ab-angle->ABCF B

Percentage Accurate: 53.3% → 56.6%

Time: 52.1s

Alternatives: 12

Speedup: 5.5×

• 32.1% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Initial Program: 53.3% 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: 56.6% accurate, 0.7× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \frac{angle_m}{180} \cdot \pi\\ t_1 := 2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;\frac{angle_m}{180} \leq 10^{+118}:\\ \;\;\;\;\left(t_1 \cdot \sin t_0\right) \cdot \cos t_0\\ \mathbf{else}:\\ \;\;\;\;\left(t_1 \cdot \sin \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle_m \cdot 0.005555555555555556\right)\right)\right)\right)\right) \cdot \cos \left(\frac{angle_m}{180} \cdot {\left(\sqrt{\pi}\right)}^{2}\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* (/ angle_m 180.0) PI)) (t_1 (* 2.0 (* (- b a) (+ b a)))))
   (*
    angle_s
    (if (<= (/ angle_m 180.0) 1e+118)
      (* (* t_1 (sin t_0)) (cos t_0))
      (*
       (* t_1 (sin (expm1 (log1p (* PI (* angle_m 0.005555555555555556))))))
       (cos (* (/ angle_m 180.0) (pow (sqrt PI) 2.0))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (angle_m / 180.0) * ((double) M_PI);
	double t_1 = 2.0 * ((b - a) * (b + a));
	double tmp;
	if ((angle_m / 180.0) <= 1e+118) {
		tmp = (t_1 * sin(t_0)) * cos(t_0);
	} else {
		tmp = (t_1 * sin(expm1(log1p((((double) M_PI) * (angle_m * 0.005555555555555556)))))) * cos(((angle_m / 180.0) * pow(sqrt(((double) M_PI)), 2.0)));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (angle_m / 180.0) * Math.PI;
	double t_1 = 2.0 * ((b - a) * (b + a));
	double tmp;
	if ((angle_m / 180.0) <= 1e+118) {
		tmp = (t_1 * Math.sin(t_0)) * Math.cos(t_0);
	} else {
		tmp = (t_1 * Math.sin(Math.expm1(Math.log1p((Math.PI * (angle_m * 0.005555555555555556)))))) * Math.cos(((angle_m / 180.0) * Math.pow(Math.sqrt(Math.PI), 2.0)));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = (angle_m / 180.0) * math.pi
	t_1 = 2.0 * ((b - a) * (b + a))
	tmp = 0
	if (angle_m / 180.0) <= 1e+118:
		tmp = (t_1 * math.sin(t_0)) * math.cos(t_0)
	else:
		tmp = (t_1 * math.sin(math.expm1(math.log1p((math.pi * (angle_m * 0.005555555555555556)))))) * math.cos(((angle_m / 180.0) * math.pow(math.sqrt(math.pi), 2.0)))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(Float64(angle_m / 180.0) * pi)
	t_1 = Float64(2.0 * Float64(Float64(b - a) * Float64(b + a)))
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 1e+118)
		tmp = Float64(Float64(t_1 * sin(t_0)) * cos(t_0));
	else
		tmp = Float64(Float64(t_1 * sin(expm1(log1p(Float64(pi * Float64(angle_m * 0.005555555555555556)))))) * cos(Float64(Float64(angle_m / 180.0) * (sqrt(pi) ^ 2.0))));
	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[(N[(angle$95$m / 180.0), $MachinePrecision] * Pi), $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[(angle$95$m / 180.0), $MachinePrecision], 1e+118], N[(N[(t$95$1 * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision], N[(N[(t$95$1 * N[Sin[N[(Exp[N[Log[1 + N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(N[(angle$95$m / 180.0), $MachinePrecision] * N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 angle 180) < 9.99999999999999967e117

   1. Initial program 60.5%

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

      1. unpow2 60.5%

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

      2. unpow2 60.5%

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

      3. difference-of-squares 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%

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

   if 9.99999999999999967e117 < (/.f64 angle 180)

   1. Initial program 33.3%

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

      1. unpow2 33.3%

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

      2. unpow2 33.3%

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

      3. difference-of-squares 38.9%

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

   3. Applied egg-rr 38.9%

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

      1. expm1-log1p-u 38.3%

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

      2. div-inv 38.3%

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

      3. metadata-eval 38.3%

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

   5. Applied egg-rr 38.3%

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

      1. add-sqr-sqrt 47.9%

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

      2. pow2 47.9%

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

   7. Applied egg-rr 47.9%

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

4. Final simplification 61.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 10^{+118}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right) \cdot \cos \left(\frac{angle}{180} \cdot \pi\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sin \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right)\right) \cdot \cos \left(\frac{angle}{180} \cdot {\left(\sqrt{\pi}\right)}^{2}\right)\\ \end{array} \]

Alternative 2: 56.8% accurate, 1.0× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;\frac{angle_m}{180} \leq 10^{+71}:\\ \;\;\;\;\left(t_0 \cdot \sin \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle_m \cdot 0.005555555555555556\right)\right)\right)\right)\right) \cdot \cos \left(angle_m \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;t_0 \cdot \sqrt{{\sin \left(\left(angle_m \cdot 0.005555555555555556\right) \cdot {\left(\sqrt{\pi}\right)}^{2}\right)}^{2}}\\ \end{array} \end{array} \end{array} \]

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 (<= (/ angle_m 180.0) 1e+71)
      (*
       (* t_0 (sin (expm1 (log1p (* PI (* angle_m 0.005555555555555556))))))
       (cos (* angle_m (* PI 0.005555555555555556))))
      (*
       t_0
       (sqrt
        (pow
         (sin (* (* angle_m 0.005555555555555556) (pow (sqrt PI) 2.0)))
         2.0)))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b - a) * (b + a));
	double tmp;
	if ((angle_m / 180.0) <= 1e+71) {
		tmp = (t_0 * sin(expm1(log1p((((double) M_PI) * (angle_m * 0.005555555555555556)))))) * cos((angle_m * (((double) M_PI) * 0.005555555555555556)));
	} else {
		tmp = t_0 * sqrt(pow(sin(((angle_m * 0.005555555555555556) * pow(sqrt(((double) M_PI)), 2.0))), 2.0));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b - a) * (b + a));
	double tmp;
	if ((angle_m / 180.0) <= 1e+71) {
		tmp = (t_0 * Math.sin(Math.expm1(Math.log1p((Math.PI * (angle_m * 0.005555555555555556)))))) * Math.cos((angle_m * (Math.PI * 0.005555555555555556)));
	} else {
		tmp = t_0 * Math.sqrt(Math.pow(Math.sin(((angle_m * 0.005555555555555556) * Math.pow(Math.sqrt(Math.PI), 2.0))), 2.0));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = 2.0 * ((b - a) * (b + a))
	tmp = 0
	if (angle_m / 180.0) <= 1e+71:
		tmp = (t_0 * math.sin(math.expm1(math.log1p((math.pi * (angle_m * 0.005555555555555556)))))) * math.cos((angle_m * (math.pi * 0.005555555555555556)))
	else:
		tmp = t_0 * math.sqrt(math.pow(math.sin(((angle_m * 0.005555555555555556) * math.pow(math.sqrt(math.pi), 2.0))), 2.0))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(2.0 * Float64(Float64(b - a) * Float64(b + a)))
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 1e+71)
		tmp = Float64(Float64(t_0 * sin(expm1(log1p(Float64(pi * Float64(angle_m * 0.005555555555555556)))))) * cos(Float64(angle_m * Float64(pi * 0.005555555555555556))));
	else
		tmp = Float64(t_0 * sqrt((sin(Float64(Float64(angle_m * 0.005555555555555556) * (sqrt(pi) ^ 2.0))) ^ 2.0)));
	end
	return Float64(angle_s * tmp)
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 1e+71], N[(N[(t$95$0 * N[Sin[N[(Exp[N[Log[1 + N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(angle$95$m * N[(Pi * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(t$95$0 * N[Sqrt[N[Power[N[Sin[N[(N[(angle$95$m * 0.005555555555555556), $MachinePrecision] * N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision], 2.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 angle 180) < 1e71

   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 65.8%

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

   3. Applied egg-rr 65.8%

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

      1. expm1-log1p-u 55.5%

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

      2. div-inv 55.5%

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

      3. metadata-eval 55.5%

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

   5. Applied egg-rr 55.5%

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

   6. Taylor expanded in angle around inf 55.6%

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

      1. *-commutative 55.6%

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

      2. associate-*l* 55.6%

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

   8. Simplified 55.6%

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

   if 1e71 < (/.f64 angle 180)

   1. Initial program 32.8%

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

      1. unpow2 32.8%

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

      2. unpow2 32.8%

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

      3. difference-of-squares 36.8%

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

   3. Applied egg-rr 36.8%

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

   4. Taylor expanded in angle around 0 30.8%

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

      1. add-sqr-sqrt 20.7%

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

      2. sqrt-unprod 39.1%

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

      3. pow2 39.1%

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

      4. div-inv 39.1%

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

      5. metadata-eval 39.1%

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

   6. Applied egg-rr 39.1%

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

      1. add-sqr-sqrt 36.9%

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

      2. pow2 36.9%

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

   8. Applied egg-rr 39.1%

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

4. Final simplification 52.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 10^{+71}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sin \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right)\right) \cdot \cos \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sqrt{{\sin \left(\left(angle \cdot 0.005555555555555556\right) \cdot {\left(\sqrt{\pi}\right)}^{2}\right)}^{2}}\\ \end{array} \]

Alternative 3: 56.8% accurate, 1.0× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \frac{1}{\frac{180}{angle_m \cdot \pi}}\\ t_1 := 2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;\frac{angle_m}{180} \leq 2 \cdot 10^{+78}:\\ \;\;\;\;\left(t_1 \cdot \sin t_0\right) \cdot \cos t_0\\ \mathbf{else}:\\ \;\;\;\;t_1 \cdot \sqrt{{\sin \left(\left(angle_m \cdot 0.005555555555555556\right) \cdot \sqrt[3]{{\pi}^{3}}\right)}^{2}}\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (/ 1.0 (/ 180.0 (* angle_m PI))))
        (t_1 (* 2.0 (* (- b a) (+ b a)))))
   (*
    angle_s
    (if (<= (/ angle_m 180.0) 2e+78)
      (* (* t_1 (sin t_0)) (cos t_0))
      (*
       t_1
       (sqrt
        (pow
         (sin (* (* angle_m 0.005555555555555556) (cbrt (pow PI 3.0))))
         2.0)))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 1.0 / (180.0 / (angle_m * ((double) M_PI)));
	double t_1 = 2.0 * ((b - a) * (b + a));
	double tmp;
	if ((angle_m / 180.0) <= 2e+78) {
		tmp = (t_1 * sin(t_0)) * cos(t_0);
	} else {
		tmp = t_1 * sqrt(pow(sin(((angle_m * 0.005555555555555556) * cbrt(pow(((double) M_PI), 3.0)))), 2.0));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 1.0 / (180.0 / (angle_m * Math.PI));
	double t_1 = 2.0 * ((b - a) * (b + a));
	double tmp;
	if ((angle_m / 180.0) <= 2e+78) {
		tmp = (t_1 * Math.sin(t_0)) * Math.cos(t_0);
	} else {
		tmp = t_1 * Math.sqrt(Math.pow(Math.sin(((angle_m * 0.005555555555555556) * Math.cbrt(Math.pow(Math.PI, 3.0)))), 2.0));
	}
	return angle_s * tmp;
}

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(1.0 / Float64(180.0 / Float64(angle_m * pi)))
	t_1 = Float64(2.0 * Float64(Float64(b - a) * Float64(b + a)))
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 2e+78)
		tmp = Float64(Float64(t_1 * sin(t_0)) * cos(t_0));
	else
		tmp = Float64(t_1 * sqrt((sin(Float64(Float64(angle_m * 0.005555555555555556) * cbrt((pi ^ 3.0)))) ^ 2.0)));
	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[(1.0 / N[(180.0 / N[(angle$95$m * Pi), $MachinePrecision]), $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[(angle$95$m / 180.0), $MachinePrecision], 2e+78], N[(N[(t$95$1 * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision], N[(t$95$1 * N[Sqrt[N[Power[N[Sin[N[(N[(angle$95$m * 0.005555555555555556), $MachinePrecision] * N[Power[N[Power[Pi, 3.0], $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]], $MachinePrecision], 2.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 angle 180) < 2.00000000000000002e78

   1. Initial program 62.1%

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

      1. unpow2 62.1%

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

      2. unpow2 62.1%

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

      3. difference-of-squares 65.5%

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

   3. Applied egg-rr 65.5%

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

      1. associate-*r/ 64.4%

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

      2. *-commutative 64.4%

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

      3. clear-num 64.3%

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

      4. *-commutative 64.3%

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

   5. Applied egg-rr 64.3%

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

      1. associate-*r/ 64.4%

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

      2. *-commutative 64.4%

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

      3. clear-num 64.3%

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

      4. *-commutative 64.3%

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

   7. Applied egg-rr 65.4%

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

   if 2.00000000000000002e78 < (/.f64 angle 180)

   1. Initial program 33.4%

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

      1. unpow2 33.4%

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

      2. unpow2 33.4%

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

      3. difference-of-squares 37.6%

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

   3. Applied egg-rr 37.6%

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

   4. Taylor expanded in angle around 0 31.4%

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

      1. add-sqr-sqrt 21.2%

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

      2. sqrt-unprod 39.5%

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

      3. pow2 39.5%

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

      4. div-inv 39.4%

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

      5. metadata-eval 39.4%

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

   6. Applied egg-rr 39.4%

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

      1. add-cbrt-cube 39.5%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sqrt{{\sin \left(\color{blue}{\sqrt[3]{\left(\pi \cdot \pi\right) \cdot \pi}} \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{2}}\right) \cdot 1 \]

      2. pow3 39.5%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sqrt{{\sin \left(\sqrt[3]{\color{blue}{{\pi}^{3}}} \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{2}}\right) \cdot 1 \]

   8. Applied egg-rr 39.5%

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

4. Final simplification 60.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 2 \cdot 10^{+78}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sin \left(\frac{1}{\frac{180}{angle \cdot \pi}}\right)\right) \cdot \cos \left(\frac{1}{\frac{180}{angle \cdot \pi}}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sqrt{{\sin \left(\left(angle \cdot 0.005555555555555556\right) \cdot \sqrt[3]{{\pi}^{3}}\right)}^{2}}\\ \end{array} \]

Alternative 4: 56.9% accurate, 1.0× speedup

Math

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

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (/ 1.0 (/ 180.0 (* angle_m PI))))
        (t_1 (* 2.0 (* (- b a) (+ b a)))))
   (*
    angle_s
    (if (<= (/ angle_m 180.0) 2e+78)
      (* (* t_1 (sin t_0)) (cos t_0))
      (*
       t_1
       (sqrt
        (pow
         (sin (* (* angle_m 0.005555555555555556) (pow (sqrt PI) 2.0)))
         2.0)))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 1.0 / (180.0 / (angle_m * ((double) M_PI)));
	double t_1 = 2.0 * ((b - a) * (b + a));
	double tmp;
	if ((angle_m / 180.0) <= 2e+78) {
		tmp = (t_1 * sin(t_0)) * cos(t_0);
	} else {
		tmp = t_1 * sqrt(pow(sin(((angle_m * 0.005555555555555556) * pow(sqrt(((double) M_PI)), 2.0))), 2.0));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 1.0 / (180.0 / (angle_m * Math.PI));
	double t_1 = 2.0 * ((b - a) * (b + a));
	double tmp;
	if ((angle_m / 180.0) <= 2e+78) {
		tmp = (t_1 * Math.sin(t_0)) * Math.cos(t_0);
	} else {
		tmp = t_1 * Math.sqrt(Math.pow(Math.sin(((angle_m * 0.005555555555555556) * Math.pow(Math.sqrt(Math.PI), 2.0))), 2.0));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = 1.0 / (180.0 / (angle_m * math.pi))
	t_1 = 2.0 * ((b - a) * (b + a))
	tmp = 0
	if (angle_m / 180.0) <= 2e+78:
		tmp = (t_1 * math.sin(t_0)) * math.cos(t_0)
	else:
		tmp = t_1 * math.sqrt(math.pow(math.sin(((angle_m * 0.005555555555555556) * math.pow(math.sqrt(math.pi), 2.0))), 2.0))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(1.0 / Float64(180.0 / Float64(angle_m * pi)))
	t_1 = Float64(2.0 * Float64(Float64(b - a) * Float64(b + a)))
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 2e+78)
		tmp = Float64(Float64(t_1 * sin(t_0)) * cos(t_0));
	else
		tmp = Float64(t_1 * sqrt((sin(Float64(Float64(angle_m * 0.005555555555555556) * (sqrt(pi) ^ 2.0))) ^ 2.0)));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = 1.0 / (180.0 / (angle_m * pi));
	t_1 = 2.0 * ((b - a) * (b + a));
	tmp = 0.0;
	if ((angle_m / 180.0) <= 2e+78)
		tmp = (t_1 * sin(t_0)) * cos(t_0);
	else
		tmp = t_1 * sqrt((sin(((angle_m * 0.005555555555555556) * (sqrt(pi) ^ 2.0))) ^ 2.0));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(1.0 / N[(180.0 / N[(angle$95$m * Pi), $MachinePrecision]), $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[(angle$95$m / 180.0), $MachinePrecision], 2e+78], N[(N[(t$95$1 * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision], N[(t$95$1 * N[Sqrt[N[Power[N[Sin[N[(N[(angle$95$m * 0.005555555555555556), $MachinePrecision] * N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision], 2.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 angle 180) < 2.00000000000000002e78

   1. Initial program 62.1%

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

      1. unpow2 62.1%

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

      2. unpow2 62.1%

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

      3. difference-of-squares 65.5%

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

   3. Applied egg-rr 65.5%

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

      1. associate-*r/ 64.4%

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

      2. *-commutative 64.4%

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

      3. clear-num 64.3%

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

      4. *-commutative 64.3%

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

   5. Applied egg-rr 64.3%

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

      1. associate-*r/ 64.4%

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

      2. *-commutative 64.4%

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

      3. clear-num 64.3%

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

      4. *-commutative 64.3%

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

   7. Applied egg-rr 65.4%

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

   if 2.00000000000000002e78 < (/.f64 angle 180)

   1. Initial program 33.4%

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

      1. unpow2 33.4%

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

      2. unpow2 33.4%

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

      3. difference-of-squares 37.6%

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

   3. Applied egg-rr 37.6%

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

   4. Taylor expanded in angle around 0 31.4%

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

      1. add-sqr-sqrt 21.2%

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

      2. sqrt-unprod 39.5%

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

      3. pow2 39.5%

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

      4. div-inv 39.4%

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

      5. metadata-eval 39.4%

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

   6. Applied egg-rr 39.4%

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

      1. add-sqr-sqrt 37.6%

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

      2. pow2 37.6%

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

   8. Applied egg-rr 39.5%

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

4. Final simplification 60.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 2 \cdot 10^{+78}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sin \left(\frac{1}{\frac{180}{angle \cdot \pi}}\right)\right) \cdot \cos \left(\frac{1}{\frac{180}{angle \cdot \pi}}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sqrt{{\sin \left(\left(angle \cdot 0.005555555555555556\right) \cdot {\left(\sqrt{\pi}\right)}^{2}\right)}^{2}}\\ \end{array} \]

Alternative 5: 56.8% accurate, 1.4× speedup

Math

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

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (/ 1.0 (/ 180.0 (* angle_m PI))))
        (t_1 (* 2.0 (* (- b a) (+ b a)))))
   (*
    angle_s
    (if (<= (/ angle_m 180.0) 2e+78)
      (* (* t_1 (sin t_0)) (cos t_0))
      (*
       t_1
       (sqrt (pow (sin (* PI (* angle_m 0.005555555555555556))) 2.0)))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 1.0 / (180.0 / (angle_m * ((double) M_PI)));
	double t_1 = 2.0 * ((b - a) * (b + a));
	double tmp;
	if ((angle_m / 180.0) <= 2e+78) {
		tmp = (t_1 * sin(t_0)) * cos(t_0);
	} else {
		tmp = t_1 * sqrt(pow(sin((((double) M_PI) * (angle_m * 0.005555555555555556))), 2.0));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 1.0 / (180.0 / (angle_m * Math.PI));
	double t_1 = 2.0 * ((b - a) * (b + a));
	double tmp;
	if ((angle_m / 180.0) <= 2e+78) {
		tmp = (t_1 * Math.sin(t_0)) * Math.cos(t_0);
	} else {
		tmp = t_1 * Math.sqrt(Math.pow(Math.sin((Math.PI * (angle_m * 0.005555555555555556))), 2.0));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = 1.0 / (180.0 / (angle_m * math.pi))
	t_1 = 2.0 * ((b - a) * (b + a))
	tmp = 0
	if (angle_m / 180.0) <= 2e+78:
		tmp = (t_1 * math.sin(t_0)) * math.cos(t_0)
	else:
		tmp = t_1 * math.sqrt(math.pow(math.sin((math.pi * (angle_m * 0.005555555555555556))), 2.0))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(1.0 / Float64(180.0 / Float64(angle_m * pi)))
	t_1 = Float64(2.0 * Float64(Float64(b - a) * Float64(b + a)))
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 2e+78)
		tmp = Float64(Float64(t_1 * sin(t_0)) * cos(t_0));
	else
		tmp = Float64(t_1 * sqrt((sin(Float64(pi * Float64(angle_m * 0.005555555555555556))) ^ 2.0)));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = 1.0 / (180.0 / (angle_m * pi));
	t_1 = 2.0 * ((b - a) * (b + a));
	tmp = 0.0;
	if ((angle_m / 180.0) <= 2e+78)
		tmp = (t_1 * sin(t_0)) * cos(t_0);
	else
		tmp = t_1 * sqrt((sin((pi * (angle_m * 0.005555555555555556))) ^ 2.0));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(1.0 / N[(180.0 / N[(angle$95$m * Pi), $MachinePrecision]), $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[(angle$95$m / 180.0), $MachinePrecision], 2e+78], N[(N[(t$95$1 * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision], N[(t$95$1 * N[Sqrt[N[Power[N[Sin[N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision], 2.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 angle 180) < 2.00000000000000002e78

   1. Initial program 62.1%

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

      1. unpow2 62.1%

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

      2. unpow2 62.1%

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

      3. difference-of-squares 65.5%

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

   3. Applied egg-rr 65.5%

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

      1. associate-*r/ 64.4%

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

      2. *-commutative 64.4%

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

      3. clear-num 64.3%

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

      4. *-commutative 64.3%

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

   5. Applied egg-rr 64.3%

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

      1. associate-*r/ 64.4%

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

      2. *-commutative 64.4%

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

      3. clear-num 64.3%

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

      4. *-commutative 64.3%

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

   7. Applied egg-rr 65.4%

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

   if 2.00000000000000002e78 < (/.f64 angle 180)

   1. Initial program 33.4%

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

      1. unpow2 33.4%

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

      2. unpow2 33.4%

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

      3. difference-of-squares 37.6%

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

   3. Applied egg-rr 37.6%

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

   4. Taylor expanded in angle around 0 31.4%

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

      1. add-sqr-sqrt 21.2%

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

      2. sqrt-unprod 39.5%

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

      3. pow2 39.5%

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

      4. div-inv 39.4%

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

      5. metadata-eval 39.4%

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

   6. Applied egg-rr 39.4%

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

4. Final simplification 60.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 2 \cdot 10^{+78}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sin \left(\frac{1}{\frac{180}{angle \cdot \pi}}\right)\right) \cdot \cos \left(\frac{1}{\frac{180}{angle \cdot \pi}}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sqrt{{\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{2}}\\ \end{array} \]

Alternative 6: 56.9% accurate, 1.5× speedup

Math

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

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 0.005555555555555556))))
   (*
    angle_s
    (if (<= (/ angle_m 180.0) 2e+78)
      (* (cos t_1) (* t_0 (sin (* 0.005555555555555556 (* angle_m PI)))))
      (* t_0 (sqrt (pow (sin t_1) 2.0)))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b - a) * (b + a));
	double t_1 = ((double) M_PI) * (angle_m * 0.005555555555555556);
	double tmp;
	if ((angle_m / 180.0) <= 2e+78) {
		tmp = cos(t_1) * (t_0 * sin((0.005555555555555556 * (angle_m * ((double) M_PI)))));
	} else {
		tmp = t_0 * sqrt(pow(sin(t_1), 2.0));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b - a) * (b + a));
	double t_1 = Math.PI * (angle_m * 0.005555555555555556);
	double tmp;
	if ((angle_m / 180.0) <= 2e+78) {
		tmp = Math.cos(t_1) * (t_0 * Math.sin((0.005555555555555556 * (angle_m * Math.PI))));
	} else {
		tmp = t_0 * Math.sqrt(Math.pow(Math.sin(t_1), 2.0));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = 2.0 * ((b - a) * (b + a))
	t_1 = math.pi * (angle_m * 0.005555555555555556)
	tmp = 0
	if (angle_m / 180.0) <= 2e+78:
		tmp = math.cos(t_1) * (t_0 * math.sin((0.005555555555555556 * (angle_m * math.pi))))
	else:
		tmp = t_0 * math.sqrt(math.pow(math.sin(t_1), 2.0))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(2.0 * Float64(Float64(b - a) * Float64(b + a)))
	t_1 = Float64(pi * Float64(angle_m * 0.005555555555555556))
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 2e+78)
		tmp = Float64(cos(t_1) * Float64(t_0 * sin(Float64(0.005555555555555556 * Float64(angle_m * pi)))));
	else
		tmp = Float64(t_0 * sqrt((sin(t_1) ^ 2.0)));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = 2.0 * ((b - a) * (b + a));
	t_1 = pi * (angle_m * 0.005555555555555556);
	tmp = 0.0;
	if ((angle_m / 180.0) <= 2e+78)
		tmp = cos(t_1) * (t_0 * sin((0.005555555555555556 * (angle_m * pi))));
	else
		tmp = t_0 * sqrt((sin(t_1) ^ 2.0));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 2e+78], N[(N[Cos[t$95$1], $MachinePrecision] * N[(t$95$0 * N[Sin[N[(0.005555555555555556 * N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(t$95$0 * N[Sqrt[N[Power[N[Sin[t$95$1], $MachinePrecision], 2.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 angle 180) < 2.00000000000000002e78

   1. Initial program 62.1%

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

      1. unpow2 62.1%

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

      2. unpow2 62.1%

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

      3. difference-of-squares 65.5%

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

   3. Applied egg-rr 65.5%

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

   4. Taylor expanded in angle around inf 64.1%

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

      1. *-commutative 64.1%

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

      2. *-commutative 64.1%

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

      3. associate-*r* 65.4%

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

   6. Simplified 65.4%

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

   7. Taylor expanded in angle around inf 65.5%

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

   if 2.00000000000000002e78 < (/.f64 angle 180)

   1. Initial program 33.4%

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

      1. unpow2 33.4%

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

      2. unpow2 33.4%

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

      3. difference-of-squares 37.6%

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

   3. Applied egg-rr 37.6%

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

   4. Taylor expanded in angle around 0 31.4%

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

      1. add-sqr-sqrt 21.2%

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

      2. sqrt-unprod 39.5%

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

      3. pow2 39.5%

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

      4. div-inv 39.4%

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

      5. metadata-eval 39.4%

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

   6. Applied egg-rr 39.4%

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

4. Final simplification 60.6%

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

Alternative 7: 56.9% accurate, 1.5× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \pi \cdot \left(angle_m \cdot 0.005555555555555556\right)\\ t_1 := 2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;\frac{angle_m}{180} \leq 5 \cdot 10^{+72}:\\ \;\;\;\;\left(t_1 \cdot \sin \left(\frac{angle_m}{180} \cdot \pi\right)\right) \cdot \cos t_0\\ \mathbf{else}:\\ \;\;\;\;t_1 \cdot \sqrt{{\sin t_0}^{2}}\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* PI (* angle_m 0.005555555555555556)))
        (t_1 (* 2.0 (* (- b a) (+ b a)))))
   (*
    angle_s
    (if (<= (/ angle_m 180.0) 5e+72)
      (* (* t_1 (sin (* (/ angle_m 180.0) PI))) (cos t_0))
      (* t_1 (sqrt (pow (sin t_0) 2.0)))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = ((double) M_PI) * (angle_m * 0.005555555555555556);
	double t_1 = 2.0 * ((b - a) * (b + a));
	double tmp;
	if ((angle_m / 180.0) <= 5e+72) {
		tmp = (t_1 * sin(((angle_m / 180.0) * ((double) M_PI)))) * cos(t_0);
	} else {
		tmp = t_1 * sqrt(pow(sin(t_0), 2.0));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = Math.PI * (angle_m * 0.005555555555555556);
	double t_1 = 2.0 * ((b - a) * (b + a));
	double tmp;
	if ((angle_m / 180.0) <= 5e+72) {
		tmp = (t_1 * Math.sin(((angle_m / 180.0) * Math.PI))) * Math.cos(t_0);
	} else {
		tmp = t_1 * Math.sqrt(Math.pow(Math.sin(t_0), 2.0));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = math.pi * (angle_m * 0.005555555555555556)
	t_1 = 2.0 * ((b - a) * (b + a))
	tmp = 0
	if (angle_m / 180.0) <= 5e+72:
		tmp = (t_1 * math.sin(((angle_m / 180.0) * math.pi))) * math.cos(t_0)
	else:
		tmp = t_1 * math.sqrt(math.pow(math.sin(t_0), 2.0))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(pi * Float64(angle_m * 0.005555555555555556))
	t_1 = Float64(2.0 * Float64(Float64(b - a) * Float64(b + a)))
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 5e+72)
		tmp = Float64(Float64(t_1 * sin(Float64(Float64(angle_m / 180.0) * pi))) * cos(t_0));
	else
		tmp = Float64(t_1 * sqrt((sin(t_0) ^ 2.0)));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = pi * (angle_m * 0.005555555555555556);
	t_1 = 2.0 * ((b - a) * (b + a));
	tmp = 0.0;
	if ((angle_m / 180.0) <= 5e+72)
		tmp = (t_1 * sin(((angle_m / 180.0) * pi))) * cos(t_0);
	else
		tmp = t_1 * sqrt((sin(t_0) ^ 2.0));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(Pi * N[(angle$95$m * 0.005555555555555556), $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[(angle$95$m / 180.0), $MachinePrecision], 5e+72], N[(N[(t$95$1 * N[Sin[N[(N[(angle$95$m / 180.0), $MachinePrecision] * Pi), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision], N[(t$95$1 * N[Sqrt[N[Power[N[Sin[t$95$0], $MachinePrecision], 2.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 angle 180) < 4.99999999999999992e72

   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 65.8%

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

   3. Applied egg-rr 65.8%

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

   4. Taylor expanded in angle around inf 64.4%

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

      1. *-commutative 64.4%

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

      2. *-commutative 64.4%

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

      3. associate-*r* 65.7%

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

   6. Simplified 65.7%

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

   if 4.99999999999999992e72 < (/.f64 angle 180)

   1. Initial program 32.8%

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

      1. unpow2 32.8%

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

      2. unpow2 32.8%

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

      3. difference-of-squares 36.8%

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

   3. Applied egg-rr 36.8%

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

   4. Taylor expanded in angle around 0 30.8%

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

      1. add-sqr-sqrt 20.7%

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

      2. sqrt-unprod 39.1%

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

      3. pow2 39.1%

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

      4. div-inv 39.1%

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

      5. metadata-eval 39.1%

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

   6. Applied egg-rr 39.1%

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

4. Final simplification 60.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 5 \cdot 10^{+72}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sqrt{{\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{2}}\\ \end{array} \]

Alternative 8: 56.9% accurate, 1.5× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \frac{angle_m}{180} \cdot \pi\\ t_1 := 2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;\frac{angle_m}{180} \leq 5 \cdot 10^{+72}:\\ \;\;\;\;\left(t_1 \cdot \sin t_0\right) \cdot \cos t_0\\ \mathbf{else}:\\ \;\;\;\;t_1 \cdot \sqrt{{\sin \left(\pi \cdot \left(angle_m \cdot 0.005555555555555556\right)\right)}^{2}}\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* (/ angle_m 180.0) PI)) (t_1 (* 2.0 (* (- b a) (+ b a)))))
   (*
    angle_s
    (if (<= (/ angle_m 180.0) 5e+72)
      (* (* t_1 (sin t_0)) (cos t_0))
      (*
       t_1
       (sqrt (pow (sin (* PI (* angle_m 0.005555555555555556))) 2.0)))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (angle_m / 180.0) * ((double) M_PI);
	double t_1 = 2.0 * ((b - a) * (b + a));
	double tmp;
	if ((angle_m / 180.0) <= 5e+72) {
		tmp = (t_1 * sin(t_0)) * cos(t_0);
	} else {
		tmp = t_1 * sqrt(pow(sin((((double) M_PI) * (angle_m * 0.005555555555555556))), 2.0));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (angle_m / 180.0) * Math.PI;
	double t_1 = 2.0 * ((b - a) * (b + a));
	double tmp;
	if ((angle_m / 180.0) <= 5e+72) {
		tmp = (t_1 * Math.sin(t_0)) * Math.cos(t_0);
	} else {
		tmp = t_1 * Math.sqrt(Math.pow(Math.sin((Math.PI * (angle_m * 0.005555555555555556))), 2.0));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = (angle_m / 180.0) * math.pi
	t_1 = 2.0 * ((b - a) * (b + a))
	tmp = 0
	if (angle_m / 180.0) <= 5e+72:
		tmp = (t_1 * math.sin(t_0)) * math.cos(t_0)
	else:
		tmp = t_1 * math.sqrt(math.pow(math.sin((math.pi * (angle_m * 0.005555555555555556))), 2.0))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(Float64(angle_m / 180.0) * pi)
	t_1 = Float64(2.0 * Float64(Float64(b - a) * Float64(b + a)))
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 5e+72)
		tmp = Float64(Float64(t_1 * sin(t_0)) * cos(t_0));
	else
		tmp = Float64(t_1 * sqrt((sin(Float64(pi * Float64(angle_m * 0.005555555555555556))) ^ 2.0)));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = (angle_m / 180.0) * pi;
	t_1 = 2.0 * ((b - a) * (b + a));
	tmp = 0.0;
	if ((angle_m / 180.0) <= 5e+72)
		tmp = (t_1 * sin(t_0)) * cos(t_0);
	else
		tmp = t_1 * sqrt((sin((pi * (angle_m * 0.005555555555555556))) ^ 2.0));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(N[(angle$95$m / 180.0), $MachinePrecision] * Pi), $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[(angle$95$m / 180.0), $MachinePrecision], 5e+72], N[(N[(t$95$1 * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision], N[(t$95$1 * N[Sqrt[N[Power[N[Sin[N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision], 2.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 angle 180) < 4.99999999999999992e72

   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 65.8%

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

   3. Applied egg-rr 65.8%

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

   if 4.99999999999999992e72 < (/.f64 angle 180)

   1. Initial program 32.8%

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

      1. unpow2 32.8%

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

      2. unpow2 32.8%

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

      3. difference-of-squares 36.8%

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

   3. Applied egg-rr 36.8%

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

   4. Taylor expanded in angle around 0 30.8%

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

      1. add-sqr-sqrt 20.7%

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

      2. sqrt-unprod 39.1%

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

      3. pow2 39.1%

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

      4. div-inv 39.1%

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

      5. metadata-eval 39.1%

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

   6. Applied egg-rr 39.1%

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

4. Final simplification 60.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 5 \cdot 10^{+72}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right) \cdot \cos \left(\frac{angle}{180} \cdot \pi\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sqrt{{\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{2}}\\ \end{array} \]

Alternative 9: 55.7% accurate, 2.9× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \left(b - a\right) \cdot \left(b + a\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;b \leq 5.5 \cdot 10^{+267}:\\ \;\;\;\;\left(2 \cdot t_0\right) \cdot \sin \left(0.005555555555555556 \cdot \left(angle_m \cdot \pi\right)\right)\\ \mathbf{else}:\\ \;\;\;\;0.011111111111111112 \cdot \left(angle_m \cdot \left(\pi \cdot t_0\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* (- b a) (+ b a))))
   (*
    angle_s
    (if (<= b 5.5e+267)
      (* (* 2.0 t_0) (sin (* 0.005555555555555556 (* angle_m PI))))
      (* 0.011111111111111112 (* angle_m (* 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 (b <= 5.5e+267) {
		tmp = (2.0 * t_0) * sin((0.005555555555555556 * (angle_m * ((double) M_PI))));
	} else {
		tmp = 0.011111111111111112 * (angle_m * (((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 (b <= 5.5e+267) {
		tmp = (2.0 * t_0) * Math.sin((0.005555555555555556 * (angle_m * Math.PI)));
	} else {
		tmp = 0.011111111111111112 * (angle_m * (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 b <= 5.5e+267:
		tmp = (2.0 * t_0) * math.sin((0.005555555555555556 * (angle_m * math.pi)))
	else:
		tmp = 0.011111111111111112 * (angle_m * (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 (b <= 5.5e+267)
		tmp = Float64(Float64(2.0 * t_0) * sin(Float64(0.005555555555555556 * Float64(angle_m * pi))));
	else
		tmp = Float64(0.011111111111111112 * Float64(angle_m * 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 (b <= 5.5e+267)
		tmp = (2.0 * t_0) * sin((0.005555555555555556 * (angle_m * pi)));
	else
		tmp = 0.011111111111111112 * (angle_m * (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[b, 5.5e+267], N[(N[(2.0 * t$95$0), $MachinePrecision] * N[Sin[N[(0.005555555555555556 * N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(0.011111111111111112 * N[(angle$95$m * N[(Pi * t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if b < 5.49999999999999985e267

   1. Initial program 56.2%

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

      1. unpow2 56.2%

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

      2. unpow2 56.2%

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

      3. difference-of-squares 59.8%

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

   3. Applied egg-rr 59.8%

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

   4. Taylor expanded in angle around 0 57.1%

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

   5. Taylor expanded in angle around inf 58.7%

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

   if 5.49999999999999985e267 < b

   1. Initial program 80.0%

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

      1. unpow2 80.0%

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

      2. unpow2 80.0%

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

      3. difference-of-squares 80.0%

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

   3. Applied egg-rr 80.0%

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

   4. Taylor expanded in angle around 0 80.0%

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

   5. Taylor expanded in angle around 0 100.0%

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

4. Final simplification 59.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq 5.5 \cdot 10^{+267}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\\ \mathbf{else}:\\ \;\;\;\;0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right)\right)\\ \end{array} \]

Alternative 10: 53.8% accurate, 5.4× speedup

Math

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

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

C

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

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

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

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	return Float64(angle_s * Float64(Float64(2.0 * Float64(Float64(b - a) * Float64(b + a))) * Float64(angle_m * Float64(pi * 0.005555555555555556))))
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))) * (angle_m * (pi * 0.005555555555555556)));
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[(angle$95$m * N[(Pi * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 56.7%

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

   1. unpow2 56.7%

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

   2. unpow2 56.7%

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

   3. difference-of-squares 60.2%

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

3. Applied egg-rr 60.2%

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

4. Taylor expanded in angle around 0 57.5%

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

5. Taylor expanded in angle around 0 57.8%

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

   1. *-commutative 57.8%

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

   2. associate-*l* 57.8%

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

7. Simplified 57.8%

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

8. Final simplification 57.8%

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

Alternative 11: 53.8% 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 56.7%

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

   1. unpow2 56.7%

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

   2. unpow2 56.7%

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

   3. difference-of-squares 60.2%

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

3. Applied egg-rr 60.2%

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

4. Taylor expanded in angle around 0 57.5%

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

5. Taylor expanded in angle around 0 57.8%

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

6. Final simplification 57.8%

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

Alternative 12: 53.8% accurate, 5.5× speedup

Math

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

FPCore

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

Derivation

1. Initial program 56.7%

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

   1. unpow2 56.7%

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

   2. unpow2 56.7%

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

   3. difference-of-squares 60.2%

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

3. Applied egg-rr 60.2%

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

4. Taylor expanded in angle around 0 57.5%

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

5. Taylor expanded in angle around 0 57.8%

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

   1. associate-*r* 57.8%

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

   2. +-commutative 57.8%

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

   3. *-commutative 57.8%

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

   4. +-commutative 57.8%

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

7. Simplified 57.8%

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

8. Final simplification 57.8%

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

Reproduce

herbie shell --seed 2023321 
(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)))))