ab-angle->ABCF B

Percentage Accurate: 54.1% → 57.1%

Time: 55.9s

Alternatives: 13

Speedup: 2.9×

• 31.5% 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: 54.1% 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: 57.1% accurate, 1.0× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\\ t_1 := {\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle_m}{180}\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;\frac{angle_m}{180} \leq 10^{+96}:\\ \;\;\;\;\cos t_1 \cdot \left(t_0 \cdot \sin \left(\frac{\pi \cdot angle_m}{180}\right)\right)\\ \mathbf{elif}\;\frac{angle_m}{180} \leq 5 \cdot 10^{+219}:\\ \;\;\;\;t_0 \cdot \sin \left(e^{\log \left(0.005555555555555556 \cdot \left(\pi \cdot angle_m\right)\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle_m}{180}\right) \cdot \left(t_0 \cdot \sin t_1\right)\\ \end{array} \end{array} \end{array} \]

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 (* (pow (sqrt PI) 2.0) (/ angle_m 180.0))))
   (*
    angle_s
    (if (<= (/ angle_m 180.0) 1e+96)
      (* (cos t_1) (* t_0 (sin (/ (* PI angle_m) 180.0))))
      (if (<= (/ angle_m 180.0) 5e+219)
        (* t_0 (sin (exp (log (* 0.005555555555555556 (* PI angle_m))))))
        (* (cos (* PI (/ angle_m 180.0))) (* t_0 (sin t_1))))))))

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 = pow(sqrt(((double) M_PI)), 2.0) * (angle_m / 180.0);
	double tmp;
	if ((angle_m / 180.0) <= 1e+96) {
		tmp = cos(t_1) * (t_0 * sin(((((double) M_PI) * angle_m) / 180.0)));
	} else if ((angle_m / 180.0) <= 5e+219) {
		tmp = t_0 * sin(exp(log((0.005555555555555556 * (((double) M_PI) * angle_m)))));
	} else {
		tmp = cos((((double) M_PI) * (angle_m / 180.0))) * (t_0 * sin(t_1));
	}
	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.pow(Math.sqrt(Math.PI), 2.0) * (angle_m / 180.0);
	double tmp;
	if ((angle_m / 180.0) <= 1e+96) {
		tmp = Math.cos(t_1) * (t_0 * Math.sin(((Math.PI * angle_m) / 180.0)));
	} else if ((angle_m / 180.0) <= 5e+219) {
		tmp = t_0 * Math.sin(Math.exp(Math.log((0.005555555555555556 * (Math.PI * angle_m)))));
	} else {
		tmp = Math.cos((Math.PI * (angle_m / 180.0))) * (t_0 * Math.sin(t_1));
	}
	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.pow(math.sqrt(math.pi), 2.0) * (angle_m / 180.0)
	tmp = 0
	if (angle_m / 180.0) <= 1e+96:
		tmp = math.cos(t_1) * (t_0 * math.sin(((math.pi * angle_m) / 180.0)))
	elif (angle_m / 180.0) <= 5e+219:
		tmp = t_0 * math.sin(math.exp(math.log((0.005555555555555556 * (math.pi * angle_m)))))
	else:
		tmp = math.cos((math.pi * (angle_m / 180.0))) * (t_0 * math.sin(t_1))
	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((sqrt(pi) ^ 2.0) * Float64(angle_m / 180.0))
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 1e+96)
		tmp = Float64(cos(t_1) * Float64(t_0 * sin(Float64(Float64(pi * angle_m) / 180.0))));
	elseif (Float64(angle_m / 180.0) <= 5e+219)
		tmp = Float64(t_0 * sin(exp(log(Float64(0.005555555555555556 * Float64(pi * angle_m))))));
	else
		tmp = Float64(cos(Float64(pi * Float64(angle_m / 180.0))) * Float64(t_0 * sin(t_1)));
	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 = (sqrt(pi) ^ 2.0) * (angle_m / 180.0);
	tmp = 0.0;
	if ((angle_m / 180.0) <= 1e+96)
		tmp = cos(t_1) * (t_0 * sin(((pi * angle_m) / 180.0)));
	elseif ((angle_m / 180.0) <= 5e+219)
		tmp = t_0 * sin(exp(log((0.005555555555555556 * (pi * angle_m)))));
	else
		tmp = cos((pi * (angle_m / 180.0))) * (t_0 * sin(t_1));
	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[(N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision] * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 1e+96], N[(N[Cos[t$95$1], $MachinePrecision] * N[(t$95$0 * N[Sin[N[(N[(Pi * angle$95$m), $MachinePrecision] / 180.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 5e+219], N[(t$95$0 * N[Sin[N[Exp[N[Log[N[(0.005555555555555556 * N[(Pi * angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[Cos[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(t$95$0 * N[Sin[t$95$1], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 angle 180) < 1.00000000000000005e96

   1. Initial program 63.1%

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

      1. unpow2 63.1%

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

      2. unpow2 63.1%

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

      3. difference-of-squares 66.0%

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

   3. Applied egg-rr 66.0%

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

      1. add-sqr-sqrt 68.3%

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

      2. pow2 68.3%

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

   5. Applied egg-rr 68.3%

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

      1. associate-*r/ 68.5%

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

   7. Applied egg-rr 68.5%

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

   if 1.00000000000000005e96 < (/.f64 angle 180) < 5e219

   1. Initial program 29.8%

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

      1. unpow2 29.8%

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

      2. unpow2 29.8%

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

      3. difference-of-squares 29.8%

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

   3. Applied egg-rr 29.8%

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

   4. Taylor expanded in angle around 0 33.2%

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

      1. *-commutative 33.2%

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

      2. add-exp-log 51.7%

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

      3. div-inv 51.7%

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

      4. metadata-eval 51.7%

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

      5. associate-*r* 51.7%

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

      6. *-commutative 51.7%

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

      7. associate-*l* 51.7%

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

   6. Applied egg-rr 51.7%

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

   if 5e219 < (/.f64 angle 180)

   1. Initial program 6.4%

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

      1. unpow2 6.4%

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

      2. unpow2 6.4%

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

      3. difference-of-squares 20.6%

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

   3. Applied egg-rr 20.6%

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

      1. add-sqr-sqrt 34.2%

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

      2. pow2 34.2%

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

   5. Applied egg-rr 33.5%

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

4. Final simplification 65.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 10^{+96}:\\ \;\;\;\;\cos \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\right)\\ \mathbf{elif}\;\frac{angle}{180} \leq 5 \cdot 10^{+219}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\right)\\ \end{array} \]

Alternative 2: 60.8% accurate, 0.6× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ angle_s \cdot \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq 2 \cdot 10^{+255}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\mathsf{expm1}\left(\mathsf{log1p}\left(0.005555555555555556 \cdot \left(\pi \cdot angle_m\right)\right)\right)\right)\right) \cdot \cos \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle_m}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;{\left(\sqrt[3]{2} \cdot \left({\left(\sqrt[3]{b}\right)}^{2} \cdot \sqrt[3]{\sin \left(\pi \cdot \left(0.005555555555555556 \cdot angle_m\right)\right)}\right)\right)}^{3} \cdot \cos \left(\pi \cdot \frac{angle_m}{180}\right)\\ \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (*
  angle_s
  (if (<= (- (pow b 2.0) (pow a 2.0)) 2e+255)
    (*
     (*
      (* 2.0 (* (+ b a) (- b a)))
      (sin (expm1 (log1p (* 0.005555555555555556 (* PI angle_m))))))
     (cos (* (pow (sqrt PI) 2.0) (/ angle_m 180.0))))
    (*
     (pow
      (*
       (cbrt 2.0)
       (*
        (pow (cbrt b) 2.0)
        (cbrt (sin (* PI (* 0.005555555555555556 angle_m))))))
      3.0)
     (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 tmp;
	if ((pow(b, 2.0) - pow(a, 2.0)) <= 2e+255) {
		tmp = ((2.0 * ((b + a) * (b - a))) * sin(expm1(log1p((0.005555555555555556 * (((double) M_PI) * angle_m)))))) * cos((pow(sqrt(((double) M_PI)), 2.0) * (angle_m / 180.0)));
	} else {
		tmp = pow((cbrt(2.0) * (pow(cbrt(b), 2.0) * cbrt(sin((((double) M_PI) * (0.005555555555555556 * angle_m)))))), 3.0) * 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 tmp;
	if ((Math.pow(b, 2.0) - Math.pow(a, 2.0)) <= 2e+255) {
		tmp = ((2.0 * ((b + a) * (b - a))) * Math.sin(Math.expm1(Math.log1p((0.005555555555555556 * (Math.PI * angle_m)))))) * Math.cos((Math.pow(Math.sqrt(Math.PI), 2.0) * (angle_m / 180.0)));
	} else {
		tmp = Math.pow((Math.cbrt(2.0) * (Math.pow(Math.cbrt(b), 2.0) * Math.cbrt(Math.sin((Math.PI * (0.005555555555555556 * angle_m)))))), 3.0) * 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)
	tmp = 0.0
	if (Float64((b ^ 2.0) - (a ^ 2.0)) <= 2e+255)
		tmp = Float64(Float64(Float64(2.0 * Float64(Float64(b + a) * Float64(b - a))) * sin(expm1(log1p(Float64(0.005555555555555556 * Float64(pi * angle_m)))))) * cos(Float64((sqrt(pi) ^ 2.0) * Float64(angle_m / 180.0))));
	else
		tmp = Float64((Float64(cbrt(2.0) * Float64((cbrt(b) ^ 2.0) * cbrt(sin(Float64(pi * Float64(0.005555555555555556 * angle_m)))))) ^ 3.0) * 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_] := N[(angle$95$s * If[LessEqual[N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision], 2e+255], N[(N[(N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(Exp[N[Log[1 + N[(0.005555555555555556 * N[(Pi * angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision] * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[Power[N[(N[Power[2.0, 1/3], $MachinePrecision] * N[(N[Power[N[Power[b, 1/3], $MachinePrecision], 2.0], $MachinePrecision] * N[Power[N[Sin[N[(Pi * N[(0.005555555555555556 * angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 3.0], $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 (-.f64 (pow.f64 b 2) (pow.f64 a 2)) < 1.99999999999999998e255

   1. Initial program 60.6%

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

      1. unpow2 60.6%

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

      2. unpow2 60.6%

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

      3. difference-of-squares 60.6%

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

   3. Applied egg-rr 60.6%

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

      1. add-sqr-sqrt 63.7%

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

      2. pow2 63.7%

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

   5. Applied egg-rr 63.7%

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

      1. *-commutative 63.7%

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

      2. div-inv 61.6%

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

      3. metadata-eval 61.6%

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

      4. associate-*r* 61.8%

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

      5. expm1-log1p-u 57.5%

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

      6. *-commutative 57.5%

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

      7. associate-*l* 57.5%

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

   7. Applied egg-rr 57.5%

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

   if 1.99999999999999998e255 < (-.f64 (pow.f64 b 2) (pow.f64 a 2))

   1. Initial program 45.8%

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

      1. add-cube-cbrt 45.6%

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

      2. pow3 45.6%

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

      3. *-commutative 45.6%

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

      4. div-inv 44.2%

         \[\leadsto {\left(\sqrt[3]{\sin \left(\pi \cdot \color{blue}{\left(angle \cdot \frac{1}{180}\right)}\right) \cdot \left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right)}\right)}^{3} \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      5. metadata-eval 44.2%

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

   3. Applied egg-rr 44.2%

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

   4. Taylor expanded in a around 0 35.3%

      \[\leadsto {\color{blue}{\left({\left(1 \cdot \left({b}^{2} \cdot \sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\right)}^{0.3333333333333333} \cdot \sqrt[3]{2}\right)}}^{3} \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   5. Step-by-step derivation

      1. *-commutative 35.3%

         \[\leadsto {\color{blue}{\left(\sqrt[3]{2} \cdot {\left(1 \cdot \left({b}^{2} \cdot \sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\right)}^{0.3333333333333333}\right)}}^{3} \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow1/3 51.6%

         \[\leadsto {\left(\sqrt[3]{2} \cdot \color{blue}{\sqrt[3]{1 \cdot \left({b}^{2} \cdot \sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)}}\right)}^{3} \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. *-lft-identity 51.6%

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

      4. *-commutative 51.6%

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

      5. *-commutative 51.6%

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

      6. *-commutative 51.6%

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

      7. associate-*r* 54.4%

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

      8. *-commutative 54.4%

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

   6. Simplified 54.4%

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

      1. *-commutative 54.4%

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

      2. *-commutative 54.4%

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

      3. cbrt-prod 54.5%

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

      4. unpow2 54.5%

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

      5. cbrt-prod 70.0%

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

      6. pow2 70.0%

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

      7. *-commutative 70.0%

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

   8. Applied egg-rr 70.0%

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

4. Final simplification 60.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq 2 \cdot 10^{+255}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\mathsf{expm1}\left(\mathsf{log1p}\left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)\right)\right) \cdot \cos \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;{\left(\sqrt[3]{2} \cdot \left({\left(\sqrt[3]{b}\right)}^{2} \cdot \sqrt[3]{\sin \left(\pi \cdot \left(0.005555555555555556 \cdot angle\right)\right)}\right)\right)}^{3} \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\\ \end{array} \]

Alternative 3: 57.3% accurate, 0.7× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 0.005555555555555556 \cdot \left(\pi \cdot angle_m\right)\\ t_1 := \left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\mathsf{expm1}\left(\mathsf{log1p}\left(t_0\right)\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;{b}^{2} \leq 2 \cdot 10^{+172}:\\ \;\;\;\;t_1 \cdot \cos \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle_m}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;t_1 \cdot \mathsf{expm1}\left(\mathsf{log1p}\left(\cos t_0\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* 0.005555555555555556 (* PI angle_m)))
        (t_1 (* (* 2.0 (* (+ b a) (- b a))) (sin (expm1 (log1p t_0))))))
   (*
    angle_s
    (if (<= (pow b 2.0) 2e+172)
      (* t_1 (cos (* (pow (sqrt PI) 2.0) (/ angle_m 180.0))))
      (* t_1 (expm1 (log1p (cos 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 * (((double) M_PI) * angle_m);
	double t_1 = (2.0 * ((b + a) * (b - a))) * sin(expm1(log1p(t_0)));
	double tmp;
	if (pow(b, 2.0) <= 2e+172) {
		tmp = t_1 * cos((pow(sqrt(((double) M_PI)), 2.0) * (angle_m / 180.0)));
	} else {
		tmp = t_1 * expm1(log1p(cos(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 * (Math.PI * angle_m);
	double t_1 = (2.0 * ((b + a) * (b - a))) * Math.sin(Math.expm1(Math.log1p(t_0)));
	double tmp;
	if (Math.pow(b, 2.0) <= 2e+172) {
		tmp = t_1 * Math.cos((Math.pow(Math.sqrt(Math.PI), 2.0) * (angle_m / 180.0)));
	} else {
		tmp = t_1 * Math.expm1(Math.log1p(Math.cos(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 * (math.pi * angle_m)
	t_1 = (2.0 * ((b + a) * (b - a))) * math.sin(math.expm1(math.log1p(t_0)))
	tmp = 0
	if math.pow(b, 2.0) <= 2e+172:
		tmp = t_1 * math.cos((math.pow(math.sqrt(math.pi), 2.0) * (angle_m / 180.0)))
	else:
		tmp = t_1 * math.expm1(math.log1p(math.cos(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(pi * angle_m))
	t_1 = Float64(Float64(2.0 * Float64(Float64(b + a) * Float64(b - a))) * sin(expm1(log1p(t_0))))
	tmp = 0.0
	if ((b ^ 2.0) <= 2e+172)
		tmp = Float64(t_1 * cos(Float64((sqrt(pi) ^ 2.0) * Float64(angle_m / 180.0))));
	else
		tmp = Float64(t_1 * expm1(log1p(cos(t_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[(0.005555555555555556 * N[(Pi * angle$95$m), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(Exp[N[Log[1 + t$95$0], $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[b, 2.0], $MachinePrecision], 2e+172], N[(t$95$1 * N[Cos[N[(N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision] * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(t$95$1 * N[(Exp[N[Log[1 + N[Cos[t$95$0], $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 b 2) < 2.0000000000000002e172

   1. Initial program 60.7%

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

      1. unpow2 60.7%

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

      2. unpow2 60.7%

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

      3. difference-of-squares 60.7%

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

   3. Applied egg-rr 60.7%

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

      1. add-sqr-sqrt 63.4%

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

      2. pow2 63.4%

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

   5. Applied egg-rr 63.4%

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

      1. *-commutative 63.4%

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

      2. div-inv 61.1%

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

      3. metadata-eval 61.1%

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

      4. associate-*r* 61.3%

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

      5. expm1-log1p-u 57.9%

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

      6. *-commutative 57.9%

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

      7. associate-*l* 57.8%

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

   7. Applied egg-rr 57.8%

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

   if 2.0000000000000002e172 < (pow.f64 b 2)

   1. Initial program 48.3%

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

      1. unpow2 48.3%

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

      2. unpow2 48.3%

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

      3. difference-of-squares 56.8%

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

   3. Applied egg-rr 56.8%

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

      1. add-sqr-sqrt 61.0%

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

      2. pow2 61.0%

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

   5. Applied egg-rr 61.0%

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

      1. *-commutative 61.0%

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

      2. div-inv 59.9%

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

      3. metadata-eval 59.9%

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

      4. associate-*r* 58.4%

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

      5. expm1-log1p-u 48.7%

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

      6. *-commutative 48.7%

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

      7. associate-*l* 48.7%

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

   7. Applied egg-rr 48.7%

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

      1. expm1-log1p-u 48.7%

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

      2. unpow2 48.7%

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

      3. add-sqr-sqrt 48.7%

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

      4. associate-*r/ 49.9%

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

      5. div-inv 50.1%

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

      6. metadata-eval 50.1%

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

      7. *-commutative 50.1%

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

   9. Applied egg-rr 50.1%

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

4. Final simplification 55.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} \leq 2 \cdot 10^{+172}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\mathsf{expm1}\left(\mathsf{log1p}\left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)\right)\right) \cdot \cos \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\mathsf{expm1}\left(\mathsf{log1p}\left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)\right)\right) \cdot \mathsf{expm1}\left(\mathsf{log1p}\left(\cos \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)\right)\\ \end{array} \]

Alternative 4: 57.2% accurate, 0.7× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 0.005555555555555556 \cdot \left(\pi \cdot angle_m\right)\\ t_1 := 2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;{b}^{2} \leq 4 \cdot 10^{+77}:\\ \;\;\;\;\cos \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle_m}{180}\right) \cdot \left(t_1 \cdot \sin \left(\pi \cdot \frac{angle_m}{180}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(t_1 \cdot \sin \left(\mathsf{expm1}\left(\mathsf{log1p}\left(t_0\right)\right)\right)\right) \cdot \mathsf{expm1}\left(\mathsf{log1p}\left(\cos t_0\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* 0.005555555555555556 (* PI angle_m)))
        (t_1 (* 2.0 (* (+ b a) (- b a)))))
   (*
    angle_s
    (if (<= (pow b 2.0) 4e+77)
      (*
       (cos (* (pow (sqrt PI) 2.0) (/ angle_m 180.0)))
       (* t_1 (sin (* PI (/ angle_m 180.0)))))
      (* (* t_1 (sin (expm1 (log1p t_0)))) (expm1 (log1p (cos 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 * (((double) M_PI) * angle_m);
	double t_1 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if (pow(b, 2.0) <= 4e+77) {
		tmp = cos((pow(sqrt(((double) M_PI)), 2.0) * (angle_m / 180.0))) * (t_1 * sin((((double) M_PI) * (angle_m / 180.0))));
	} else {
		tmp = (t_1 * sin(expm1(log1p(t_0)))) * expm1(log1p(cos(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 * (Math.PI * angle_m);
	double t_1 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if (Math.pow(b, 2.0) <= 4e+77) {
		tmp = Math.cos((Math.pow(Math.sqrt(Math.PI), 2.0) * (angle_m / 180.0))) * (t_1 * Math.sin((Math.PI * (angle_m / 180.0))));
	} else {
		tmp = (t_1 * Math.sin(Math.expm1(Math.log1p(t_0)))) * Math.expm1(Math.log1p(Math.cos(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 * (math.pi * angle_m)
	t_1 = 2.0 * ((b + a) * (b - a))
	tmp = 0
	if math.pow(b, 2.0) <= 4e+77:
		tmp = math.cos((math.pow(math.sqrt(math.pi), 2.0) * (angle_m / 180.0))) * (t_1 * math.sin((math.pi * (angle_m / 180.0))))
	else:
		tmp = (t_1 * math.sin(math.expm1(math.log1p(t_0)))) * math.expm1(math.log1p(math.cos(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(pi * angle_m))
	t_1 = Float64(2.0 * Float64(Float64(b + a) * Float64(b - a)))
	tmp = 0.0
	if ((b ^ 2.0) <= 4e+77)
		tmp = Float64(cos(Float64((sqrt(pi) ^ 2.0) * Float64(angle_m / 180.0))) * Float64(t_1 * sin(Float64(pi * Float64(angle_m / 180.0)))));
	else
		tmp = Float64(Float64(t_1 * sin(expm1(log1p(t_0)))) * expm1(log1p(cos(t_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[(0.005555555555555556 * N[(Pi * angle$95$m), $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], 4e+77], N[(N[Cos[N[(N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision] * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(t$95$1 * N[Sin[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(t$95$1 * N[Sin[N[(Exp[N[Log[1 + t$95$0], $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[(Exp[N[Log[1 + N[Cos[t$95$0], $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 b 2) < 3.99999999999999993e77

   1. Initial program 62.4%

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

      1. unpow2 62.4%

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

      2. unpow2 62.4%

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

      3. difference-of-squares 62.4%

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

   3. Applied egg-rr 62.4%

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

      1. add-sqr-sqrt 65.3%

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

      2. pow2 65.3%

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

   5. Applied egg-rr 65.3%

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

   if 3.99999999999999993e77 < (pow.f64 b 2)

   1. Initial program 47.0%

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

      1. unpow2 47.0%

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

      2. unpow2 47.0%

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

      3. difference-of-squares 54.5%

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

   3. Applied egg-rr 54.5%

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

      1. add-sqr-sqrt 58.3%

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

      2. pow2 58.3%

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

   5. Applied egg-rr 58.3%

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

      1. *-commutative 58.3%

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

      2. div-inv 57.2%

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

      3. metadata-eval 57.2%

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

      4. associate-*r* 57.1%

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

      5. expm1-log1p-u 47.1%

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

      6. *-commutative 47.1%

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

      7. associate-*l* 47.0%

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

   7. Applied egg-rr 47.0%

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

      1. expm1-log1p-u 47.0%

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

      2. unpow2 47.0%

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

      3. add-sqr-sqrt 47.0%

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

      4. associate-*r/ 47.9%

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

      5. div-inv 48.0%

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

      6. metadata-eval 48.0%

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

      7. *-commutative 48.0%

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

   9. Applied egg-rr 48.0%

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

4. Final simplification 58.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} \leq 4 \cdot 10^{+77}:\\ \;\;\;\;\cos \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\mathsf{expm1}\left(\mathsf{log1p}\left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)\right)\right) \cdot \mathsf{expm1}\left(\mathsf{log1p}\left(\cos \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)\right)\\ \end{array} \]

Alternative 5: 57.0% accurate, 1.0× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;\frac{angle_m}{180} \leq 5 \cdot 10^{+78} \lor \neg \left(\frac{angle_m}{180} \leq 5 \cdot 10^{+186}\right):\\ \;\;\;\;\cos \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle_m}{180}\right) \cdot \left(t_0 \cdot \sin \left(\pi \cdot \frac{angle_m}{180}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;t_0 \cdot \sin \left(e^{\log \left(0.005555555555555556 \cdot \left(\pi \cdot angle_m\right)\right)}\right)\\ \end{array} \end{array} \end{array} \]

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 (or (<= (/ angle_m 180.0) 5e+78) (not (<= (/ angle_m 180.0) 5e+186)))
      (*
       (cos (* (pow (sqrt PI) 2.0) (/ angle_m 180.0)))
       (* t_0 (sin (* PI (/ angle_m 180.0)))))
      (* t_0 (sin (exp (log (* 0.005555555555555556 (* PI angle_m))))))))))

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 (((angle_m / 180.0) <= 5e+78) || !((angle_m / 180.0) <= 5e+186)) {
		tmp = cos((pow(sqrt(((double) M_PI)), 2.0) * (angle_m / 180.0))) * (t_0 * sin((((double) M_PI) * (angle_m / 180.0))));
	} else {
		tmp = t_0 * sin(exp(log((0.005555555555555556 * (((double) M_PI) * angle_m)))));
	}
	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 (((angle_m / 180.0) <= 5e+78) || !((angle_m / 180.0) <= 5e+186)) {
		tmp = Math.cos((Math.pow(Math.sqrt(Math.PI), 2.0) * (angle_m / 180.0))) * (t_0 * Math.sin((Math.PI * (angle_m / 180.0))));
	} else {
		tmp = t_0 * Math.sin(Math.exp(Math.log((0.005555555555555556 * (Math.PI * angle_m)))));
	}
	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 ((angle_m / 180.0) <= 5e+78) or not ((angle_m / 180.0) <= 5e+186):
		tmp = math.cos((math.pow(math.sqrt(math.pi), 2.0) * (angle_m / 180.0))) * (t_0 * math.sin((math.pi * (angle_m / 180.0))))
	else:
		tmp = t_0 * math.sin(math.exp(math.log((0.005555555555555556 * (math.pi * angle_m)))))
	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(angle_m / 180.0) <= 5e+78) || !(Float64(angle_m / 180.0) <= 5e+186))
		tmp = Float64(cos(Float64((sqrt(pi) ^ 2.0) * Float64(angle_m / 180.0))) * Float64(t_0 * sin(Float64(pi * Float64(angle_m / 180.0)))));
	else
		tmp = Float64(t_0 * sin(exp(log(Float64(0.005555555555555556 * Float64(pi * angle_m))))));
	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 (((angle_m / 180.0) <= 5e+78) || ~(((angle_m / 180.0) <= 5e+186)))
		tmp = cos(((sqrt(pi) ^ 2.0) * (angle_m / 180.0))) * (t_0 * sin((pi * (angle_m / 180.0))));
	else
		tmp = t_0 * sin(exp(log((0.005555555555555556 * (pi * angle_m)))));
	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[Or[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 5e+78], N[Not[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 5e+186]], $MachinePrecision]], N[(N[Cos[N[(N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision] * N[(angle$95$m / 180.0), $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[N[(0.005555555555555556 * N[(Pi * angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 angle 180) < 4.99999999999999984e78 or 4.99999999999999954e186 < (/.f64 angle 180)

   1. Initial program 60.6%

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

      1. unpow2 60.6%

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

      2. unpow2 60.6%

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

      3. difference-of-squares 63.7%

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

   3. Applied egg-rr 63.7%

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

      1. add-sqr-sqrt 67.6%

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

      2. pow2 67.6%

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

   5. Applied egg-rr 67.6%

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

   if 4.99999999999999984e78 < (/.f64 angle 180) < 4.99999999999999954e186

   1. Initial program 26.9%

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

      1. unpow2 26.9%

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

      2. unpow2 26.9%

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

      3. difference-of-squares 26.9%

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

   3. Applied egg-rr 26.9%

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

   4. Taylor expanded in angle around 0 27.4%

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

      1. *-commutative 27.4%

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

      2. add-exp-log 48.3%

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

      3. div-inv 48.3%

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

      4. metadata-eval 48.3%

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

      5. associate-*r* 48.3%

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

      6. *-commutative 48.3%

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

      7. associate-*l* 48.3%

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

   6. Applied egg-rr 48.3%

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

4. Final simplification 65.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 5 \cdot 10^{+78} \lor \neg \left(\frac{angle}{180} \leq 5 \cdot 10^{+186}\right):\\ \;\;\;\;\cos \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)}\right)\\ \end{array} \]

Alternative 6: 57.0% accurate, 1.0× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\\ t_1 := \pi \cdot \frac{angle_m}{180}\\ t_2 := {\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle_m}{180}\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;\frac{angle_m}{180} \leq 5 \cdot 10^{+78}:\\ \;\;\;\;\cos t_2 \cdot \left(t_0 \cdot \sin t_1\right)\\ \mathbf{elif}\;\frac{angle_m}{180} \leq 5 \cdot 10^{+219}:\\ \;\;\;\;t_0 \cdot \sin \left(e^{\log \left(0.005555555555555556 \cdot \left(\pi \cdot angle_m\right)\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;\cos t_1 \cdot \left(t_0 \cdot \sin t_2\right)\\ \end{array} \end{array} \end{array} \]

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)))
        (t_2 (* (pow (sqrt PI) 2.0) (/ angle_m 180.0))))
   (*
    angle_s
    (if (<= (/ angle_m 180.0) 5e+78)
      (* (cos t_2) (* t_0 (sin t_1)))
      (if (<= (/ angle_m 180.0) 5e+219)
        (* t_0 (sin (exp (log (* 0.005555555555555556 (* PI angle_m))))))
        (* (cos t_1) (* t_0 (sin t_2))))))))

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 t_2 = pow(sqrt(((double) M_PI)), 2.0) * (angle_m / 180.0);
	double tmp;
	if ((angle_m / 180.0) <= 5e+78) {
		tmp = cos(t_2) * (t_0 * sin(t_1));
	} else if ((angle_m / 180.0) <= 5e+219) {
		tmp = t_0 * sin(exp(log((0.005555555555555556 * (((double) M_PI) * angle_m)))));
	} else {
		tmp = cos(t_1) * (t_0 * sin(t_2));
	}
	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 t_2 = Math.pow(Math.sqrt(Math.PI), 2.0) * (angle_m / 180.0);
	double tmp;
	if ((angle_m / 180.0) <= 5e+78) {
		tmp = Math.cos(t_2) * (t_0 * Math.sin(t_1));
	} else if ((angle_m / 180.0) <= 5e+219) {
		tmp = t_0 * Math.sin(Math.exp(Math.log((0.005555555555555556 * (Math.PI * angle_m)))));
	} else {
		tmp = Math.cos(t_1) * (t_0 * Math.sin(t_2));
	}
	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)
	t_2 = math.pow(math.sqrt(math.pi), 2.0) * (angle_m / 180.0)
	tmp = 0
	if (angle_m / 180.0) <= 5e+78:
		tmp = math.cos(t_2) * (t_0 * math.sin(t_1))
	elif (angle_m / 180.0) <= 5e+219:
		tmp = t_0 * math.sin(math.exp(math.log((0.005555555555555556 * (math.pi * angle_m)))))
	else:
		tmp = math.cos(t_1) * (t_0 * math.sin(t_2))
	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))
	t_2 = Float64((sqrt(pi) ^ 2.0) * Float64(angle_m / 180.0))
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 5e+78)
		tmp = Float64(cos(t_2) * Float64(t_0 * sin(t_1)));
	elseif (Float64(angle_m / 180.0) <= 5e+219)
		tmp = Float64(t_0 * sin(exp(log(Float64(0.005555555555555556 * Float64(pi * angle_m))))));
	else
		tmp = Float64(cos(t_1) * Float64(t_0 * sin(t_2)));
	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);
	t_2 = (sqrt(pi) ^ 2.0) * (angle_m / 180.0);
	tmp = 0.0;
	if ((angle_m / 180.0) <= 5e+78)
		tmp = cos(t_2) * (t_0 * sin(t_1));
	elseif ((angle_m / 180.0) <= 5e+219)
		tmp = t_0 * sin(exp(log((0.005555555555555556 * (pi * angle_m)))));
	else
		tmp = cos(t_1) * (t_0 * sin(t_2));
	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]}, Block[{t$95$2 = N[(N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision] * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 5e+78], N[(N[Cos[t$95$2], $MachinePrecision] * N[(t$95$0 * N[Sin[t$95$1], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 5e+219], N[(t$95$0 * N[Sin[N[Exp[N[Log[N[(0.005555555555555556 * N[(Pi * angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[Cos[t$95$1], $MachinePrecision] * N[(t$95$0 * N[Sin[t$95$2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 angle 180) < 4.99999999999999984e78

   1. Initial program 63.2%

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

      1. unpow2 63.2%

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

      2. unpow2 63.2%

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

      3. difference-of-squares 66.1%

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

   3. Applied egg-rr 66.1%

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

      1. add-sqr-sqrt 69.0%

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

      2. pow2 69.0%

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

   5. Applied egg-rr 69.0%

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

   if 4.99999999999999984e78 < (/.f64 angle 180) < 5e219

   1. Initial program 31.3%

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

      1. unpow2 31.3%

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

      2. unpow2 31.3%

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

      3. difference-of-squares 31.3%

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

   3. Applied egg-rr 31.3%

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

   4. Taylor expanded in angle around 0 31.5%

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

      1. *-commutative 31.5%

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

      2. add-exp-log 49.6%

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

      3. div-inv 49.6%

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

      4. metadata-eval 49.6%

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

      5. associate-*r* 49.6%

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

      6. *-commutative 49.6%

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

      7. associate-*l* 49.6%

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

   6. Applied egg-rr 49.6%

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

   if 5e219 < (/.f64 angle 180)

   1. Initial program 6.4%

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

      1. unpow2 6.4%

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

      2. unpow2 6.4%

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

      3. difference-of-squares 20.6%

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

   3. Applied egg-rr 20.6%

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

      1. add-sqr-sqrt 34.2%

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

      2. pow2 34.2%

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

   5. Applied egg-rr 33.5%

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

4. Final simplification 65.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 5 \cdot 10^{+78}:\\ \;\;\;\;\cos \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)\\ \mathbf{elif}\;\frac{angle}{180} \leq 5 \cdot 10^{+219}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\right)\\ \end{array} \]

Alternative 7: 55.6% accurate, 1.5× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 0.005555555555555556 \cdot \left(\pi \cdot angle_m\right)\\ t_1 := 2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;a \leq 1.35 \cdot 10^{+17}:\\ \;\;\;\;t_1 \cdot \sin \left(\frac{\pi \cdot angle_m}{180}\right)\\ \mathbf{elif}\;a \leq 2.5 \cdot 10^{+189}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle_m}{180}\right) \cdot \left(t_1 \cdot \sin t_0\right)\\ \mathbf{else}:\\ \;\;\;\;t_1 \cdot \sin \left(e^{\log t_0}\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* 0.005555555555555556 (* PI angle_m)))
        (t_1 (* 2.0 (* (+ b a) (- b a)))))
   (*
    angle_s
    (if (<= a 1.35e+17)
      (* t_1 (sin (/ (* PI angle_m) 180.0)))
      (if (<= a 2.5e+189)
        (* (cos (* PI (/ angle_m 180.0))) (* t_1 (sin t_0)))
        (* t_1 (sin (exp (log 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 * (((double) M_PI) * angle_m);
	double t_1 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if (a <= 1.35e+17) {
		tmp = t_1 * sin(((((double) M_PI) * angle_m) / 180.0));
	} else if (a <= 2.5e+189) {
		tmp = cos((((double) M_PI) * (angle_m / 180.0))) * (t_1 * sin(t_0));
	} else {
		tmp = t_1 * sin(exp(log(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 * (Math.PI * angle_m);
	double t_1 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if (a <= 1.35e+17) {
		tmp = t_1 * Math.sin(((Math.PI * angle_m) / 180.0));
	} else if (a <= 2.5e+189) {
		tmp = Math.cos((Math.PI * (angle_m / 180.0))) * (t_1 * Math.sin(t_0));
	} else {
		tmp = t_1 * Math.sin(Math.exp(Math.log(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 * (math.pi * angle_m)
	t_1 = 2.0 * ((b + a) * (b - a))
	tmp = 0
	if a <= 1.35e+17:
		tmp = t_1 * math.sin(((math.pi * angle_m) / 180.0))
	elif a <= 2.5e+189:
		tmp = math.cos((math.pi * (angle_m / 180.0))) * (t_1 * math.sin(t_0))
	else:
		tmp = t_1 * math.sin(math.exp(math.log(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(pi * angle_m))
	t_1 = Float64(2.0 * Float64(Float64(b + a) * Float64(b - a)))
	tmp = 0.0
	if (a <= 1.35e+17)
		tmp = Float64(t_1 * sin(Float64(Float64(pi * angle_m) / 180.0)));
	elseif (a <= 2.5e+189)
		tmp = Float64(cos(Float64(pi * Float64(angle_m / 180.0))) * Float64(t_1 * sin(t_0)));
	else
		tmp = Float64(t_1 * sin(exp(log(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 * (pi * angle_m);
	t_1 = 2.0 * ((b + a) * (b - a));
	tmp = 0.0;
	if (a <= 1.35e+17)
		tmp = t_1 * sin(((pi * angle_m) / 180.0));
	elseif (a <= 2.5e+189)
		tmp = cos((pi * (angle_m / 180.0))) * (t_1 * sin(t_0));
	else
		tmp = t_1 * sin(exp(log(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[(Pi * angle$95$m), $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, 1.35e+17], N[(t$95$1 * N[Sin[N[(N[(Pi * angle$95$m), $MachinePrecision] / 180.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[a, 2.5e+189], N[(N[Cos[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(t$95$1 * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(t$95$1 * N[Sin[N[Exp[N[Log[t$95$0], $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if a < 1.35e17

   1. Initial program 60.4%

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

      1. unpow2 60.4%

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

      2. unpow2 60.4%

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

      3. difference-of-squares 63.0%

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

   3. Applied egg-rr 63.0%

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

   4. Taylor expanded in angle around 0 65.7%

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

      1. associate-*r/ 64.7%

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

   6. Applied egg-rr 66.4%

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

   if 1.35e17 < a < 2.5000000000000002e189

   1. Initial program 44.4%

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

      1. unpow2 44.4%

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

      2. unpow2 44.4%

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

      3. difference-of-squares 49.1%

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

   3. Applied egg-rr 49.1%

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

   4. Taylor expanded in angle around inf 49.1%

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

   if 2.5000000000000002e189 < a

   1. Initial program 47.6%

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

      1. unpow2 47.6%

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

      2. unpow2 47.6%

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

      3. difference-of-squares 48.0%

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

   3. Applied egg-rr 48.0%

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

   4. Taylor expanded in angle around 0 57.6%

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

      1. *-commutative 57.6%

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

      2. add-exp-log 38.1%

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

      3. div-inv 38.1%

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

      4. metadata-eval 38.1%

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

      5. associate-*r* 38.1%

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

      6. *-commutative 38.1%

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

      7. associate-*l* 38.1%

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

   6. Applied egg-rr 38.1%

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

4. Final simplification 61.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 1.35 \cdot 10^{+17}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\\ \mathbf{elif}\;a \leq 2.5 \cdot 10^{+189}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)}\right)\\ \end{array} \]

Alternative 8: 56.9% accurate, 1.5× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\\ t_1 := t_0 \cdot \sin \left(\pi \cdot \left(0.005555555555555556 \cdot angle_m\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;\frac{angle_m}{180} \leq 5 \cdot 10^{+74}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle_m}{180}\right) \cdot t_1\\ \mathbf{elif}\;\frac{angle_m}{180} \leq 5 \cdot 10^{+190}:\\ \;\;\;\;t_0 \cdot \sin \left(e^{\log \left(0.005555555555555556 \cdot \left(\pi \cdot angle_m\right)\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;t_1\\ \end{array} \end{array} \end{array} \]

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 (* t_0 (sin (* PI (* 0.005555555555555556 angle_m))))))
   (*
    angle_s
    (if (<= (/ angle_m 180.0) 5e+74)
      (* (cos (* PI (/ angle_m 180.0))) t_1)
      (if (<= (/ angle_m 180.0) 5e+190)
        (* t_0 (sin (exp (log (* 0.005555555555555556 (* PI angle_m))))))
        t_1)))))

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 = t_0 * sin((((double) M_PI) * (0.005555555555555556 * angle_m)));
	double tmp;
	if ((angle_m / 180.0) <= 5e+74) {
		tmp = cos((((double) M_PI) * (angle_m / 180.0))) * t_1;
	} else if ((angle_m / 180.0) <= 5e+190) {
		tmp = t_0 * sin(exp(log((0.005555555555555556 * (((double) M_PI) * angle_m)))));
	} else {
		tmp = t_1;
	}
	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 = t_0 * Math.sin((Math.PI * (0.005555555555555556 * angle_m)));
	double tmp;
	if ((angle_m / 180.0) <= 5e+74) {
		tmp = Math.cos((Math.PI * (angle_m / 180.0))) * t_1;
	} else if ((angle_m / 180.0) <= 5e+190) {
		tmp = t_0 * Math.sin(Math.exp(Math.log((0.005555555555555556 * (Math.PI * angle_m)))));
	} else {
		tmp = t_1;
	}
	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 = t_0 * math.sin((math.pi * (0.005555555555555556 * angle_m)))
	tmp = 0
	if (angle_m / 180.0) <= 5e+74:
		tmp = math.cos((math.pi * (angle_m / 180.0))) * t_1
	elif (angle_m / 180.0) <= 5e+190:
		tmp = t_0 * math.sin(math.exp(math.log((0.005555555555555556 * (math.pi * angle_m)))))
	else:
		tmp = t_1
	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(t_0 * sin(Float64(pi * Float64(0.005555555555555556 * angle_m))))
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 5e+74)
		tmp = Float64(cos(Float64(pi * Float64(angle_m / 180.0))) * t_1);
	elseif (Float64(angle_m / 180.0) <= 5e+190)
		tmp = Float64(t_0 * sin(exp(log(Float64(0.005555555555555556 * Float64(pi * angle_m))))));
	else
		tmp = t_1;
	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 = t_0 * sin((pi * (0.005555555555555556 * angle_m)));
	tmp = 0.0;
	if ((angle_m / 180.0) <= 5e+74)
		tmp = cos((pi * (angle_m / 180.0))) * t_1;
	elseif ((angle_m / 180.0) <= 5e+190)
		tmp = t_0 * sin(exp(log((0.005555555555555556 * (pi * angle_m)))));
	else
		tmp = t_1;
	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[(t$95$0 * N[Sin[N[(Pi * N[(0.005555555555555556 * angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 5e+74], N[(N[Cos[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * t$95$1), $MachinePrecision], If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 5e+190], N[(t$95$0 * N[Sin[N[Exp[N[Log[N[(0.005555555555555556 * N[(Pi * angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision], t$95$1]]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 angle 180) < 4.99999999999999963e74

   1. Initial program 64.1%

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

      1. unpow2 64.1%

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

      2. unpow2 64.1%

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

      3. difference-of-squares 67.0%

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

   3. Applied egg-rr 67.0%

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

   4. Taylor expanded in angle around inf 67.2%

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

      1. associate-*r* 68.9%

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

      2. *-commutative 68.9%

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

      3. *-commutative 68.9%

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

      4. *-commutative 68.9%

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

   6. Simplified 66.3%

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

   if 4.99999999999999963e74 < (/.f64 angle 180) < 5.00000000000000036e190

   1. Initial program 26.8%

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

      1. unpow2 26.8%

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

      2. unpow2 26.8%

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

      3. difference-of-squares 26.8%

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

   3. Applied egg-rr 26.8%

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

   4. Taylor expanded in angle around 0 30.7%

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

      1. *-commutative 30.7%

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

      2. add-exp-log 48.6%

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

      3. div-inv 48.6%

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

      4. metadata-eval 48.6%

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

      5. associate-*r* 48.6%

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

      6. *-commutative 48.6%

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

      7. associate-*l* 48.6%

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

   6. Applied egg-rr 48.6%

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

   if 5.00000000000000036e190 < (/.f64 angle 180)

   1. Initial program 24.3%

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

      1. unpow2 24.3%

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

      2. unpow2 24.3%

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

      3. difference-of-squares 30.5%

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

   3. Applied egg-rr 30.5%

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

   4. Taylor expanded in angle around 0 22.6%

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

   5. Taylor expanded in angle around inf 22.5%

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

      1. associate-*r* 28.9%

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

      2. *-commutative 28.9%

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

      3. *-commutative 28.9%

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

      4. *-commutative 28.9%

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

   7. Simplified 28.9%

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

4. Final simplification 61.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 5 \cdot 10^{+74}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(0.005555555555555556 \cdot angle\right)\right)\right)\\ \mathbf{elif}\;\frac{angle}{180} \leq 5 \cdot 10^{+190}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(e^{\log \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(0.005555555555555556 \cdot angle\right)\right)\\ \end{array} \]

Alternative 9: 54.6% accurate, 1.9× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \left(b + a\right) \cdot \left(b - a\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 4 \cdot 10^{-18}:\\ \;\;\;\;2 \cdot \left(\sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle_m\right)\right) \cdot t_0\right)\\ \mathbf{else}:\\ \;\;\;\;\left(angle_m \cdot 0.011111111111111112\right) \cdot \left(\pi \cdot t_0\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* (+ b a) (- b a))))
   (*
    angle_s
    (if (<= (pow a 2.0) 4e-18)
      (* 2.0 (* (sin (* 0.005555555555555556 (* PI angle_m))) t_0))
      (* (* angle_m 0.011111111111111112) (* PI 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 = (b + a) * (b - a);
	double tmp;
	if (pow(a, 2.0) <= 4e-18) {
		tmp = 2.0 * (sin((0.005555555555555556 * (((double) M_PI) * angle_m))) * t_0);
	} else {
		tmp = (angle_m * 0.011111111111111112) * (((double) M_PI) * 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 = (b + a) * (b - a);
	double tmp;
	if (Math.pow(a, 2.0) <= 4e-18) {
		tmp = 2.0 * (Math.sin((0.005555555555555556 * (Math.PI * angle_m))) * t_0);
	} else {
		tmp = (angle_m * 0.011111111111111112) * (Math.PI * 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 = (b + a) * (b - a)
	tmp = 0
	if math.pow(a, 2.0) <= 4e-18:
		tmp = 2.0 * (math.sin((0.005555555555555556 * (math.pi * angle_m))) * t_0)
	else:
		tmp = (angle_m * 0.011111111111111112) * (math.pi * 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(Float64(b + a) * Float64(b - a))
	tmp = 0.0
	if ((a ^ 2.0) <= 4e-18)
		tmp = Float64(2.0 * Float64(sin(Float64(0.005555555555555556 * Float64(pi * angle_m))) * t_0));
	else
		tmp = Float64(Float64(angle_m * 0.011111111111111112) * Float64(pi * 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 = (b + a) * (b - a);
	tmp = 0.0;
	if ((a ^ 2.0) <= 4e-18)
		tmp = 2.0 * (sin((0.005555555555555556 * (pi * angle_m))) * t_0);
	else
		tmp = (angle_m * 0.011111111111111112) * (pi * 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[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 4e-18], N[(2.0 * N[(N[Sin[N[(0.005555555555555556 * N[(Pi * angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * t$95$0), $MachinePrecision]), $MachinePrecision], N[(N[(angle$95$m * 0.011111111111111112), $MachinePrecision] * N[(Pi * t$95$0), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a 2) < 4.0000000000000003e-18

   1. Initial program 66.4%

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

      1. unpow2 66.4%

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

      2. unpow2 66.4%

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

      3. difference-of-squares 66.4%

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

   3. Applied egg-rr 66.4%

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

   4. Taylor expanded in angle around 0 66.0%

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

   5. Taylor expanded in angle around inf 66.3%

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

   if 4.0000000000000003e-18 < (pow.f64 a 2)

   1. Initial program 45.6%

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

      1. unpow2 45.6%

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

      2. unpow2 45.6%

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

      3. difference-of-squares 51.5%

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

   3. Applied egg-rr 51.5%

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

   4. Taylor expanded in angle around 0 53.4%

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

   5. Taylor expanded in angle around 0 55.1%

      \[\leadsto \color{blue}{\left(0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right)\right)\right)} \cdot 1 \]
   6. Step-by-step derivation

      1. associate-*r* 55.2%

         \[\leadsto \color{blue}{\left(\left(0.011111111111111112 \cdot angle\right) \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right)\right)} \cdot 1 \]

   7. Simplified 55.2%

      \[\leadsto \color{blue}{\left(\left(0.011111111111111112 \cdot angle\right) \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right)\right)} \cdot 1 \]
3. Recombined 2 regimes into one program.

4. Final simplification 61.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 4 \cdot 10^{-18}:\\ \;\;\;\;2 \cdot \left(\sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right) \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(angle \cdot 0.011111111111111112\right) \cdot \left(\pi \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right)\\ \end{array} \]

Alternative 10: 54.9% accurate, 2.9× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \left(b + a\right) \cdot \left(b - a\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;a \leq 0.0013:\\ \;\;\;\;\left(2 \cdot t_0\right) \cdot \sin \left(\frac{\pi \cdot angle_m}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(angle_m \cdot 0.011111111111111112\right) \cdot \left(\pi \cdot t_0\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* (+ b a) (- b a))))
   (*
    angle_s
    (if (<= a 0.0013)
      (* (* 2.0 t_0) (sin (/ (* PI angle_m) 180.0)))
      (* (* angle_m 0.011111111111111112) (* PI 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 = (b + a) * (b - a);
	double tmp;
	if (a <= 0.0013) {
		tmp = (2.0 * t_0) * sin(((((double) M_PI) * angle_m) / 180.0));
	} else {
		tmp = (angle_m * 0.011111111111111112) * (((double) M_PI) * 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 = (b + a) * (b - a);
	double tmp;
	if (a <= 0.0013) {
		tmp = (2.0 * t_0) * Math.sin(((Math.PI * angle_m) / 180.0));
	} else {
		tmp = (angle_m * 0.011111111111111112) * (Math.PI * 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 = (b + a) * (b - a)
	tmp = 0
	if a <= 0.0013:
		tmp = (2.0 * t_0) * math.sin(((math.pi * angle_m) / 180.0))
	else:
		tmp = (angle_m * 0.011111111111111112) * (math.pi * 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(Float64(b + a) * Float64(b - a))
	tmp = 0.0
	if (a <= 0.0013)
		tmp = Float64(Float64(2.0 * t_0) * sin(Float64(Float64(pi * angle_m) / 180.0)));
	else
		tmp = Float64(Float64(angle_m * 0.011111111111111112) * Float64(pi * 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 = (b + a) * (b - a);
	tmp = 0.0;
	if (a <= 0.0013)
		tmp = (2.0 * t_0) * sin(((pi * angle_m) / 180.0));
	else
		tmp = (angle_m * 0.011111111111111112) * (pi * 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[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[a, 0.0013], N[(N[(2.0 * t$95$0), $MachinePrecision] * N[Sin[N[(N[(Pi * angle$95$m), $MachinePrecision] / 180.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[(angle$95$m * 0.011111111111111112), $MachinePrecision] * N[(Pi * t$95$0), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if a < 0.0012999999999999999

   1. Initial program 60.9%

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

      1. unpow2 60.9%

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

      2. unpow2 60.9%

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

      3. difference-of-squares 63.6%

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

   3. Applied egg-rr 63.6%

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

   4. Taylor expanded in angle around 0 65.7%

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

      1. associate-*r/ 64.7%

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

   6. Applied egg-rr 66.4%

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

   if 0.0012999999999999999 < a

   1. Initial program 45.2%

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

      1. unpow2 45.2%

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

      2. unpow2 45.2%

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

      3. difference-of-squares 48.2%

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

   3. Applied egg-rr 48.2%

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

   4. Taylor expanded in angle around 0 45.3%

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

   5. Taylor expanded in angle around 0 50.1%

      \[\leadsto \color{blue}{\left(0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right)\right)\right)} \cdot 1 \]
   6. Step-by-step derivation

      1. associate-*r* 50.2%

         \[\leadsto \color{blue}{\left(\left(0.011111111111111112 \cdot angle\right) \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right)\right)} \cdot 1 \]

   7. Simplified 50.2%

      \[\leadsto \color{blue}{\left(\left(0.011111111111111112 \cdot angle\right) \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right)\right)} \cdot 1 \]
3. Recombined 2 regimes into one program.

4. Final simplification 62.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 0.0013:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi \cdot angle}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(angle \cdot 0.011111111111111112\right) \cdot \left(\pi \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right)\\ \end{array} \]

Alternative 11: 55.3% accurate, 2.9× speedup

Math

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

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 (* PI (* 0.005555555555555556 angle_m))))))

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((((double) M_PI) * (0.005555555555555556 * angle_m))));
}

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((Math.PI * (0.005555555555555556 * angle_m))));
}

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((math.pi * (0.005555555555555556 * angle_m))))

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(pi * Float64(0.005555555555555556 * angle_m)))))
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((pi * (0.005555555555555556 * angle_m))));
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[(Pi * N[(0.005555555555555556 * angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

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. 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 59.4%

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

3. Applied egg-rr 59.4%

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

4. Taylor expanded in angle around 0 60.1%

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

5. Taylor expanded in angle around inf 58.7%

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

   1. associate-*r* 60.6%

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

   2. *-commutative 60.6%

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

   3. *-commutative 60.6%

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

   4. *-commutative 60.6%

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

7. Simplified 60.6%

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

8. Final simplification 60.6%

   \[\leadsto \left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(0.005555555555555556 \cdot angle\right)\right) \]

Alternative 12: 53.7% accurate, 5.5× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ angle_s \cdot \left(0.011111111111111112 \cdot \left(angle_m \cdot \left(\pi \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right)\right)\right) \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (* angle_s (* 0.011111111111111112 (* angle_m (* PI (* (+ b 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 * (0.011111111111111112 * (angle_m * (((double) M_PI) * ((b + 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 * (0.011111111111111112 * (angle_m * (Math.PI * ((b + 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 * (0.011111111111111112 * (angle_m * (math.pi * ((b + 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(0.011111111111111112 * Float64(angle_m * Float64(pi * Float64(Float64(b + 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 * (0.011111111111111112 * (angle_m * (pi * ((b + 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[(0.011111111111111112 * N[(angle$95$m * N[(Pi * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

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. 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 59.4%

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

3. Applied egg-rr 59.4%

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

4. Taylor expanded in angle around 0 60.1%

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

5. Taylor expanded in angle around 0 59.3%

   \[\leadsto \color{blue}{\left(0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right)\right)\right)} \cdot 1 \]

6. Final simplification 59.3%

   \[\leadsto 0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right)\right) \]

Alternative 13: 53.7% accurate, 5.5× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ angle_s \cdot \left(\left(angle_m \cdot 0.011111111111111112\right) \cdot \left(\pi \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right)\right) \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (* angle_s (* (* angle_m 0.011111111111111112) (* PI (* (+ b 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 * ((angle_m * 0.011111111111111112) * (((double) M_PI) * ((b + 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 * ((angle_m * 0.011111111111111112) * (Math.PI * ((b + 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 * ((angle_m * 0.011111111111111112) * (math.pi * ((b + 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(Float64(angle_m * 0.011111111111111112) * Float64(pi * Float64(Float64(b + 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 * ((angle_m * 0.011111111111111112) * (pi * ((b + 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[(angle$95$m * 0.011111111111111112), $MachinePrecision] * N[(Pi * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

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. 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 59.4%

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

3. Applied egg-rr 59.4%

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

4. Taylor expanded in angle around 0 60.1%

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

5. Taylor expanded in angle around 0 59.3%

   \[\leadsto \color{blue}{\left(0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right)\right)\right)} \cdot 1 \]
6. Step-by-step derivation

   1. associate-*r* 59.4%

      \[\leadsto \color{blue}{\left(\left(0.011111111111111112 \cdot angle\right) \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right)\right)} \cdot 1 \]

7. Simplified 59.4%

   \[\leadsto \color{blue}{\left(\left(0.011111111111111112 \cdot angle\right) \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right)\right)} \cdot 1 \]

8. Final simplification 59.4%

   \[\leadsto \left(angle \cdot 0.011111111111111112\right) \cdot \left(\pi \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \]

Reproduce

herbie shell --seed 2023322 
(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)))))