\pi \cdot \pi\right) \cdot \pi}} \cdot \frac{angle}{180}\right) \]

      2. pow3 54.3%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\sqrt[3]{\color{blue}{{\pi}^{3}}} \cdot \frac{angle}{180}\right) \]

   6. Applied egg-rr 54.3%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\color{blue}{\sqrt[3]{{\pi}^{3}}} \cdot \frac{angle}{180}\right) \]

   if 7.20000000000000042e160 < a

   1. Initial program 71.9%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 71.9%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 71.9%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 79.3%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 79.3%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   5. Step-by-step derivation

      1. div-inv 71.9%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \color{blue}{\left(angle \cdot \frac{1}{180}\right)}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. metadata-eval 71.9%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot \color{blue}{0.005555555555555556}\right)\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. add-exp-log 45.2%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(e^{\log \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   6. Applied egg-rr 45.2%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(e^{\log \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
3. Recombined 3 regimes into one program.

4. Final simplification 53.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 8.2 \cdot 10^{-91}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{angle \cdot \pi}{180}\right)\right) \cdot \cos \left(\frac{1}{\frac{180}{angle \cdot \pi}}\right)\\ \mathbf{elif}\;a \leq 7.2 \cdot 10^{+160}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\frac{angle}{180} \cdot \sqrt[3]{{\pi}^{3}}\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 57.4% accurate, 1.0× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\\ t_1 := \pi \cdot \frac{angle\_m}{180}\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq 10^{-17}:\\ \;\;\;\;\cos \left(\frac{1}{\frac{180}{angle\_m \cdot \pi}}\right) \cdot \left(t\_0 \cdot \sin t\_1\right)\\ \mathbf{else}:\\ \;\;\;\;\cos t\_1 \cdot \left(t\_0 \cdot \sin \left(0.005555555555555556 \cdot \left(angle\_m \cdot \pi\right)\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* 2.0 (* (+ b a) (- b a)))) (t_1 (* PI (/ angle_m 180.0))))
   (*
    angle_s
    (if (<= (- (pow b 2.0) (pow a 2.0)) 1e-17)
      (* (cos (/ 1.0 (/ 180.0 (* angle_m PI)))) (* t_0 (sin t_1)))
      (* (cos t_1) (* t_0 (sin (* 0.005555555555555556 (* angle_m PI)))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b + a) * (b - a));
	double t_1 = ((double) M_PI) * (angle_m / 180.0);
	double tmp;
	if ((pow(b, 2.0) - pow(a, 2.0)) <= 1e-17) {
		tmp = cos((1.0 / (180.0 / (angle_m * ((double) M_PI))))) * (t_0 * sin(t_1));
	} else {
		tmp = cos(t_1) * (t_0 * sin((0.005555555555555556 * (angle_m * ((double) M_PI)))));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b + a) * (b - a));
	double t_1 = Math.PI * (angle_m / 180.0);
	double tmp;
	if ((Math.pow(b, 2.0) - Math.pow(a, 2.0)) <= 1e-17) {
		tmp = Math.cos((1.0 / (180.0 / (angle_m * Math.PI)))) * (t_0 * Math.sin(t_1));
	} else {
		tmp = Math.cos(t_1) * (t_0 * Math.sin((0.005555555555555556 * (angle_m * Math.PI))));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = 2.0 * ((b + a) * (b - a))
	t_1 = math.pi * (angle_m / 180.0)
	tmp = 0
	if (math.pow(b, 2.0) - math.pow(a, 2.0)) <= 1e-17:
		tmp = math.cos((1.0 / (180.0 / (angle_m * math.pi)))) * (t_0 * math.sin(t_1))
	else:
		tmp = math.cos(t_1) * (t_0 * math.sin((0.005555555555555556 * (angle_m * math.pi))))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(2.0 * Float64(Float64(b + a) * Float64(b - a)))
	t_1 = Float64(pi * Float64(angle_m / 180.0))
	tmp = 0.0
	if (Float64((b ^ 2.0) - (a ^ 2.0)) <= 1e-17)
		tmp = Float64(cos(Float64(1.0 / Float64(180.0 / Float64(angle_m * pi)))) * Float64(t_0 * sin(t_1)));
	else
		tmp = Float64(cos(t_1) * Float64(t_0 * sin(Float64(0.005555555555555556 * Float64(angle_m * pi)))));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = 2.0 * ((b + a) * (b - a));
	t_1 = pi * (angle_m / 180.0);
	tmp = 0.0;
	if (((b ^ 2.0) - (a ^ 2.0)) <= 1e-17)
		tmp = cos((1.0 / (180.0 / (angle_m * pi)))) * (t_0 * sin(t_1));
	else
		tmp = cos(t_1) * (t_0 * sin((0.005555555555555556 * (angle_m * pi))));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision], 1e-17], N[(N[Cos[N[(1.0 / N[(180.0 / N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(t$95$0 * N[Sin[t$95$1], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Cos[t$95$1], $MachinePrecision] * N[(t$95$0 * N[Sin[N[(0.005555555555555556 * N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (-.f64 (pow.f64 b 2) (pow.f64 a 2)) < 1.00000000000000007e-17

   1. Initial program 56.1%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 56.1%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 56.1%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 56.1%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 56.1%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   5. Step-by-step derivation

      1. associate-*r/ 56.1%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)} \]

      2. *-commutative 56.1%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \left(\frac{\color{blue}{angle \cdot \pi}}{180}\right) \]

      3. clear-num 57.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{1}{\frac{180}{angle \cdot \pi}}\right)} \]

      4. *-commutative 57.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \left(\frac{1}{\frac{180}{\color{blue}{\pi \cdot angle}}}\right) \]

   6. Applied egg-rr 58.0%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{1}{\frac{180}{\pi \cdot angle}}\right)} \]

   if 1.00000000000000007e-17 < (-.f64 (pow.f64 b 2) (pow.f64 a 2))

   1. Initial program 42.8%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 42.8%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 42.8%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 50.4%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 50.4%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   5. Taylor expanded in angle around 0 54.4%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 56.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq 10^{-17}:\\ \;\;\;\;\cos \left(\frac{1}{\frac{180}{angle \cdot \pi}}\right) \cdot \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 8: 56.8% accurate, 1.0× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq -1 \cdot 10^{+258}:\\ \;\;\;\;t\_0 \cdot \sin \left(\frac{\pi}{\frac{180}{angle\_m}}\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle\_m}{180}\right) \cdot \left(t\_0 \cdot \sin \left(0.005555555555555556 \cdot \left(angle\_m \cdot \pi\right)\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* 2.0 (* (+ b a) (- b a)))))
   (*
    angle_s
    (if (<= (- (pow b 2.0) (pow a 2.0)) -1e+258)
      (* t_0 (sin (/ PI (/ 180.0 angle_m))))
      (*
       (cos (* PI (/ angle_m 180.0)))
       (* t_0 (sin (* 0.005555555555555556 (* angle_m PI)))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if ((pow(b, 2.0) - pow(a, 2.0)) <= -1e+258) {
		tmp = t_0 * sin((((double) M_PI) / (180.0 / angle_m)));
	} else {
		tmp = cos((((double) M_PI) * (angle_m / 180.0))) * (t_0 * sin((0.005555555555555556 * (angle_m * ((double) M_PI)))));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if ((Math.pow(b, 2.0) - Math.pow(a, 2.0)) <= -1e+258) {
		tmp = t_0 * Math.sin((Math.PI / (180.0 / angle_m)));
	} else {
		tmp = Math.cos((Math.PI * (angle_m / 180.0))) * (t_0 * Math.sin((0.005555555555555556 * (angle_m * Math.PI))));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = 2.0 * ((b + a) * (b - a))
	tmp = 0
	if (math.pow(b, 2.0) - math.pow(a, 2.0)) <= -1e+258:
		tmp = t_0 * math.sin((math.pi / (180.0 / angle_m)))
	else:
		tmp = math.cos((math.pi * (angle_m / 180.0))) * (t_0 * math.sin((0.005555555555555556 * (angle_m * math.pi))))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(2.0 * Float64(Float64(b + a) * Float64(b - a)))
	tmp = 0.0
	if (Float64((b ^ 2.0) - (a ^ 2.0)) <= -1e+258)
		tmp = Float64(t_0 * sin(Float64(pi / Float64(180.0 / angle_m))));
	else
		tmp = Float64(cos(Float64(pi * Float64(angle_m / 180.0))) * Float64(t_0 * sin(Float64(0.005555555555555556 * Float64(angle_m * pi)))));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = 2.0 * ((b + a) * (b - a));
	tmp = 0.0;
	if (((b ^ 2.0) - (a ^ 2.0)) <= -1e+258)
		tmp = t_0 * sin((pi / (180.0 / angle_m)));
	else
		tmp = cos((pi * (angle_m / 180.0))) * (t_0 * sin((0.005555555555555556 * (angle_m * pi))));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision], -1e+258], N[(t$95$0 * N[Sin[N[(Pi / N[(180.0 / angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[Cos[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(t$95$0 * N[Sin[N[(0.005555555555555556 * N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (-.f64 (pow.f64 b 2) (pow.f64 a 2)) < -1.00000000000000006e258

   1. Initial program 52.7%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 52.7%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 52.7%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 52.7%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 52.7%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   5. Taylor expanded in angle around 0 53.5%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \color{blue}{1} \]
   6. Step-by-step derivation

      1. clear-num 54.9%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \color{blue}{\frac{1}{\frac{180}{angle}}}\right)\right) \cdot 1 \]

      2. un-div-inv 56.1%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi}{\frac{180}{angle}}\right)}\right) \cdot 1 \]

   7. Applied egg-rr 56.1%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi}{\frac{180}{angle}}\right)}\right) \cdot 1 \]

   if -1.00000000000000006e258 < (-.f64 (pow.f64 b 2) (pow.f64 a 2))

   1. Initial program 50.6%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 50.6%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 50.6%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 54.4%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 54.4%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   5. Taylor expanded in angle around 0 56.3%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 56.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq -1 \cdot 10^{+258}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi}{\frac{180}{angle}}\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 9: 57.7% accurate, 1.3× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \pi \cdot \frac{angle\_m}{180}\\ t_1 := 2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;{b}^{2} \leq 10^{+47}:\\ \;\;\;\;\left(t\_1 \cdot \sin t\_0\right) \cdot \cos \left(\frac{\pi}{\frac{180}{angle\_m}}\right)\\ \mathbf{else}:\\ \;\;\;\;\cos t\_0 \cdot \left(t\_1 \cdot \sin \left(0.005555555555555556 \cdot \left(angle\_m \cdot \pi\right)\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* PI (/ angle_m 180.0))) (t_1 (* 2.0 (* (+ b a) (- b a)))))
   (*
    angle_s
    (if (<= (pow b 2.0) 1e+47)
      (* (* t_1 (sin t_0)) (cos (/ PI (/ 180.0 angle_m))))
      (* (cos t_0) (* t_1 (sin (* 0.005555555555555556 (* angle_m PI)))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = ((double) M_PI) * (angle_m / 180.0);
	double t_1 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if (pow(b, 2.0) <= 1e+47) {
		tmp = (t_1 * sin(t_0)) * cos((((double) M_PI) / (180.0 / angle_m)));
	} else {
		tmp = cos(t_0) * (t_1 * sin((0.005555555555555556 * (angle_m * ((double) M_PI)))));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = Math.PI * (angle_m / 180.0);
	double t_1 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if (Math.pow(b, 2.0) <= 1e+47) {
		tmp = (t_1 * Math.sin(t_0)) * Math.cos((Math.PI / (180.0 / angle_m)));
	} else {
		tmp = Math.cos(t_0) * (t_1 * Math.sin((0.005555555555555556 * (angle_m * Math.PI))));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = math.pi * (angle_m / 180.0)
	t_1 = 2.0 * ((b + a) * (b - a))
	tmp = 0
	if math.pow(b, 2.0) <= 1e+47:
		tmp = (t_1 * math.sin(t_0)) * math.cos((math.pi / (180.0 / angle_m)))
	else:
		tmp = math.cos(t_0) * (t_1 * math.sin((0.005555555555555556 * (angle_m * math.pi))))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(pi * Float64(angle_m / 180.0))
	t_1 = Float64(2.0 * Float64(Float64(b + a) * Float64(b - a)))
	tmp = 0.0
	if ((b ^ 2.0) <= 1e+47)
		tmp = Float64(Float64(t_1 * sin(t_0)) * cos(Float64(pi / Float64(180.0 / angle_m))));
	else
		tmp = Float64(cos(t_0) * Float64(t_1 * sin(Float64(0.005555555555555556 * Float64(angle_m * pi)))));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = pi * (angle_m / 180.0);
	t_1 = 2.0 * ((b + a) * (b - a));
	tmp = 0.0;
	if ((b ^ 2.0) <= 1e+47)
		tmp = (t_1 * sin(t_0)) * cos((pi / (180.0 / angle_m)));
	else
		tmp = cos(t_0) * (t_1 * sin((0.005555555555555556 * (angle_m * pi))));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[b, 2.0], $MachinePrecision], 1e+47], N[(N[(t$95$1 * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(Pi / N[(180.0 / angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[Cos[t$95$0], $MachinePrecision] * N[(t$95$1 * N[Sin[N[(0.005555555555555556 * N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 b 2) < 1e47

   1. Initial program 55.4%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 55.4%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 55.4%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 55.4%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 55.4%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   5. Step-by-step derivation

      1. clear-num 53.8%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \color{blue}{\frac{1}{\frac{180}{angle}}}\right)\right) \cdot 1 \]

      2. un-div-inv 55.3%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi}{\frac{180}{angle}}\right)}\right) \cdot 1 \]

   6. Applied egg-rr 56.9%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{\pi}{\frac{180}{angle}}\right)} \]

   if 1e47 < (pow.f64 b 2)

   1. Initial program 45.5%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 45.5%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 45.5%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 52.1%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 52.1%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   5. Taylor expanded in angle around 0 55.7%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 56.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} \leq 10^{+47}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 10: 53.9% accurate, 1.8× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ angle\_s \cdot \left(\cos \left(\frac{1}{\frac{180}{angle\_m \cdot \pi}}\right) \cdot \left(\sin \left(\frac{angle\_m \cdot \pi}{180}\right) \cdot \left(2 \cdot \left(b \cdot \left(b - a\right) + a \cdot \left(b - a\right)\right)\right)\right)\right) \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (*
  angle_s
  (*
   (cos (/ 1.0 (/ 180.0 (* angle_m PI))))
   (*
    (sin (/ (* angle_m PI) 180.0))
    (* 2.0 (+ (* b (- b a)) (* a (- b a))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	return angle_s * (cos((1.0 / (180.0 / (angle_m * ((double) M_PI))))) * (sin(((angle_m * ((double) M_PI)) / 180.0)) * (2.0 * ((b * (b - a)) + (a * (b - a))))));
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	return angle_s * (Math.cos((1.0 / (180.0 / (angle_m * Math.PI)))) * (Math.sin(((angle_m * Math.PI) / 180.0)) * (2.0 * ((b * (b - a)) + (a * (b - a))))));
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	return angle_s * (math.cos((1.0 / (180.0 / (angle_m * math.pi)))) * (math.sin(((angle_m * math.pi) / 180.0)) * (2.0 * ((b * (b - a)) + (a * (b - a))))))

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	return Float64(angle_s * Float64(cos(Float64(1.0 / Float64(180.0 / Float64(angle_m * pi)))) * Float64(sin(Float64(Float64(angle_m * pi) / 180.0)) * Float64(2.0 * Float64(Float64(b * Float64(b - a)) + Float64(a * Float64(b - a)))))))
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp = code(angle_s, a, b, angle_m)
	tmp = angle_s * (cos((1.0 / (180.0 / (angle_m * pi)))) * (sin(((angle_m * pi) / 180.0)) * (2.0 * ((b * (b - a)) + (a * (b - a))))));
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := N[(angle$95$s * N[(N[Cos[N[(1.0 / N[(180.0 / N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sin[N[(N[(angle$95$m * Pi), $MachinePrecision] / 180.0), $MachinePrecision]], $MachinePrecision] * N[(2.0 * N[(N[(b * N[(b - a), $MachinePrecision]), $MachinePrecision] + N[(a * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 51.1%

   \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
2. Add Preprocessing
3. Step-by-step derivation

   1. unpow2 51.1%

      \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   2. unpow2 51.1%

      \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   3. difference-of-squares 53.9%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

4. Applied egg-rr 53.9%

   \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
5. Step-by-step derivation

   1. associate-*r/ 54.2%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

6. Applied egg-rr 54.2%

   \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
7. Step-by-step derivation

   1. associate-*r/ 54.7%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)} \]

   2. *-commutative 54.7%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \left(\frac{\color{blue}{angle \cdot \pi}}{180}\right) \]

   3. clear-num 56.0%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{1}{\frac{180}{angle \cdot \pi}}\right)} \]

   4. *-commutative 56.0%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \left(\frac{1}{\frac{180}{\color{blue}{\pi \cdot angle}}}\right) \]

8. Applied egg-rr 56.0%

   \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{1}{\frac{180}{\pi \cdot angle}}\right)} \]
9. Step-by-step derivation

   1. *-commutative 56.0%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b - a\right) \cdot \left(b + a\right)\right)}\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \left(\frac{1}{\frac{180}{\pi \cdot angle}}\right) \]

   2. distribute-lft-in 53.2%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b - a\right) \cdot b + \left(b - a\right) \cdot a\right)}\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \left(\frac{1}{\frac{180}{\pi \cdot angle}}\right) \]

10. Applied egg-rr 53.2%

    \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b - a\right) \cdot b + \left(b - a\right) \cdot a\right)}\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \left(\frac{1}{\frac{180}{\pi \cdot angle}}\right) \]

11. Final simplification 53.2%

    \[\leadsto \cos \left(\frac{1}{\frac{180}{angle \cdot \pi}}\right) \cdot \left(\sin \left(\frac{angle \cdot \pi}{180}\right) \cdot \left(2 \cdot \left(b \cdot \left(b - a\right) + a \cdot \left(b - a\right)\right)\right)\right) \]
12. Add Preprocessing

Alternative 11: 57.7% accurate, 1.9× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ angle\_s \cdot \left(\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{angle\_m \cdot \pi}{180}\right)\right) \cdot \cos \left(\frac{1}{\frac{180}{angle\_m \cdot \pi}}\right)\right) \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (*
  angle_s
  (*
   (* (* 2.0 (* (+ b a) (- b a))) (sin (/ (* angle_m PI) 180.0)))
   (cos (/ 1.0 (/ 180.0 (* angle_m PI)))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	return angle_s * (((2.0 * ((b + a) * (b - a))) * sin(((angle_m * ((double) M_PI)) / 180.0))) * cos((1.0 / (180.0 / (angle_m * ((double) M_PI))))));
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	return angle_s * (((2.0 * ((b + a) * (b - a))) * Math.sin(((angle_m * Math.PI) / 180.0))) * Math.cos((1.0 / (180.0 / (angle_m * Math.PI)))));
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	return angle_s * (((2.0 * ((b + a) * (b - a))) * math.sin(((angle_m * math.pi) / 180.0))) * math.cos((1.0 / (180.0 / (angle_m * math.pi)))))

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	return Float64(angle_s * Float64(Float64(Float64(2.0 * Float64(Float64(b + a) * Float64(b - a))) * sin(Float64(Float64(angle_m * pi) / 180.0))) * cos(Float64(1.0 / Float64(180.0 / Float64(angle_m * pi))))))
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp = code(angle_s, a, b, angle_m)
	tmp = angle_s * (((2.0 * ((b + a) * (b - a))) * sin(((angle_m * pi) / 180.0))) * cos((1.0 / (180.0 / (angle_m * pi)))));
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := N[(angle$95$s * N[(N[(N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(N[(angle$95$m * Pi), $MachinePrecision] / 180.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(1.0 / N[(180.0 / N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 51.1%

   \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
2. Add Preprocessing
3. Step-by-step derivation

   1. unpow2 51.1%

      \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   2. unpow2 51.1%

      \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   3. difference-of-squares 53.9%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

4. Applied egg-rr 53.9%

   \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
5. Step-by-step derivation

   1. associate-*r/ 54.2%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

6. Applied egg-rr 54.2%

   \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
7. Step-by-step derivation

   1. associate-*r/ 54.7%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)} \]

   2. *-commutative 54.7%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \left(\frac{\color{blue}{angle \cdot \pi}}{180}\right) \]

   3. clear-num 56.0%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{1}{\frac{180}{angle \cdot \pi}}\right)} \]

   4. *-commutative 56.0%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \left(\frac{1}{\frac{180}{\color{blue}{\pi \cdot angle}}}\right) \]

8. Applied egg-rr 56.0%

   \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \cos \color{blue}{\left(\frac{1}{\frac{180}{\pi \cdot angle}}\right)} \]

9. Final simplification 56.0%

   \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{angle \cdot \pi}{180}\right)\right) \cdot \cos \left(\frac{1}{\frac{180}{angle \cdot \pi}}\right) \]
10. Add Preprocessing

Alternative 12: 55.4% accurate, 1.9× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 0.005555555555555556 \cdot \left(angle\_m \cdot \pi\right)\\ t_1 := 2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{-139}:\\ \;\;\;\;t\_1 \cdot \sin t\_0\\ \mathbf{else}:\\ \;\;\;\;t\_1 \cdot t\_0\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* 0.005555555555555556 (* angle_m PI)))
        (t_1 (* 2.0 (* (+ b a) (- b a)))))
   (* angle_s (if (<= (pow a 2.0) 5e-139) (* t_1 (sin t_0)) (* t_1 t_0)))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 0.005555555555555556 * (angle_m * ((double) M_PI));
	double t_1 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if (pow(a, 2.0) <= 5e-139) {
		tmp = t_1 * sin(t_0);
	} else {
		tmp = t_1 * t_0;
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 0.005555555555555556 * (angle_m * Math.PI);
	double t_1 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if (Math.pow(a, 2.0) <= 5e-139) {
		tmp = t_1 * Math.sin(t_0);
	} else {
		tmp = t_1 * t_0;
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = 0.005555555555555556 * (angle_m * math.pi)
	t_1 = 2.0 * ((b + a) * (b - a))
	tmp = 0
	if math.pow(a, 2.0) <= 5e-139:
		tmp = t_1 * math.sin(t_0)
	else:
		tmp = t_1 * t_0
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(0.005555555555555556 * Float64(angle_m * pi))
	t_1 = Float64(2.0 * Float64(Float64(b + a) * Float64(b - a)))
	tmp = 0.0
	if ((a ^ 2.0) <= 5e-139)
		tmp = Float64(t_1 * sin(t_0));
	else
		tmp = Float64(t_1 * t_0);
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = 0.005555555555555556 * (angle_m * pi);
	t_1 = 2.0 * ((b + a) * (b - a));
	tmp = 0.0;
	if ((a ^ 2.0) <= 5e-139)
		tmp = t_1 * sin(t_0);
	else
		tmp = t_1 * t_0;
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(0.005555555555555556 * N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 5e-139], N[(t$95$1 * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision], N[(t$95$1 * t$95$0), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a 2) < 5.00000000000000034e-139

   1. Initial program 54.7%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 54.7%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 54.7%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 54.7%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 54.7%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   5. Taylor expanded in angle around 0 52.0%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \color{blue}{1} \]

   6. Taylor expanded in angle around inf 54.3%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \color{blue}{\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}\right) \cdot 1 \]

   if 5.00000000000000034e-139 < (pow.f64 a 2)

   1. Initial program 48.9%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 48.9%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 48.9%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 53.5%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 53.5%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   5. Taylor expanded in angle around 0 51.4%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \color{blue}{1} \]

   6. Taylor expanded in angle around 0 52.5%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \color{blue}{\left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}\right) \cdot 1 \]
3. Recombined 2 regimes into one program.

4. Final simplification 53.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{-139}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 13: 55.4% accurate, 1.9× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 2 \cdot 10^{-165}:\\ \;\;\;\;t\_0 \cdot \sin \left(\pi \cdot \left(angle\_m \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;t\_0 \cdot \left(0.005555555555555556 \cdot \left(angle\_m \cdot \pi\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* 2.0 (* (+ b a) (- b a)))))
   (*
    angle_s
    (if (<= (pow a 2.0) 2e-165)
      (* t_0 (sin (* PI (* angle_m 0.005555555555555556))))
      (* t_0 (* 0.005555555555555556 (* angle_m PI)))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if (pow(a, 2.0) <= 2e-165) {
		tmp = t_0 * sin((((double) M_PI) * (angle_m * 0.005555555555555556)));
	} else {
		tmp = t_0 * (0.005555555555555556 * (angle_m * ((double) M_PI)));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if (Math.pow(a, 2.0) <= 2e-165) {
		tmp = t_0 * Math.sin((Math.PI * (angle_m * 0.005555555555555556)));
	} else {
		tmp = t_0 * (0.005555555555555556 * (angle_m * Math.PI));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = 2.0 * ((b + a) * (b - a))
	tmp = 0
	if math.pow(a, 2.0) <= 2e-165:
		tmp = t_0 * math.sin((math.pi * (angle_m * 0.005555555555555556)))
	else:
		tmp = t_0 * (0.005555555555555556 * (angle_m * math.pi))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(2.0 * Float64(Float64(b + a) * Float64(b - a)))
	tmp = 0.0
	if ((a ^ 2.0) <= 2e-165)
		tmp = Float64(t_0 * sin(Float64(pi * Float64(angle_m * 0.005555555555555556))));
	else
		tmp = Float64(t_0 * Float64(0.005555555555555556 * Float64(angle_m * pi)));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = 2.0 * ((b + a) * (b - a));
	tmp = 0.0;
	if ((a ^ 2.0) <= 2e-165)
		tmp = t_0 * sin((pi * (angle_m * 0.005555555555555556)));
	else
		tmp = t_0 * (0.005555555555555556 * (angle_m * pi));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 2e-165], N[(t$95$0 * N[Sin[N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(t$95$0 * N[(0.005555555555555556 * N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a 2) < 2e-165

   1. Initial program 56.6%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 56.6%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 56.6%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 56.6%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 56.6%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   5. Taylor expanded in angle around 0 53.8%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \color{blue}{1} \]

   6. Taylor expanded in angle around inf 55.0%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \color{blue}{\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}\right) \cdot 1 \]
   7. Step-by-step derivation

      1. associate-*r* 56.4%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\left(0.005555555555555556 \cdot angle\right) \cdot \pi\right)}\right) \cdot 1 \]

      2. *-commutative 56.4%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\color{blue}{\left(angle \cdot 0.005555555555555556\right)} \cdot \pi\right)\right) \cdot 1 \]

      3. *-commutative 56.4%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}\right) \cdot 1 \]

      4. *-commutative 56.4%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \color{blue}{\left(0.005555555555555556 \cdot angle\right)}\right)\right) \cdot 1 \]

   8. Simplified 56.4%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \color{blue}{\sin \left(\pi \cdot \left(0.005555555555555556 \cdot angle\right)\right)}\right) \cdot 1 \]

   if 2e-165 < (pow.f64 a 2)

   1. Initial program 47.9%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 47.9%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 47.9%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 52.4%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 52.4%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   5. Taylor expanded in angle around 0 50.4%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \color{blue}{1} \]

   6. Taylor expanded in angle around 0 51.9%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \color{blue}{\left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}\right) \cdot 1 \]
3. Recombined 2 regimes into one program.

4. Final simplification 53.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 2 \cdot 10^{-165}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 14: 57.3% accurate, 1.9× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ angle\_s \cdot \left(\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{angle\_m \cdot \pi}{180}\right)\right) \cdot \cos \left(0.005555555555555556 \cdot \left(angle\_m \cdot \pi\right)\right)\right) \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (*
  angle_s
  (*
   (* (* 2.0 (* (+ b a) (- b a))) (sin (/ (* angle_m PI) 180.0)))
   (cos (* 0.005555555555555556 (* angle_m PI))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	return angle_s * (((2.0 * ((b + a) * (b - a))) * sin(((angle_m * ((double) M_PI)) / 180.0))) * cos((0.005555555555555556 * (angle_m * ((double) M_PI)))));
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	return angle_s * (((2.0 * ((b + a) * (b - a))) * Math.sin(((angle_m * Math.PI) / 180.0))) * Math.cos((0.005555555555555556 * (angle_m * Math.PI))));
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	return angle_s * (((2.0 * ((b + a) * (b - a))) * math.sin(((angle_m * math.pi) / 180.0))) * math.cos((0.005555555555555556 * (angle_m * math.pi))))

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	return Float64(angle_s * Float64(Float64(Float64(2.0 * Float64(Float64(b + a) * Float64(b - a))) * sin(Float64(Float64(angle_m * pi) / 180.0))) * cos(Float64(0.005555555555555556 * Float64(angle_m * pi)))))
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp = code(angle_s, a, b, angle_m)
	tmp = angle_s * (((2.0 * ((b + a) * (b - a))) * sin(((angle_m * pi) / 180.0))) * cos((0.005555555555555556 * (angle_m * pi))));
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := N[(angle$95$s * N[(N[(N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(N[(angle$95$m * Pi), $MachinePrecision] / 180.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(0.005555555555555556 * N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 51.1%

   \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
2. Add Preprocessing
3. Step-by-step derivation

   1. unpow2 51.1%

      \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   2. unpow2 51.1%

      \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   3. difference-of-squares 53.9%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

4. Applied egg-rr 53.9%

   \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
5. Step-by-step derivation

   1. associate-*r/ 54.2%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

6. Applied egg-rr 54.2%

   \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

7. Taylor expanded in angle around inf 54.9%

   \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right) \cdot \color{blue}{\cos \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)} \]

8. Final simplification 54.9%

   \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{angle \cdot \pi}{180}\right)\right) \cdot \cos \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right) \]
9. Add Preprocessing

Alternative 15: 56.0% accurate, 3.6× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ angle\_s \cdot \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{angle\_m}{\frac{180}{\pi}}\right)\right) \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (* angle_s (* (* 2.0 (* (+ b a) (- b a))) (sin (/ angle_m (/ 180.0 PI))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	return angle_s * ((2.0 * ((b + a) * (b - a))) * sin((angle_m / (180.0 / ((double) M_PI)))));
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	return angle_s * ((2.0 * ((b + a) * (b - a))) * Math.sin((angle_m / (180.0 / Math.PI))));
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	return angle_s * ((2.0 * ((b + a) * (b - a))) * math.sin((angle_m / (180.0 / math.pi))))

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	return Float64(angle_s * Float64(Float64(2.0 * Float64(Float64(b + a) * Float64(b - a))) * sin(Float64(angle_m / Float64(180.0 / pi)))))
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp = code(angle_s, a, b, angle_m)
	tmp = angle_s * ((2.0 * ((b + a) * (b - a))) * sin((angle_m / (180.0 / pi))));
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := N[(angle$95$s * N[(N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(angle$95$m / N[(180.0 / Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 51.1%

   \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
2. Add Preprocessing
3. Step-by-step derivation

   1. unpow2 51.1%

      \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   2. unpow2 51.1%

      \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   3. difference-of-squares 53.9%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

4. Applied egg-rr 53.9%

   \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

5. Taylor expanded in angle around 0 51.6%

   \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \color{blue}{1} \]
6. Step-by-step derivation

   1. associate-*r/ 51.9%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right) \cdot 1 \]

   2. *-commutative 51.9%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\color{blue}{angle \cdot \pi}}{180}\right)\right) \cdot 1 \]

   3. associate-/l* 52.8%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{angle}{\frac{180}{\pi}}\right)}\right) \cdot 1 \]

7. Applied egg-rr 52.8%

   \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left(\frac{angle}{\frac{180}{\pi}}\right)}\right) \cdot 1 \]

8. Final simplification 52.8%

   \[\leadsto \left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{angle}{\frac{180}{\pi}}\right) \]
9. Add Preprocessing

Alternative 16: 54.6% accurate, 27.9× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ angle\_s \cdot \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \left(0.005555555555555556 \cdot \left(angle\_m \cdot \pi\right)\right)\right) \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (*
  angle_s
  (* (* 2.0 (* (+ b a) (- b a))) (* 0.005555555555555556 (* angle_m PI)))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	return angle_s * ((2.0 * ((b + a) * (b - a))) * (0.005555555555555556 * (angle_m * ((double) M_PI))));
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	return angle_s * ((2.0 * ((b + a) * (b - a))) * (0.005555555555555556 * (angle_m * Math.PI)));
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	return angle_s * ((2.0 * ((b + a) * (b - a))) * (0.005555555555555556 * (angle_m * math.pi)))

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	return Float64(angle_s * Float64(Float64(2.0 * Float64(Float64(b + a) * Float64(b - a))) * Float64(0.005555555555555556 * Float64(angle_m * pi))))
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp = code(angle_s, a, b, angle_m)
	tmp = angle_s * ((2.0 * ((b + a) * (b - a))) * (0.005555555555555556 * (angle_m * pi)));
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := N[(angle$95$s * N[(N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(0.005555555555555556 * N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 51.1%

   \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
2. Add Preprocessing
3. Step-by-step derivation

   1. unpow2 51.1%

      \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   2. unpow2 51.1%

      \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   3. difference-of-squares 53.9%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

4. Applied egg-rr 53.9%

   \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

5. Taylor expanded in angle around 0 51.6%

   \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \color{blue}{1} \]

6. Taylor expanded in angle around 0 51.2%

   \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \color{blue}{\left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}\right) \cdot 1 \]

7. Final simplification 51.2%

   \[\leadsto \left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right) \]
8. Add Preprocessing

Reproduce

herbie shell --seed 2024040 
(FPCore (a b angle)
  :name "ab-angle->ABCF B"
  :precision binary64
  (* (* (* 2.0 (- (pow b 2.0) (pow a 2.0))) (sin (* PI (/ angle 180.0)))) (cos (* PI (/ angle 180.0)))))