ab-angle->ABCF B

Percentage Accurate: 54.0% → 57.3%

Time: 31.2s

Alternatives: 12

Speedup: 3.6×

• 32.8% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Initial Program: 54.0% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Alternative 1: 57.3% accurate, 1.0× speedup

Math

\[\begin{array}{l} a_m = \left|a\right| \\ angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \pi \cdot \left(angle_m \cdot 0.005555555555555556\right)\\ t_1 := 2 \cdot \left(\left(a_m + b\right) \cdot \left(b - a_m\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;a_m \leq 3.6 \cdot 10^{+66}:\\ \;\;\;\;\left(t_1 \cdot \sin \left(\frac{\pi}{\frac{180}{angle_m}}\right)\right) \cdot \cos \left({\left(\sqrt[3]{t_0}\right)}^{3}\right)\\ \mathbf{elif}\;a_m \leq 2 \cdot 10^{+219}:\\ \;\;\;\;t_1 \cdot \sin t_0\\ \mathbf{else}:\\ \;\;\;\;\left(t_1 \cdot \sin \left(\pi \cdot \frac{angle_m}{180}\right)\right) \cdot \cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(t_0\right)\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

a_m = (fabs.f64 a)
angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a_m b angle_m)
 :precision binary64
 (let* ((t_0 (* PI (* angle_m 0.005555555555555556)))
        (t_1 (* 2.0 (* (+ a_m b) (- b a_m)))))
   (*
    angle_s
    (if (<= a_m 3.6e+66)
      (* (* t_1 (sin (/ PI (/ 180.0 angle_m)))) (cos (pow (cbrt t_0) 3.0)))
      (if (<= a_m 2e+219)
        (* t_1 (sin t_0))
        (*
         (* t_1 (sin (* PI (/ angle_m 180.0))))
         (cos (expm1 (log1p t_0)))))))))

C

a_m = fabs(a);
angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a_m, double b, double angle_m) {
	double t_0 = ((double) M_PI) * (angle_m * 0.005555555555555556);
	double t_1 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (a_m <= 3.6e+66) {
		tmp = (t_1 * sin((((double) M_PI) / (180.0 / angle_m)))) * cos(pow(cbrt(t_0), 3.0));
	} else if (a_m <= 2e+219) {
		tmp = t_1 * sin(t_0);
	} else {
		tmp = (t_1 * sin((((double) M_PI) * (angle_m / 180.0)))) * cos(expm1(log1p(t_0)));
	}
	return angle_s * tmp;
}

Java

a_m = Math.abs(a);
angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a_m, double b, double angle_m) {
	double t_0 = Math.PI * (angle_m * 0.005555555555555556);
	double t_1 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (a_m <= 3.6e+66) {
		tmp = (t_1 * Math.sin((Math.PI / (180.0 / angle_m)))) * Math.cos(Math.pow(Math.cbrt(t_0), 3.0));
	} else if (a_m <= 2e+219) {
		tmp = t_1 * Math.sin(t_0);
	} else {
		tmp = (t_1 * Math.sin((Math.PI * (angle_m / 180.0)))) * Math.cos(Math.expm1(Math.log1p(t_0)));
	}
	return angle_s * tmp;
}

Julia

a_m = abs(a)
angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a_m, b, angle_m)
	t_0 = Float64(pi * Float64(angle_m * 0.005555555555555556))
	t_1 = Float64(2.0 * Float64(Float64(a_m + b) * Float64(b - a_m)))
	tmp = 0.0
	if (a_m <= 3.6e+66)
		tmp = Float64(Float64(t_1 * sin(Float64(pi / Float64(180.0 / angle_m)))) * cos((cbrt(t_0) ^ 3.0)));
	elseif (a_m <= 2e+219)
		tmp = Float64(t_1 * sin(t_0));
	else
		tmp = Float64(Float64(t_1 * sin(Float64(pi * Float64(angle_m / 180.0)))) * cos(expm1(log1p(t_0))));
	end
	return Float64(angle_s * tmp)
end

Wolfram

a_m = N[Abs[a], $MachinePrecision]
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$95$m_, 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[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[a$95$m, 3.6e+66], N[(N[(t$95$1 * N[Sin[N[(Pi / N[(180.0 / angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[Power[N[Power[t$95$0, 1/3], $MachinePrecision], 3.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[a$95$m, 2e+219], N[(t$95$1 * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision], N[(N[(t$95$1 * N[Sin[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(Exp[N[Log[1 + t$95$0], $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if a < 3.6e66

   1. Initial program 53.2%

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

      1. unpow2 53.2%

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

      2. unpow2 53.2%

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

      3. difference-of-squares 55.2%

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

   4. Applied egg-rr 55.2%

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

      1. div-inv 55.5%

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

      2. metadata-eval 55.5%

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

      3. add-cube-cbrt 56.3%

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

      4. pow3 57.3%

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

   6. Applied egg-rr 57.3%

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

      1. associate-*r/ 59.4%

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

      2. associate-/l* 59.2%

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

   8. Applied egg-rr 59.2%

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

   if 3.6e66 < a < 1.99999999999999993e219

   1. Initial program 35.3%

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

      1. unpow2 35.3%

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

      2. unpow2 35.3%

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

      3. difference-of-squares 38.7%

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

   4. Applied egg-rr 38.7%

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

   5. Taylor expanded in angle around 0 45.1%

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

   6. Taylor expanded in angle around inf 38.8%

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

      1. *-commutative 38.8%

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

      2. *-commutative 38.8%

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

      3. associate-*r* 45.5%

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

   8. Simplified 45.5%

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

   if 1.99999999999999993e219 < a

   1. Initial program 43.7%

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

      1. unpow2 43.7%

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

      2. unpow2 43.7%

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

      3. difference-of-squares 77.0%

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

   4. Applied egg-rr 77.0%

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

      1. div-inv 77.0%

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

      2. metadata-eval 77.0%

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

      3. expm1-log1p-u 72.2%

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

   6. Applied egg-rr 72.2%

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

4. Final simplification 58.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 3.6 \cdot 10^{+66}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi}{\frac{180}{angle}}\right)\right) \cdot \cos \left({\left(\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right)}^{3}\right)\\ \mathbf{elif}\;a \leq 2 \cdot 10^{+219}:\\ \;\;\;\;\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 57.2% accurate, 0.8× speedup

Math

\[\begin{array}{l} a_m = \left|a\right| \\ angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle_m \cdot 0.005555555555555556\right)\right)\right)\\ t_1 := 2 \cdot \left(\left(a_m + b\right) \cdot \left(b - a_m\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;{b}^{2} \leq 10^{+280}:\\ \;\;\;\;\left(t_1 \cdot \sin \left(\pi \cdot \frac{angle_m}{180}\right)\right) \cdot \cos t_0\\ \mathbf{else}:\\ \;\;\;\;\left(t_1 \cdot \sin t_0\right) \cdot \cos \left(\left(\pi \cdot angle_m\right) \cdot -0.005555555555555556\right)\\ \end{array} \end{array} \end{array} \]

FPCore

a_m = (fabs.f64 a)
angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a_m b angle_m)
 :precision binary64
 (let* ((t_0 (expm1 (log1p (* PI (* angle_m 0.005555555555555556)))))
        (t_1 (* 2.0 (* (+ a_m b) (- b a_m)))))
   (*
    angle_s
    (if (<= (pow b 2.0) 1e+280)
      (* (* t_1 (sin (* PI (/ angle_m 180.0)))) (cos t_0))
      (* (* t_1 (sin t_0)) (cos (* (* PI angle_m) -0.005555555555555556)))))))

C

a_m = fabs(a);
angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a_m, double b, double angle_m) {
	double t_0 = expm1(log1p((((double) M_PI) * (angle_m * 0.005555555555555556))));
	double t_1 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (pow(b, 2.0) <= 1e+280) {
		tmp = (t_1 * sin((((double) M_PI) * (angle_m / 180.0)))) * cos(t_0);
	} else {
		tmp = (t_1 * sin(t_0)) * cos(((((double) M_PI) * angle_m) * -0.005555555555555556));
	}
	return angle_s * tmp;
}

Java

a_m = Math.abs(a);
angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a_m, double b, double angle_m) {
	double t_0 = Math.expm1(Math.log1p((Math.PI * (angle_m * 0.005555555555555556))));
	double t_1 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (Math.pow(b, 2.0) <= 1e+280) {
		tmp = (t_1 * Math.sin((Math.PI * (angle_m / 180.0)))) * Math.cos(t_0);
	} else {
		tmp = (t_1 * Math.sin(t_0)) * Math.cos(((Math.PI * angle_m) * -0.005555555555555556));
	}
	return angle_s * tmp;
}

Python

a_m = math.fabs(a)
angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a_m, b, angle_m):
	t_0 = math.expm1(math.log1p((math.pi * (angle_m * 0.005555555555555556))))
	t_1 = 2.0 * ((a_m + b) * (b - a_m))
	tmp = 0
	if math.pow(b, 2.0) <= 1e+280:
		tmp = (t_1 * math.sin((math.pi * (angle_m / 180.0)))) * math.cos(t_0)
	else:
		tmp = (t_1 * math.sin(t_0)) * math.cos(((math.pi * angle_m) * -0.005555555555555556))
	return angle_s * tmp

Julia

a_m = abs(a)
angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a_m, b, angle_m)
	t_0 = expm1(log1p(Float64(pi * Float64(angle_m * 0.005555555555555556))))
	t_1 = Float64(2.0 * Float64(Float64(a_m + b) * Float64(b - a_m)))
	tmp = 0.0
	if ((b ^ 2.0) <= 1e+280)
		tmp = Float64(Float64(t_1 * sin(Float64(pi * Float64(angle_m / 180.0)))) * cos(t_0));
	else
		tmp = Float64(Float64(t_1 * sin(t_0)) * cos(Float64(Float64(pi * angle_m) * -0.005555555555555556)));
	end
	return Float64(angle_s * tmp)
end

Wolfram

a_m = N[Abs[a], $MachinePrecision]
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$95$m_, b_, angle$95$m_] := Block[{t$95$0 = N[(Exp[N[Log[1 + N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision]}, Block[{t$95$1 = N[(2.0 * N[(N[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[b, 2.0], $MachinePrecision], 1e+280], N[(N[(t$95$1 * N[Sin[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision], N[(N[(t$95$1 * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(N[(Pi * angle$95$m), $MachinePrecision] * -0.005555555555555556), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 b 2) < 1e280

   1. Initial program 55.9%

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

      1. unpow2 55.9%

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

      2. unpow2 55.9%

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

      3. difference-of-squares 55.9%

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

   4. Applied egg-rr 55.9%

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

      1. div-inv 55.7%

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

      2. metadata-eval 55.7%

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

      3. expm1-log1p-u 51.0%

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

   6. Applied egg-rr 51.0%

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

   if 1e280 < (pow.f64 b 2)

   1. Initial program 36.1%

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

      1. unpow2 36.1%

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

      2. unpow2 36.1%

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

      3. difference-of-squares 52.5%

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

   4. Applied egg-rr 52.5%

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

      1. div-inv 53.8%

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

      2. metadata-eval 53.8%

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

      3. expm1-log1p-u 47.0%

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

   6. Applied egg-rr 47.0%

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

      1. div-inv 53.8%

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

      2. metadata-eval 53.8%

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

      3. expm1-log1p-u 47.0%

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

   8. Applied egg-rr 51.1%

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

   9. Taylor expanded in angle around inf 55.2%

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

       1. *-commutative 55.2%

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

       2. metadata-eval 55.2%

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

       3. associate-/r/ 51.1%

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

       4. metadata-eval 51.1%

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

       5. distribute-neg-frac 51.1%

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

       6. cos-neg 51.1%

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

       7. associate-/r/ 55.2%

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

       8. metadata-eval 55.2%

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

       9. *-commutative 55.2%

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

       10. *-commutative 55.2%

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

   11. Simplified 55.2%

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

4. Final simplification 52.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} \leq 10^{+280}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right)\right) \cdot \cos \left(\left(\pi \cdot angle\right) \cdot -0.005555555555555556\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 57.0% accurate, 1.0× speedup

Math

\[\begin{array}{l} a_m = \left|a\right| \\ angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \pi \cdot \left(angle_m \cdot 0.005555555555555556\right)\\ t_1 := 2 \cdot \left(\left(a_m + b\right) \cdot \left(b - a_m\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;a_m \leq 2.95 \cdot 10^{-189}:\\ \;\;\;\;\left(t_1 \cdot \sin \left(\pi \cdot \frac{angle_m}{180}\right)\right) \cdot \cos \left(angle_m \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\\ \mathbf{elif}\;a_m \leq 1.5 \cdot 10^{+60}:\\ \;\;\;\;\left(t_1 \cdot \sin \left(\mathsf{expm1}\left(\mathsf{log1p}\left(t_0\right)\right)\right)\right) \cdot \cos \left(\left(\pi \cdot angle_m\right) \cdot -0.005555555555555556\right)\\ \mathbf{else}:\\ \;\;\;\;t_1 \cdot \sin t_0\\ \end{array} \end{array} \end{array} \]

FPCore

a_m = (fabs.f64 a)
angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a_m b angle_m)
 :precision binary64
 (let* ((t_0 (* PI (* angle_m 0.005555555555555556)))
        (t_1 (* 2.0 (* (+ a_m b) (- b a_m)))))
   (*
    angle_s
    (if (<= a_m 2.95e-189)
      (*
       (* t_1 (sin (* PI (/ angle_m 180.0))))
       (cos (* angle_m (* PI 0.005555555555555556))))
      (if (<= a_m 1.5e+60)
        (*
         (* t_1 (sin (expm1 (log1p t_0))))
         (cos (* (* PI angle_m) -0.005555555555555556)))
        (* t_1 (sin t_0)))))))

C

a_m = fabs(a);
angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a_m, double b, double angle_m) {
	double t_0 = ((double) M_PI) * (angle_m * 0.005555555555555556);
	double t_1 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (a_m <= 2.95e-189) {
		tmp = (t_1 * sin((((double) M_PI) * (angle_m / 180.0)))) * cos((angle_m * (((double) M_PI) * 0.005555555555555556)));
	} else if (a_m <= 1.5e+60) {
		tmp = (t_1 * sin(expm1(log1p(t_0)))) * cos(((((double) M_PI) * angle_m) * -0.005555555555555556));
	} else {
		tmp = t_1 * sin(t_0);
	}
	return angle_s * tmp;
}

Java

a_m = Math.abs(a);
angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a_m, double b, double angle_m) {
	double t_0 = Math.PI * (angle_m * 0.005555555555555556);
	double t_1 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (a_m <= 2.95e-189) {
		tmp = (t_1 * Math.sin((Math.PI * (angle_m / 180.0)))) * Math.cos((angle_m * (Math.PI * 0.005555555555555556)));
	} else if (a_m <= 1.5e+60) {
		tmp = (t_1 * Math.sin(Math.expm1(Math.log1p(t_0)))) * Math.cos(((Math.PI * angle_m) * -0.005555555555555556));
	} else {
		tmp = t_1 * Math.sin(t_0);
	}
	return angle_s * tmp;
}

Python

a_m = math.fabs(a)
angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a_m, b, angle_m):
	t_0 = math.pi * (angle_m * 0.005555555555555556)
	t_1 = 2.0 * ((a_m + b) * (b - a_m))
	tmp = 0
	if a_m <= 2.95e-189:
		tmp = (t_1 * math.sin((math.pi * (angle_m / 180.0)))) * math.cos((angle_m * (math.pi * 0.005555555555555556)))
	elif a_m <= 1.5e+60:
		tmp = (t_1 * math.sin(math.expm1(math.log1p(t_0)))) * math.cos(((math.pi * angle_m) * -0.005555555555555556))
	else:
		tmp = t_1 * math.sin(t_0)
	return angle_s * tmp

Julia

a_m = abs(a)
angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a_m, b, angle_m)
	t_0 = Float64(pi * Float64(angle_m * 0.005555555555555556))
	t_1 = Float64(2.0 * Float64(Float64(a_m + b) * Float64(b - a_m)))
	tmp = 0.0
	if (a_m <= 2.95e-189)
		tmp = Float64(Float64(t_1 * sin(Float64(pi * Float64(angle_m / 180.0)))) * cos(Float64(angle_m * Float64(pi * 0.005555555555555556))));
	elseif (a_m <= 1.5e+60)
		tmp = Float64(Float64(t_1 * sin(expm1(log1p(t_0)))) * cos(Float64(Float64(pi * angle_m) * -0.005555555555555556)));
	else
		tmp = Float64(t_1 * sin(t_0));
	end
	return Float64(angle_s * tmp)
end

Wolfram

a_m = N[Abs[a], $MachinePrecision]
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$95$m_, 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[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[a$95$m, 2.95e-189], N[(N[(t$95$1 * N[Sin[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(angle$95$m * N[(Pi * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[a$95$m, 1.5e+60], N[(N[(t$95$1 * N[Sin[N[(Exp[N[Log[1 + t$95$0], $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(N[(Pi * angle$95$m), $MachinePrecision] * -0.005555555555555556), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(t$95$1 * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if a < 2.95e-189

   1. Initial program 51.6%

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

      1. unpow2 51.6%

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

      2. unpow2 51.6%

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

      3. difference-of-squares 54.1%

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

   4. Applied egg-rr 54.1%

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

   5. Taylor expanded in angle around inf 54.9%

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

      1. *-commutative 54.9%

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

      2. associate-*l* 55.4%

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

   7. Simplified 55.4%

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

   if 2.95e-189 < a < 1.4999999999999999e60

   1. Initial program 60.1%

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

      1. unpow2 60.1%

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

      2. unpow2 60.1%

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

      3. difference-of-squares 60.1%

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

   4. Applied egg-rr 60.1%

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

      1. div-inv 62.3%

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

      2. metadata-eval 62.3%

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

      3. expm1-log1p-u 49.8%

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

   6. Applied egg-rr 49.8%

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

      1. div-inv 62.3%

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

      2. metadata-eval 62.3%

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

      3. expm1-log1p-u 49.8%

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

   8. Applied egg-rr 52.3%

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

   9. Taylor expanded in angle around inf 53.4%

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

       1. *-commutative 53.4%

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

       2. metadata-eval 53.4%

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

       3. associate-/r/ 51.1%

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

       4. metadata-eval 51.1%

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

       5. distribute-neg-frac 51.1%

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

       6. cos-neg 51.1%

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

       7. associate-/r/ 53.4%

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

       8. metadata-eval 53.4%

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

       9. *-commutative 53.4%

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

       10. *-commutative 53.4%

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

   11. Simplified 53.4%

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

   if 1.4999999999999999e60 < a

   1. Initial program 38.5%

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

      1. unpow2 38.5%

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

      2. unpow2 38.5%

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

      3. difference-of-squares 53.0%

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

   4. Applied egg-rr 53.0%

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

   5. Taylor expanded in angle around 0 56.6%

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

   6. Taylor expanded in angle around inf 51.2%

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

      1. *-commutative 51.2%

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

      2. *-commutative 51.2%

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

      3. associate-*r* 56.8%

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

   8. Simplified 56.8%

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

4. Final simplification 55.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 2.95 \cdot 10^{-189}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\\ \mathbf{elif}\;a \leq 1.5 \cdot 10^{+60}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right)\right) \cdot \cos \left(\left(\pi \cdot angle\right) \cdot -0.005555555555555556\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 57.6% accurate, 1.0× speedup

Math

\[\begin{array}{l} a_m = \left|a\right| \\ angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \pi \cdot \left(angle_m \cdot 0.005555555555555556\right)\\ t_1 := 2 \cdot \left(\left(a_m + b\right) \cdot \left(b - a_m\right)\right)\\ t_2 := t_1 \cdot \sin \left(\pi \cdot \frac{angle_m}{180}\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;a_m \leq 2.8 \cdot 10^{+59}:\\ \;\;\;\;t_2 \cdot \cos \left(\frac{angle_m}{180} \cdot \sqrt[3]{{\pi}^{3}}\right)\\ \mathbf{elif}\;a_m \leq 5 \cdot 10^{+225}:\\ \;\;\;\;t_1 \cdot \sin t_0\\ \mathbf{else}:\\ \;\;\;\;t_2 \cdot \cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(t_0\right)\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

a_m = (fabs.f64 a)
angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a_m b angle_m)
 :precision binary64
 (let* ((t_0 (* PI (* angle_m 0.005555555555555556)))
        (t_1 (* 2.0 (* (+ a_m b) (- b a_m))))
        (t_2 (* t_1 (sin (* PI (/ angle_m 180.0))))))
   (*
    angle_s
    (if (<= a_m 2.8e+59)
      (* t_2 (cos (* (/ angle_m 180.0) (cbrt (pow PI 3.0)))))
      (if (<= a_m 5e+225)
        (* t_1 (sin t_0))
        (* t_2 (cos (expm1 (log1p t_0)))))))))

C

a_m = fabs(a);
angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a_m, double b, double angle_m) {
	double t_0 = ((double) M_PI) * (angle_m * 0.005555555555555556);
	double t_1 = 2.0 * ((a_m + b) * (b - a_m));
	double t_2 = t_1 * sin((((double) M_PI) * (angle_m / 180.0)));
	double tmp;
	if (a_m <= 2.8e+59) {
		tmp = t_2 * cos(((angle_m / 180.0) * cbrt(pow(((double) M_PI), 3.0))));
	} else if (a_m <= 5e+225) {
		tmp = t_1 * sin(t_0);
	} else {
		tmp = t_2 * cos(expm1(log1p(t_0)));
	}
	return angle_s * tmp;
}

Java

a_m = Math.abs(a);
angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a_m, double b, double angle_m) {
	double t_0 = Math.PI * (angle_m * 0.005555555555555556);
	double t_1 = 2.0 * ((a_m + b) * (b - a_m));
	double t_2 = t_1 * Math.sin((Math.PI * (angle_m / 180.0)));
	double tmp;
	if (a_m <= 2.8e+59) {
		tmp = t_2 * Math.cos(((angle_m / 180.0) * Math.cbrt(Math.pow(Math.PI, 3.0))));
	} else if (a_m <= 5e+225) {
		tmp = t_1 * Math.sin(t_0);
	} else {
		tmp = t_2 * Math.cos(Math.expm1(Math.log1p(t_0)));
	}
	return angle_s * tmp;
}

Julia

a_m = abs(a)
angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a_m, b, angle_m)
	t_0 = Float64(pi * Float64(angle_m * 0.005555555555555556))
	t_1 = Float64(2.0 * Float64(Float64(a_m + b) * Float64(b - a_m)))
	t_2 = Float64(t_1 * sin(Float64(pi * Float64(angle_m / 180.0))))
	tmp = 0.0
	if (a_m <= 2.8e+59)
		tmp = Float64(t_2 * cos(Float64(Float64(angle_m / 180.0) * cbrt((pi ^ 3.0)))));
	elseif (a_m <= 5e+225)
		tmp = Float64(t_1 * sin(t_0));
	else
		tmp = Float64(t_2 * cos(expm1(log1p(t_0))));
	end
	return Float64(angle_s * tmp)
end

Wolfram

a_m = N[Abs[a], $MachinePrecision]
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$95$m_, 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[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(t$95$1 * N[Sin[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[a$95$m, 2.8e+59], N[(t$95$2 * N[Cos[N[(N[(angle$95$m / 180.0), $MachinePrecision] * N[Power[N[Power[Pi, 3.0], $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[a$95$m, 5e+225], N[(t$95$1 * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision], N[(t$95$2 * N[Cos[N[(Exp[N[Log[1 + t$95$0], $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if a < 2.7999999999999998e59

   1. Initial program 53.5%

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

      1. unpow2 53.5%

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

      2. unpow2 53.5%

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

      3. difference-of-squares 55.5%

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

   4. Applied egg-rr 55.5%

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

      1. add-cbrt-cube 58.6%

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

      2. pow3 58.6%

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

   6. Applied egg-rr 58.6%

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

   if 2.7999999999999998e59 < a < 4.99999999999999981e225

   1. Initial program 35.4%

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

      1. unpow2 35.4%

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

      2. unpow2 35.4%

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

      3. difference-of-squares 38.5%

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

   4. Applied egg-rr 38.5%

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

   5. Taylor expanded in angle around 0 44.4%

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

   6. Taylor expanded in angle around inf 38.6%

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

      1. *-commutative 38.6%

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

      2. *-commutative 38.6%

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

      3. associate-*r* 44.7%

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

   8. Simplified 44.7%

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

   if 4.99999999999999981e225 < a

   1. Initial program 43.7%

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

      1. unpow2 43.7%

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

      2. unpow2 43.7%

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

      3. difference-of-squares 77.0%

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

   4. Applied egg-rr 77.0%

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

      1. div-inv 77.0%

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

      2. metadata-eval 77.0%

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

      3. expm1-log1p-u 72.2%

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

   6. Applied egg-rr 72.2%

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

4. Final simplification 57.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 2.8 \cdot 10^{+59}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\frac{angle}{180} \cdot \sqrt[3]{{\pi}^{3}}\right)\\ \mathbf{elif}\;a \leq 5 \cdot 10^{+225}:\\ \;\;\;\;\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 57.1% accurate, 1.3× speedup

Math

\[\begin{array}{l} a_m = \left|a\right| \\ angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(a_m + b\right) \cdot \left(b - a_m\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;{a_m}^{2} \leq 5 \cdot 10^{+118}:\\ \;\;\;\;\left(t_0 \cdot \sin \left(\pi \cdot \frac{angle_m}{180}\right)\right) \cdot \cos \left(\frac{0.005555555555555556}{\frac{1}{\pi \cdot angle_m}}\right)\\ \mathbf{else}:\\ \;\;\;\;t_0 \cdot \sin \left(\pi \cdot \left(angle_m \cdot 0.005555555555555556\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

a_m = (fabs.f64 a)
angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a_m b angle_m)
 :precision binary64
 (let* ((t_0 (* 2.0 (* (+ a_m b) (- b a_m)))))
   (*
    angle_s
    (if (<= (pow a_m 2.0) 5e+118)
      (*
       (* t_0 (sin (* PI (/ angle_m 180.0))))
       (cos (/ 0.005555555555555556 (/ 1.0 (* PI angle_m)))))
      (* t_0 (sin (* PI (* angle_m 0.005555555555555556))))))))

C

a_m = fabs(a);
angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a_m, double b, double angle_m) {
	double t_0 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (pow(a_m, 2.0) <= 5e+118) {
		tmp = (t_0 * sin((((double) M_PI) * (angle_m / 180.0)))) * cos((0.005555555555555556 / (1.0 / (((double) M_PI) * angle_m))));
	} else {
		tmp = t_0 * sin((((double) M_PI) * (angle_m * 0.005555555555555556)));
	}
	return angle_s * tmp;
}

Java

a_m = Math.abs(a);
angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a_m, double b, double angle_m) {
	double t_0 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (Math.pow(a_m, 2.0) <= 5e+118) {
		tmp = (t_0 * Math.sin((Math.PI * (angle_m / 180.0)))) * Math.cos((0.005555555555555556 / (1.0 / (Math.PI * angle_m))));
	} else {
		tmp = t_0 * Math.sin((Math.PI * (angle_m * 0.005555555555555556)));
	}
	return angle_s * tmp;
}

Python

a_m = math.fabs(a)
angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a_m, b, angle_m):
	t_0 = 2.0 * ((a_m + b) * (b - a_m))
	tmp = 0
	if math.pow(a_m, 2.0) <= 5e+118:
		tmp = (t_0 * math.sin((math.pi * (angle_m / 180.0)))) * math.cos((0.005555555555555556 / (1.0 / (math.pi * angle_m))))
	else:
		tmp = t_0 * math.sin((math.pi * (angle_m * 0.005555555555555556)))
	return angle_s * tmp

Julia

a_m = abs(a)
angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a_m, b, angle_m)
	t_0 = Float64(2.0 * Float64(Float64(a_m + b) * Float64(b - a_m)))
	tmp = 0.0
	if ((a_m ^ 2.0) <= 5e+118)
		tmp = Float64(Float64(t_0 * sin(Float64(pi * Float64(angle_m / 180.0)))) * cos(Float64(0.005555555555555556 / Float64(1.0 / Float64(pi * angle_m)))));
	else
		tmp = Float64(t_0 * sin(Float64(pi * Float64(angle_m * 0.005555555555555556))));
	end
	return Float64(angle_s * tmp)
end

MATLAB

a_m = abs(a);
angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a_m, b, angle_m)
	t_0 = 2.0 * ((a_m + b) * (b - a_m));
	tmp = 0.0;
	if ((a_m ^ 2.0) <= 5e+118)
		tmp = (t_0 * sin((pi * (angle_m / 180.0)))) * cos((0.005555555555555556 / (1.0 / (pi * angle_m))));
	else
		tmp = t_0 * sin((pi * (angle_m * 0.005555555555555556)));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

a_m = N[Abs[a], $MachinePrecision]
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$95$m_, b_, angle$95$m_] := Block[{t$95$0 = N[(2.0 * N[(N[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[a$95$m, 2.0], $MachinePrecision], 5e+118], N[(N[(t$95$0 * N[Sin[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(0.005555555555555556 / N[(1.0 / N[(Pi * angle$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(t$95$0 * N[Sin[N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a 2) < 4.99999999999999972e118

   1. Initial program 58.7%

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

      1. unpow2 58.7%

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

      2. unpow2 58.7%

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

      3. difference-of-squares 58.7%

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

   4. Applied egg-rr 58.7%

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

      1. div-inv 59.5%

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

      2. metadata-eval 59.5%

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

      3. expm1-log1p-u 51.2%

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

   6. Applied egg-rr 51.2%

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

      1. expm1-log1p-u 59.5%

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

      2. metadata-eval 59.5%

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

      3. div-inv 58.7%

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

      4. associate-*r/ 61.8%

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

      5. clear-num 61.1%

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

      6. div-inv 62.2%

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

      7. associate-/r* 62.8%

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

      8. metadata-eval 62.8%

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

   8. Applied egg-rr 62.8%

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

   if 4.99999999999999972e118 < (pow.f64 a 2)

   1. Initial program 39.1%

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

      1. unpow2 39.1%

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

      2. unpow2 39.1%

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

      3. difference-of-squares 50.0%

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

   4. Applied egg-rr 50.0%

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

   5. Taylor expanded in angle around 0 55.3%

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

   6. Taylor expanded in angle around inf 52.5%

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

      1. *-commutative 52.5%

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

      2. *-commutative 52.5%

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

      3. associate-*r* 59.4%

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

   8. Simplified 59.4%

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

4. Final simplification 61.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{+118}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\frac{0.005555555555555556}{\frac{1}{\pi \cdot angle}}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 57.1% accurate, 1.3× speedup

Math

\[\begin{array}{l} a_m = \left|a\right| \\ angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(a_m + b\right) \cdot \left(b - a_m\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;{a_m}^{2} \leq 5 \cdot 10^{+120}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle_m}{180}\right) \cdot \left(t_0 \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle_m\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;t_0 \cdot \sin \left(\pi \cdot \left(angle_m \cdot 0.005555555555555556\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

a_m = (fabs.f64 a)
angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a_m b angle_m)
 :precision binary64
 (let* ((t_0 (* 2.0 (* (+ a_m b) (- b a_m)))))
   (*
    angle_s
    (if (<= (pow a_m 2.0) 5e+120)
      (*
       (cos (* PI (/ angle_m 180.0)))
       (* t_0 (sin (* 0.005555555555555556 (* PI angle_m)))))
      (* t_0 (sin (* PI (* angle_m 0.005555555555555556))))))))

C

a_m = fabs(a);
angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a_m, double b, double angle_m) {
	double t_0 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (pow(a_m, 2.0) <= 5e+120) {
		tmp = cos((((double) M_PI) * (angle_m / 180.0))) * (t_0 * sin((0.005555555555555556 * (((double) M_PI) * angle_m))));
	} else {
		tmp = t_0 * sin((((double) M_PI) * (angle_m * 0.005555555555555556)));
	}
	return angle_s * tmp;
}

Java

a_m = Math.abs(a);
angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a_m, double b, double angle_m) {
	double t_0 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (Math.pow(a_m, 2.0) <= 5e+120) {
		tmp = Math.cos((Math.PI * (angle_m / 180.0))) * (t_0 * Math.sin((0.005555555555555556 * (Math.PI * angle_m))));
	} else {
		tmp = t_0 * Math.sin((Math.PI * (angle_m * 0.005555555555555556)));
	}
	return angle_s * tmp;
}

Python

a_m = math.fabs(a)
angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a_m, b, angle_m):
	t_0 = 2.0 * ((a_m + b) * (b - a_m))
	tmp = 0
	if math.pow(a_m, 2.0) <= 5e+120:
		tmp = math.cos((math.pi * (angle_m / 180.0))) * (t_0 * math.sin((0.005555555555555556 * (math.pi * angle_m))))
	else:
		tmp = t_0 * math.sin((math.pi * (angle_m * 0.005555555555555556)))
	return angle_s * tmp

Julia

a_m = abs(a)
angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a_m, b, angle_m)
	t_0 = Float64(2.0 * Float64(Float64(a_m + b) * Float64(b - a_m)))
	tmp = 0.0
	if ((a_m ^ 2.0) <= 5e+120)
		tmp = Float64(cos(Float64(pi * Float64(angle_m / 180.0))) * Float64(t_0 * sin(Float64(0.005555555555555556 * Float64(pi * angle_m)))));
	else
		tmp = Float64(t_0 * sin(Float64(pi * Float64(angle_m * 0.005555555555555556))));
	end
	return Float64(angle_s * tmp)
end

MATLAB

a_m = abs(a);
angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a_m, b, angle_m)
	t_0 = 2.0 * ((a_m + b) * (b - a_m));
	tmp = 0.0;
	if ((a_m ^ 2.0) <= 5e+120)
		tmp = cos((pi * (angle_m / 180.0))) * (t_0 * sin((0.005555555555555556 * (pi * angle_m))));
	else
		tmp = t_0 * sin((pi * (angle_m * 0.005555555555555556)));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

a_m = N[Abs[a], $MachinePrecision]
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$95$m_, b_, angle$95$m_] := Block[{t$95$0 = N[(2.0 * N[(N[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[a$95$m, 2.0], $MachinePrecision], 5e+120], N[(N[Cos[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(t$95$0 * N[Sin[N[(0.005555555555555556 * N[(Pi * angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(t$95$0 * N[Sin[N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a 2) < 5.00000000000000019e120

   1. Initial program 58.3%

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

      1. unpow2 58.3%

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

      2. unpow2 58.3%

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

      3. difference-of-squares 58.3%

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

   4. Applied egg-rr 58.3%

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

   5. Taylor expanded in angle around 0 60.8%

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

   if 5.00000000000000019e120 < (pow.f64 a 2)

   1. Initial program 39.4%

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

      1. unpow2 39.4%

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

      2. unpow2 39.4%

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

      3. difference-of-squares 50.4%

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

   4. Applied egg-rr 50.4%

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

   5. Taylor expanded in angle around 0 55.8%

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

   6. Taylor expanded in angle around inf 53.0%

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

      1. *-commutative 53.0%

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

      2. *-commutative 53.0%

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

      3. associate-*r* 60.0%

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

   8. Simplified 60.0%

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

4. Final simplification 60.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{+120}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 57.1% accurate, 1.3× speedup

Math

\[\begin{array}{l} a_m = \left|a\right| \\ angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(a_m + b\right) \cdot \left(b - a_m\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;{a_m}^{2} \leq 10^{+124}:\\ \;\;\;\;\left(t_0 \cdot \sin \left(angle_m \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\right) \cdot \cos \left(\pi \cdot \frac{angle_m}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;t_0 \cdot \sin \left(\pi \cdot \left(angle_m \cdot 0.005555555555555556\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

a_m = (fabs.f64 a)
angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a_m b angle_m)
 :precision binary64
 (let* ((t_0 (* 2.0 (* (+ a_m b) (- b a_m)))))
   (*
    angle_s
    (if (<= (pow a_m 2.0) 1e+124)
      (*
       (* t_0 (sin (* angle_m (* PI 0.005555555555555556))))
       (cos (* PI (/ angle_m 180.0))))
      (* t_0 (sin (* PI (* angle_m 0.005555555555555556))))))))

C

a_m = fabs(a);
angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a_m, double b, double angle_m) {
	double t_0 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (pow(a_m, 2.0) <= 1e+124) {
		tmp = (t_0 * sin((angle_m * (((double) M_PI) * 0.005555555555555556)))) * cos((((double) M_PI) * (angle_m / 180.0)));
	} else {
		tmp = t_0 * sin((((double) M_PI) * (angle_m * 0.005555555555555556)));
	}
	return angle_s * tmp;
}

Java

a_m = Math.abs(a);
angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a_m, double b, double angle_m) {
	double t_0 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (Math.pow(a_m, 2.0) <= 1e+124) {
		tmp = (t_0 * Math.sin((angle_m * (Math.PI * 0.005555555555555556)))) * Math.cos((Math.PI * (angle_m / 180.0)));
	} else {
		tmp = t_0 * Math.sin((Math.PI * (angle_m * 0.005555555555555556)));
	}
	return angle_s * tmp;
}

Python

a_m = math.fabs(a)
angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a_m, b, angle_m):
	t_0 = 2.0 * ((a_m + b) * (b - a_m))
	tmp = 0
	if math.pow(a_m, 2.0) <= 1e+124:
		tmp = (t_0 * math.sin((angle_m * (math.pi * 0.005555555555555556)))) * math.cos((math.pi * (angle_m / 180.0)))
	else:
		tmp = t_0 * math.sin((math.pi * (angle_m * 0.005555555555555556)))
	return angle_s * tmp

Julia

a_m = abs(a)
angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a_m, b, angle_m)
	t_0 = Float64(2.0 * Float64(Float64(a_m + b) * Float64(b - a_m)))
	tmp = 0.0
	if ((a_m ^ 2.0) <= 1e+124)
		tmp = Float64(Float64(t_0 * sin(Float64(angle_m * Float64(pi * 0.005555555555555556)))) * cos(Float64(pi * Float64(angle_m / 180.0))));
	else
		tmp = Float64(t_0 * sin(Float64(pi * Float64(angle_m * 0.005555555555555556))));
	end
	return Float64(angle_s * tmp)
end

MATLAB

a_m = abs(a);
angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a_m, b, angle_m)
	t_0 = 2.0 * ((a_m + b) * (b - a_m));
	tmp = 0.0;
	if ((a_m ^ 2.0) <= 1e+124)
		tmp = (t_0 * sin((angle_m * (pi * 0.005555555555555556)))) * cos((pi * (angle_m / 180.0)));
	else
		tmp = t_0 * sin((pi * (angle_m * 0.005555555555555556)));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

a_m = N[Abs[a], $MachinePrecision]
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$95$m_, b_, angle$95$m_] := Block[{t$95$0 = N[(2.0 * N[(N[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[a$95$m, 2.0], $MachinePrecision], 1e+124], N[(N[(t$95$0 * N[Sin[N[(angle$95$m * N[(Pi * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(t$95$0 * N[Sin[N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a 2) < 9.99999999999999948e123

   1. Initial program 58.2%

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

      1. unpow2 58.2%

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

      2. unpow2 58.2%

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

      3. difference-of-squares 58.2%

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

   4. Applied egg-rr 58.2%

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

   5. Taylor expanded in angle around inf 60.7%

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

      1. *-commutative 60.7%

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

      2. associate-*l* 60.8%

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

   7. Simplified 60.8%

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

   if 9.99999999999999948e123 < (pow.f64 a 2)

   1. Initial program 39.2%

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

      1. unpow2 39.2%

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

      2. unpow2 39.2%

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

      3. difference-of-squares 50.4%

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

   4. Applied egg-rr 50.4%

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

   5. Taylor expanded in angle around 0 55.8%

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

   6. Taylor expanded in angle around inf 53.0%

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

      1. *-commutative 53.0%

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

      2. *-commutative 53.0%

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

      3. associate-*r* 60.1%

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

   8. Simplified 60.1%

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

4. Final simplification 60.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 10^{+124}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 8: 57.2% accurate, 1.3× speedup

Math

\[\begin{array}{l} a_m = \left|a\right| \\ angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(a_m + b\right) \cdot \left(b - a_m\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;{a_m}^{2} \leq 5 \cdot 10^{+120}:\\ \;\;\;\;\left(t_0 \cdot \sin \left(\pi \cdot \frac{angle_m}{180}\right)\right) \cdot \cos \left(angle_m \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;t_0 \cdot \sin \left(\pi \cdot \left(angle_m \cdot 0.005555555555555556\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

a_m = (fabs.f64 a)
angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a_m b angle_m)
 :precision binary64
 (let* ((t_0 (* 2.0 (* (+ a_m b) (- b a_m)))))
   (*
    angle_s
    (if (<= (pow a_m 2.0) 5e+120)
      (*
       (* t_0 (sin (* PI (/ angle_m 180.0))))
       (cos (* angle_m (* PI 0.005555555555555556))))
      (* t_0 (sin (* PI (* angle_m 0.005555555555555556))))))))

C

a_m = fabs(a);
angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a_m, double b, double angle_m) {
	double t_0 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (pow(a_m, 2.0) <= 5e+120) {
		tmp = (t_0 * sin((((double) M_PI) * (angle_m / 180.0)))) * cos((angle_m * (((double) M_PI) * 0.005555555555555556)));
	} else {
		tmp = t_0 * sin((((double) M_PI) * (angle_m * 0.005555555555555556)));
	}
	return angle_s * tmp;
}

Java

a_m = Math.abs(a);
angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a_m, double b, double angle_m) {
	double t_0 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (Math.pow(a_m, 2.0) <= 5e+120) {
		tmp = (t_0 * Math.sin((Math.PI * (angle_m / 180.0)))) * Math.cos((angle_m * (Math.PI * 0.005555555555555556)));
	} else {
		tmp = t_0 * Math.sin((Math.PI * (angle_m * 0.005555555555555556)));
	}
	return angle_s * tmp;
}

Python

a_m = math.fabs(a)
angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a_m, b, angle_m):
	t_0 = 2.0 * ((a_m + b) * (b - a_m))
	tmp = 0
	if math.pow(a_m, 2.0) <= 5e+120:
		tmp = (t_0 * math.sin((math.pi * (angle_m / 180.0)))) * math.cos((angle_m * (math.pi * 0.005555555555555556)))
	else:
		tmp = t_0 * math.sin((math.pi * (angle_m * 0.005555555555555556)))
	return angle_s * tmp

Julia

a_m = abs(a)
angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a_m, b, angle_m)
	t_0 = Float64(2.0 * Float64(Float64(a_m + b) * Float64(b - a_m)))
	tmp = 0.0
	if ((a_m ^ 2.0) <= 5e+120)
		tmp = Float64(Float64(t_0 * sin(Float64(pi * Float64(angle_m / 180.0)))) * cos(Float64(angle_m * Float64(pi * 0.005555555555555556))));
	else
		tmp = Float64(t_0 * sin(Float64(pi * Float64(angle_m * 0.005555555555555556))));
	end
	return Float64(angle_s * tmp)
end

MATLAB

a_m = abs(a);
angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a_m, b, angle_m)
	t_0 = 2.0 * ((a_m + b) * (b - a_m));
	tmp = 0.0;
	if ((a_m ^ 2.0) <= 5e+120)
		tmp = (t_0 * sin((pi * (angle_m / 180.0)))) * cos((angle_m * (pi * 0.005555555555555556)));
	else
		tmp = t_0 * sin((pi * (angle_m * 0.005555555555555556)));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

a_m = N[Abs[a], $MachinePrecision]
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$95$m_, b_, angle$95$m_] := Block[{t$95$0 = N[(2.0 * N[(N[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[a$95$m, 2.0], $MachinePrecision], 5e+120], N[(N[(t$95$0 * N[Sin[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(angle$95$m * N[(Pi * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(t$95$0 * N[Sin[N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a 2) < 5.00000000000000019e120

   1. Initial program 58.3%

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

      1. unpow2 58.3%

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

      2. unpow2 58.3%

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

      3. difference-of-squares 58.3%

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

   4. Applied egg-rr 58.3%

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

   5. Taylor expanded in angle around inf 60.4%

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

      1. *-commutative 60.4%

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

      2. associate-*l* 61.5%

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

   7. Simplified 61.5%

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

   if 5.00000000000000019e120 < (pow.f64 a 2)

   1. Initial program 39.4%

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

      1. unpow2 39.4%

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

      2. unpow2 39.4%

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

      3. difference-of-squares 50.4%

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

   4. Applied egg-rr 50.4%

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

   5. Taylor expanded in angle around 0 55.8%

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

   6. Taylor expanded in angle around inf 53.0%

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

      1. *-commutative 53.0%

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

      2. *-commutative 53.0%

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

      3. associate-*r* 60.0%

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

   8. Simplified 60.0%

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

4. Final simplification 60.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{+120}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 9: 55.9% accurate, 3.6× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

a_m = N[Abs[a], $MachinePrecision]
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$95$m_, b_, angle$95$m_] := N[(angle$95$s * N[(N[(2.0 * N[(N[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(0.005555555555555556 * N[(Pi * angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 50.2%

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

   1. unpow2 50.2%

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

   2. unpow2 50.2%

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

   3. difference-of-squares 54.9%

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

4. Applied egg-rr 54.9%

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

5. Taylor expanded in angle around 0 57.1%

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

6. Taylor expanded in angle around inf 55.6%

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

7. Final simplification 55.6%

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

Alternative 10: 56.0% accurate, 3.6× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

a_m = N[Abs[a], $MachinePrecision]
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$95$m_, b_, angle$95$m_] := N[(angle$95$s * N[(N[(2.0 * N[(N[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 50.2%

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

   1. unpow2 50.2%

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

   2. unpow2 50.2%

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

   3. difference-of-squares 54.9%

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

4. Applied egg-rr 54.9%

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

5. Taylor expanded in angle around 0 57.1%

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

6. Taylor expanded in angle around inf 55.6%

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

   1. *-commutative 55.6%

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

   2. *-commutative 55.6%

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

   3. associate-*r* 57.8%

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

8. Simplified 57.8%

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

9. Final simplification 57.8%

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

Alternative 11: 53.9% accurate, 27.9× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

a_m = N[Abs[a], $MachinePrecision]
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$95$m_, b_, angle$95$m_] := N[(angle$95$s * N[(N[(2.0 * N[(N[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(angle$95$m * N[(Pi * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 50.2%

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

   1. unpow2 50.2%

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

   2. unpow2 50.2%

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

   3. difference-of-squares 54.9%

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

4. Applied egg-rr 54.9%

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

5. Taylor expanded in angle around 0 57.1%

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

6. Taylor expanded in angle around 0 51.6%

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

   1. *-commutative 51.6%

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

   2. associate-*l* 51.6%

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

8. Simplified 51.6%

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

9. Final simplification 51.6%

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

Alternative 12: 53.9% accurate, 32.2× speedup

Math

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

FPCore

a_m = (fabs.f64 a)
angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a_m b angle_m)
 :precision binary64
 (*
  angle_s
  (* 0.011111111111111112 (* angle_m (* (* (+ a_m b) (- b a_m)) PI)))))

C

a_m = fabs(a);
angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a_m, double b, double angle_m) {
	return angle_s * (0.011111111111111112 * (angle_m * (((a_m + b) * (b - a_m)) * ((double) M_PI))));
}

Java

a_m = Math.abs(a);
angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a_m, double b, double angle_m) {
	return angle_s * (0.011111111111111112 * (angle_m * (((a_m + b) * (b - a_m)) * Math.PI)));
}

Python

a_m = math.fabs(a)
angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a_m, b, angle_m):
	return angle_s * (0.011111111111111112 * (angle_m * (((a_m + b) * (b - a_m)) * math.pi)))

Julia

a_m = abs(a)
angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a_m, b, angle_m)
	return Float64(angle_s * Float64(0.011111111111111112 * Float64(angle_m * Float64(Float64(Float64(a_m + b) * Float64(b - a_m)) * pi))))
end

MATLAB

a_m = abs(a);
angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp = code(angle_s, a_m, b, angle_m)
	tmp = angle_s * (0.011111111111111112 * (angle_m * (((a_m + b) * (b - a_m)) * pi)));
end

Wolfram

a_m = N[Abs[a], $MachinePrecision]
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$95$m_, b_, angle$95$m_] := N[(angle$95$s * N[(0.011111111111111112 * N[(angle$95$m * N[(N[(N[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision] * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 50.2%

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

   1. unpow2 50.2%

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

   2. unpow2 50.2%

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

   3. difference-of-squares 54.9%

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

4. Applied egg-rr 54.9%

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

5. Taylor expanded in angle around 0 57.1%

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

6. Taylor expanded in angle around 0 51.6%

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

7. Final simplification 51.6%

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

Reproduce

herbie shell --seed 2024018 
(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)))))