ab-angle->ABCF B

Percentage Accurate: 53.9% → 60.6%

Time: 16.0s

Alternatives: 12

Speedup: 32.2×

• 33.3% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Initial Program: 53.9% 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: 60.6% accurate, 0.7× speedup

Math

\[\begin{array}{l} angle\_m = \left|angle\right| \\ angle\_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \pi \cdot \left(angle\_m \cdot 0.005555555555555556\right)\\ t_1 := \sin t\_0\\ t_2 := \cos t\_0\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;\frac{angle\_m}{180} \leq 10^{+63}:\\ \;\;\;\;\mathsf{fma}\left(b, 2 \cdot \left(b \cdot \left(t\_1 \cdot t\_2\right)\right), {a}^{2} \cdot \left(t\_2 \cdot \left(t\_1 \cdot -2\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \left(2 \cdot \left(\sin \left(\frac{angle\_m}{180} \cdot {\left(\sqrt{\pi}\right)}^{2}\right) \cdot \cos \left(\frac{angle\_m}{180} \cdot \pi\right)\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle\_m = (fabs.f64 angle)
angle\_s = (copysign.f64 #s(literal 1 binary64) angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* PI (* angle_m 0.005555555555555556)))
        (t_1 (sin t_0))
        (t_2 (cos t_0)))
   (*
    angle_s
    (if (<= (/ angle_m 180.0) 1e+63)
      (fma b (* 2.0 (* b (* t_1 t_2))) (* (pow a 2.0) (* t_2 (* t_1 -2.0))))
      (*
       (* (- b a) (+ b a))
       (*
        2.0
        (*
         (sin (* (/ angle_m 180.0) (pow (sqrt PI) 2.0)))
         (cos (* (/ angle_m 180.0) PI)))))))))

C

angle\_m = fabs(angle);
angle\_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = ((double) M_PI) * (angle_m * 0.005555555555555556);
	double t_1 = sin(t_0);
	double t_2 = cos(t_0);
	double tmp;
	if ((angle_m / 180.0) <= 1e+63) {
		tmp = fma(b, (2.0 * (b * (t_1 * t_2))), (pow(a, 2.0) * (t_2 * (t_1 * -2.0))));
	} else {
		tmp = ((b - a) * (b + a)) * (2.0 * (sin(((angle_m / 180.0) * pow(sqrt(((double) M_PI)), 2.0))) * cos(((angle_m / 180.0) * ((double) M_PI)))));
	}
	return angle_s * tmp;
}

Julia

angle\_m = abs(angle)
angle\_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(pi * Float64(angle_m * 0.005555555555555556))
	t_1 = sin(t_0)
	t_2 = cos(t_0)
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 1e+63)
		tmp = fma(b, Float64(2.0 * Float64(b * Float64(t_1 * t_2))), Float64((a ^ 2.0) * Float64(t_2 * Float64(t_1 * -2.0))));
	else
		tmp = Float64(Float64(Float64(b - a) * Float64(b + a)) * Float64(2.0 * Float64(sin(Float64(Float64(angle_m / 180.0) * (sqrt(pi) ^ 2.0))) * cos(Float64(Float64(angle_m / 180.0) * pi)))));
	end
	return Float64(angle_s * tmp)
end

Wolfram

angle\_m = N[Abs[angle], $MachinePrecision]
angle\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[Sin[t$95$0], $MachinePrecision]}, Block[{t$95$2 = N[Cos[t$95$0], $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 1e+63], N[(b * N[(2.0 * N[(b * N[(t$95$1 * t$95$2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(N[Power[a, 2.0], $MachinePrecision] * N[(t$95$2 * N[(t$95$1 * -2.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision] * N[(2.0 * N[(N[Sin[N[(N[(angle$95$m / 180.0), $MachinePrecision] * N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[Cos[N[(N[(angle$95$m / 180.0), $MachinePrecision] * Pi), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 angle #s(literal 180 binary64)) < 1.00000000000000006e63

   1. Initial program 58.8%

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

      1. associate-*l* 58.8%

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

      2. *-commutative 58.8%

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

      3. associate-*l* 58.8%

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

   3. Simplified 58.8%

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

      1. unpow2 58.8%

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

      2. unpow2 58.8%

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

      3. difference-of-squares 64.2%

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

   6. Applied egg-rr 64.2%

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

      1. add-sqr-sqrt 65.4%

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

      2. pow2 65.4%

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

   8. Applied egg-rr 65.4%

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

   9. Taylor expanded in b around 0 63.3%

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

       1. +-commutative 63.3%

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

       2. fma-define 67.1%

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

   11. Simplified 71.5%

       \[\leadsto \color{blue}{\mathsf{fma}\left(b, 2 \cdot \left(\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \left(b + 0\right)\right), {a}^{2} \cdot \left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot -2\right)\right)\right)} \]

   if 1.00000000000000006e63 < (/.f64 angle #s(literal 180 binary64))

   1. Initial program 25.8%

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

      1. associate-*l* 25.8%

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

      2. *-commutative 25.8%

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

      3. associate-*l* 25.8%

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

   3. Simplified 25.8%

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

      1. unpow2 25.8%

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

      2. unpow2 25.8%

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

      3. difference-of-squares 29.8%

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

   6. Applied egg-rr 29.8%

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

      1. add-sqr-sqrt 35.7%

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

      2. pow2 35.7%

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

   8. Applied egg-rr 56.2%

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

4. Final simplification 68.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 10^{+63}:\\ \;\;\;\;\mathsf{fma}\left(b, 2 \cdot \left(b \cdot \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right), {a}^{2} \cdot \left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot -2\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \left(2 \cdot \left(\sin \left(\frac{angle}{180} \cdot {\left(\sqrt{\pi}\right)}^{2}\right) \cdot \cos \left(\frac{angle}{180} \cdot \pi\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 61.6% accurate, 0.8× speedup

Math

\[\begin{array}{l} angle\_m = \left|angle\right| \\ angle\_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \pi \cdot \left(angle\_m \cdot 0.005555555555555556\right)\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{+276}:\\ \;\;\;\;b \cdot \left(b \cdot \sin \left(2 \cdot t\_0\right)\right) + \left(\sin t\_0 \cdot -2\right) \cdot \left(\cos t\_0 \cdot {a}^{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \left(2 \cdot \left(\cos \left(\frac{angle\_m}{180} \cdot \pi\right) \cdot \sin \left(\frac{\sqrt{\pi}}{180} \cdot \frac{\sqrt{\pi}}{\frac{1}{angle\_m}}\right)\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle\_m = (fabs.f64 angle)
angle\_s = (copysign.f64 #s(literal 1 binary64) angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* PI (* angle_m 0.005555555555555556))))
   (*
    angle_s
    (if (<= (pow a 2.0) 5e+276)
      (+
       (* b (* b (sin (* 2.0 t_0))))
       (* (* (sin t_0) -2.0) (* (cos t_0) (pow a 2.0))))
      (*
       (* (- b a) (+ b a))
       (*
        2.0
        (*
         (cos (* (/ angle_m 180.0) PI))
         (sin (* (/ (sqrt PI) 180.0) (/ (sqrt PI) (/ 1.0 angle_m)))))))))))

C

angle\_m = fabs(angle);
angle\_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = ((double) M_PI) * (angle_m * 0.005555555555555556);
	double tmp;
	if (pow(a, 2.0) <= 5e+276) {
		tmp = (b * (b * sin((2.0 * t_0)))) + ((sin(t_0) * -2.0) * (cos(t_0) * pow(a, 2.0)));
	} else {
		tmp = ((b - a) * (b + a)) * (2.0 * (cos(((angle_m / 180.0) * ((double) M_PI))) * sin(((sqrt(((double) M_PI)) / 180.0) * (sqrt(((double) M_PI)) / (1.0 / angle_m))))));
	}
	return angle_s * tmp;
}

Java

angle\_m = Math.abs(angle);
angle\_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = Math.PI * (angle_m * 0.005555555555555556);
	double tmp;
	if (Math.pow(a, 2.0) <= 5e+276) {
		tmp = (b * (b * Math.sin((2.0 * t_0)))) + ((Math.sin(t_0) * -2.0) * (Math.cos(t_0) * Math.pow(a, 2.0)));
	} else {
		tmp = ((b - a) * (b + a)) * (2.0 * (Math.cos(((angle_m / 180.0) * Math.PI)) * Math.sin(((Math.sqrt(Math.PI) / 180.0) * (Math.sqrt(Math.PI) / (1.0 / angle_m))))));
	}
	return angle_s * tmp;
}

Python

angle\_m = math.fabs(angle)
angle\_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = math.pi * (angle_m * 0.005555555555555556)
	tmp = 0
	if math.pow(a, 2.0) <= 5e+276:
		tmp = (b * (b * math.sin((2.0 * t_0)))) + ((math.sin(t_0) * -2.0) * (math.cos(t_0) * math.pow(a, 2.0)))
	else:
		tmp = ((b - a) * (b + a)) * (2.0 * (math.cos(((angle_m / 180.0) * math.pi)) * math.sin(((math.sqrt(math.pi) / 180.0) * (math.sqrt(math.pi) / (1.0 / angle_m))))))
	return angle_s * tmp

Julia

angle\_m = abs(angle)
angle\_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(pi * Float64(angle_m * 0.005555555555555556))
	tmp = 0.0
	if ((a ^ 2.0) <= 5e+276)
		tmp = Float64(Float64(b * Float64(b * sin(Float64(2.0 * t_0)))) + Float64(Float64(sin(t_0) * -2.0) * Float64(cos(t_0) * (a ^ 2.0))));
	else
		tmp = Float64(Float64(Float64(b - a) * Float64(b + a)) * Float64(2.0 * Float64(cos(Float64(Float64(angle_m / 180.0) * pi)) * sin(Float64(Float64(sqrt(pi) / 180.0) * Float64(sqrt(pi) / Float64(1.0 / angle_m)))))));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle\_m = abs(angle);
angle\_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = pi * (angle_m * 0.005555555555555556);
	tmp = 0.0;
	if ((a ^ 2.0) <= 5e+276)
		tmp = (b * (b * sin((2.0 * t_0)))) + ((sin(t_0) * -2.0) * (cos(t_0) * (a ^ 2.0)));
	else
		tmp = ((b - a) * (b + a)) * (2.0 * (cos(((angle_m / 180.0) * pi)) * sin(((sqrt(pi) / 180.0) * (sqrt(pi) / (1.0 / angle_m))))));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle\_m = N[Abs[angle], $MachinePrecision]
angle\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 5e+276], N[(N[(b * N[(b * N[Sin[N[(2.0 * t$95$0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(N[(N[Sin[t$95$0], $MachinePrecision] * -2.0), $MachinePrecision] * N[(N[Cos[t$95$0], $MachinePrecision] * N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision] * N[(2.0 * N[(N[Cos[N[(N[(angle$95$m / 180.0), $MachinePrecision] * Pi), $MachinePrecision]], $MachinePrecision] * N[Sin[N[(N[(N[Sqrt[Pi], $MachinePrecision] / 180.0), $MachinePrecision] * N[(N[Sqrt[Pi], $MachinePrecision] / N[(1.0 / angle$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a #s(literal 2 binary64)) < 5.00000000000000001e276

   1. Initial program 59.4%

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

      1. associate-*l* 59.4%

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

      2. *-commutative 59.4%

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

      3. associate-*l* 59.4%

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

   3. Simplified 59.4%

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

      1. unpow2 59.4%

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

      2. unpow2 59.4%

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

      3. difference-of-squares 59.4%

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

   6. Applied egg-rr 59.4%

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

      1. add-sqr-sqrt 59.1%

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

      2. pow2 59.1%

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

   8. Applied egg-rr 59.1%

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

   9. Taylor expanded in b around 0 65.3%

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

       1. +-commutative 65.3%

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

       2. fma-define 65.3%

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

   11. Simplified 67.5%

       \[\leadsto \color{blue}{\mathsf{fma}\left(b, 2 \cdot \left(\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \left(b + 0\right)\right), {a}^{2} \cdot \left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot -2\right)\right)\right)} \]
   12. Step-by-step derivation

       1. fma-undefine 67.5%

          \[\leadsto \color{blue}{b \cdot \left(2 \cdot \left(\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \left(b + 0\right)\right)\right) + {a}^{2} \cdot \left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot -2\right)\right)} \]

   13. Applied egg-rr 67.5%

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

   if 5.00000000000000001e276 < (pow.f64 a #s(literal 2 binary64))

   1. Initial program 34.3%

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

      1. associate-*l* 34.3%

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

      2. *-commutative 34.3%

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

      3. associate-*l* 34.3%

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

   3. Simplified 34.3%

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

      1. unpow2 34.3%

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

      2. unpow2 34.3%

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

      3. difference-of-squares 52.5%

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

   6. Applied egg-rr 52.5%

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

      1. add-sqr-sqrt 60.9%

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

      2. pow2 60.9%

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

   8. Applied egg-rr 67.8%

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

      1. unpow2 67.8%

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

      2. add-sqr-sqrt 52.5%

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

      3. clear-num 55.3%

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

      4. un-div-inv 58.1%

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

   10. Applied egg-rr 58.1%

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

       1. add-sqr-sqrt 66.4%

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

       2. div-inv 66.4%

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

       3. times-frac 70.6%

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

   12. Applied egg-rr 70.6%

       \[\leadsto \left(\left(b + a\right) \cdot \left(b - a\right)\right) \cdot \left(2 \cdot \left(\sin \color{blue}{\left(\frac{\sqrt{\pi}}{180} \cdot \frac{\sqrt{\pi}}{\frac{1}{angle}}\right)} \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 68.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{+276}:\\ \;\;\;\;b \cdot \left(b \cdot \sin \left(2 \cdot \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right) + \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot -2\right) \cdot \left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot {a}^{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \left(2 \cdot \left(\cos \left(\frac{angle}{180} \cdot \pi\right) \cdot \sin \left(\frac{\sqrt{\pi}}{180} \cdot \frac{\sqrt{\pi}}{\frac{1}{angle}}\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 57.2% accurate, 1.0× speedup

Math

\[\begin{array}{l} angle\_m = \left|angle\right| \\ angle\_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \frac{angle\_m}{180} \cdot \pi\\ t_1 := \left(b - a\right) \cdot \left(b + a\right)\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;\frac{angle\_m}{180} \leq 10^{+63}:\\ \;\;\;\;t\_1 \cdot \left(2 \cdot \left(\sin t\_0 \cdot \cos \left({\left({\left(angle\_m \cdot \left(\pi \cdot 0.005555555555555556\right)\right)}^{3}\right)}^{0.3333333333333333}\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;t\_1 \cdot \left(2 \cdot \left(\sin \left(\frac{angle\_m}{180} \cdot {\left(\sqrt{\pi}\right)}^{2}\right) \cdot \cos t\_0\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle\_m = (fabs.f64 angle)
angle\_s = (copysign.f64 #s(literal 1 binary64) angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* (/ angle_m 180.0) PI)) (t_1 (* (- b a) (+ b a))))
   (*
    angle_s
    (if (<= (/ angle_m 180.0) 1e+63)
      (*
       t_1
       (*
        2.0
        (*
         (sin t_0)
         (cos
          (pow
           (pow (* angle_m (* PI 0.005555555555555556)) 3.0)
           0.3333333333333333)))))
      (*
       t_1
       (*
        2.0
        (* (sin (* (/ angle_m 180.0) (pow (sqrt PI) 2.0))) (cos t_0))))))))

C

angle\_m = fabs(angle);
angle\_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (angle_m / 180.0) * ((double) M_PI);
	double t_1 = (b - a) * (b + a);
	double tmp;
	if ((angle_m / 180.0) <= 1e+63) {
		tmp = t_1 * (2.0 * (sin(t_0) * cos(pow(pow((angle_m * (((double) M_PI) * 0.005555555555555556)), 3.0), 0.3333333333333333))));
	} else {
		tmp = t_1 * (2.0 * (sin(((angle_m / 180.0) * pow(sqrt(((double) M_PI)), 2.0))) * cos(t_0)));
	}
	return angle_s * tmp;
}

Java

angle\_m = Math.abs(angle);
angle\_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (angle_m / 180.0) * Math.PI;
	double t_1 = (b - a) * (b + a);
	double tmp;
	if ((angle_m / 180.0) <= 1e+63) {
		tmp = t_1 * (2.0 * (Math.sin(t_0) * Math.cos(Math.pow(Math.pow((angle_m * (Math.PI * 0.005555555555555556)), 3.0), 0.3333333333333333))));
	} else {
		tmp = t_1 * (2.0 * (Math.sin(((angle_m / 180.0) * Math.pow(Math.sqrt(Math.PI), 2.0))) * Math.cos(t_0)));
	}
	return angle_s * tmp;
}

Python

angle\_m = math.fabs(angle)
angle\_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = (angle_m / 180.0) * math.pi
	t_1 = (b - a) * (b + a)
	tmp = 0
	if (angle_m / 180.0) <= 1e+63:
		tmp = t_1 * (2.0 * (math.sin(t_0) * math.cos(math.pow(math.pow((angle_m * (math.pi * 0.005555555555555556)), 3.0), 0.3333333333333333))))
	else:
		tmp = t_1 * (2.0 * (math.sin(((angle_m / 180.0) * math.pow(math.sqrt(math.pi), 2.0))) * math.cos(t_0)))
	return angle_s * tmp

Julia

angle\_m = abs(angle)
angle\_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(Float64(angle_m / 180.0) * pi)
	t_1 = Float64(Float64(b - a) * Float64(b + a))
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 1e+63)
		tmp = Float64(t_1 * Float64(2.0 * Float64(sin(t_0) * cos(((Float64(angle_m * Float64(pi * 0.005555555555555556)) ^ 3.0) ^ 0.3333333333333333)))));
	else
		tmp = Float64(t_1 * Float64(2.0 * Float64(sin(Float64(Float64(angle_m / 180.0) * (sqrt(pi) ^ 2.0))) * cos(t_0))));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle\_m = abs(angle);
angle\_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = (angle_m / 180.0) * pi;
	t_1 = (b - a) * (b + a);
	tmp = 0.0;
	if ((angle_m / 180.0) <= 1e+63)
		tmp = t_1 * (2.0 * (sin(t_0) * cos((((angle_m * (pi * 0.005555555555555556)) ^ 3.0) ^ 0.3333333333333333))));
	else
		tmp = t_1 * (2.0 * (sin(((angle_m / 180.0) * (sqrt(pi) ^ 2.0))) * cos(t_0)));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle\_m = N[Abs[angle], $MachinePrecision]
angle\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(N[(angle$95$m / 180.0), $MachinePrecision] * Pi), $MachinePrecision]}, Block[{t$95$1 = N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 1e+63], N[(t$95$1 * N[(2.0 * N[(N[Sin[t$95$0], $MachinePrecision] * N[Cos[N[Power[N[Power[N[(angle$95$m * N[(Pi * 0.005555555555555556), $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision], 0.3333333333333333], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(t$95$1 * N[(2.0 * N[(N[Sin[N[(N[(angle$95$m / 180.0), $MachinePrecision] * N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 angle #s(literal 180 binary64)) < 1.00000000000000006e63

   1. Initial program 58.8%

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

      1. associate-*l* 58.8%

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

      2. *-commutative 58.8%

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

      3. associate-*l* 58.8%

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

   3. Simplified 58.8%

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

      1. unpow2 58.8%

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

      2. unpow2 58.8%

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

      3. difference-of-squares 64.2%

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

   6. Applied egg-rr 64.2%

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

      1. add-sqr-sqrt 65.4%

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

      2. pow2 65.4%

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

   8. Applied egg-rr 65.4%

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

      1. unpow2 65.4%

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

      2. add-sqr-sqrt 64.2%

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

      3. add-cbrt-cube 58.3%

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

      4. pow1/3 39.0%

         \[\leadsto \left(\left(b + a\right) \cdot \left(b - a\right)\right) \cdot \left(2 \cdot \left(\sin \left(\pi \cdot \frac{angle}{180}\right) \cdot \cos \color{blue}{\left({\left(\left(\left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \left(\pi \cdot \frac{angle}{180}\right)\right)}^{0.3333333333333333}\right)}\right)\right) \]

      5. pow3 39.0%

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

      6. *-commutative 39.0%

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

      7. div-inv 39.0%

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

      8. metadata-eval 39.0%

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

      9. associate-*l* 39.5%

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

   10. Applied egg-rr 39.5%

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

   if 1.00000000000000006e63 < (/.f64 angle #s(literal 180 binary64))

   1. Initial program 25.8%

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

      1. associate-*l* 25.8%

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

      2. *-commutative 25.8%

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

      3. associate-*l* 25.8%

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

   3. Simplified 25.8%

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

      1. unpow2 25.8%

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

      2. unpow2 25.8%

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

      3. difference-of-squares 29.8%

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

   6. Applied egg-rr 29.8%

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

      1. add-sqr-sqrt 35.7%

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

      2. pow2 35.7%

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

   8. Applied egg-rr 56.2%

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

4. Final simplification 42.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 10^{+63}:\\ \;\;\;\;\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \left(2 \cdot \left(\sin \left(\frac{angle}{180} \cdot \pi\right) \cdot \cos \left({\left({\left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)}^{3}\right)}^{0.3333333333333333}\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \left(2 \cdot \left(\sin \left(\frac{angle}{180} \cdot {\left(\sqrt{\pi}\right)}^{2}\right) \cdot \cos \left(\frac{angle}{180} \cdot \pi\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 57.3% accurate, 1.0× speedup

Math

\[\begin{array}{l} angle\_m = \left|angle\right| \\ angle\_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \left(b - a\right) \cdot \left(b + a\right)\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 10^{+303}:\\ \;\;\;\;t\_0 \cdot \left(2 \cdot \left(\cos \left(\frac{angle\_m}{180} \cdot \pi\right) \cdot \sin \left({\left(\frac{\frac{180}{angle\_m}}{\pi}\right)}^{-1}\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;t\_0 \cdot \left(angle\_m \cdot \left(-2.2862368541380886 \cdot 10^{-7} \cdot \left({angle\_m}^{2} \cdot {\pi}^{3}\right) + \pi \cdot 0.011111111111111112\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle\_m = (fabs.f64 angle)
angle\_s = (copysign.f64 #s(literal 1 binary64) angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* (- b a) (+ b a))))
   (*
    angle_s
    (if (<= (pow a 2.0) 1e+303)
      (*
       t_0
       (*
        2.0
        (*
         (cos (* (/ angle_m 180.0) PI))
         (sin (pow (/ (/ 180.0 angle_m) PI) -1.0)))))
      (*
       t_0
       (*
        angle_m
        (+
         (* -2.2862368541380886e-7 (* (pow angle_m 2.0) (pow PI 3.0)))
         (* PI 0.011111111111111112))))))))

C

angle\_m = fabs(angle);
angle\_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (b - a) * (b + a);
	double tmp;
	if (pow(a, 2.0) <= 1e+303) {
		tmp = t_0 * (2.0 * (cos(((angle_m / 180.0) * ((double) M_PI))) * sin(pow(((180.0 / angle_m) / ((double) M_PI)), -1.0))));
	} else {
		tmp = t_0 * (angle_m * ((-2.2862368541380886e-7 * (pow(angle_m, 2.0) * pow(((double) M_PI), 3.0))) + (((double) M_PI) * 0.011111111111111112)));
	}
	return angle_s * tmp;
}

Java

angle\_m = Math.abs(angle);
angle\_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (b - a) * (b + a);
	double tmp;
	if (Math.pow(a, 2.0) <= 1e+303) {
		tmp = t_0 * (2.0 * (Math.cos(((angle_m / 180.0) * Math.PI)) * Math.sin(Math.pow(((180.0 / angle_m) / Math.PI), -1.0))));
	} else {
		tmp = t_0 * (angle_m * ((-2.2862368541380886e-7 * (Math.pow(angle_m, 2.0) * Math.pow(Math.PI, 3.0))) + (Math.PI * 0.011111111111111112)));
	}
	return angle_s * tmp;
}

Python

angle\_m = math.fabs(angle)
angle\_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = (b - a) * (b + a)
	tmp = 0
	if math.pow(a, 2.0) <= 1e+303:
		tmp = t_0 * (2.0 * (math.cos(((angle_m / 180.0) * math.pi)) * math.sin(math.pow(((180.0 / angle_m) / math.pi), -1.0))))
	else:
		tmp = t_0 * (angle_m * ((-2.2862368541380886e-7 * (math.pow(angle_m, 2.0) * math.pow(math.pi, 3.0))) + (math.pi * 0.011111111111111112)))
	return angle_s * tmp

Julia

angle\_m = abs(angle)
angle\_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(Float64(b - a) * Float64(b + a))
	tmp = 0.0
	if ((a ^ 2.0) <= 1e+303)
		tmp = Float64(t_0 * Float64(2.0 * Float64(cos(Float64(Float64(angle_m / 180.0) * pi)) * sin((Float64(Float64(180.0 / angle_m) / pi) ^ -1.0)))));
	else
		tmp = Float64(t_0 * Float64(angle_m * Float64(Float64(-2.2862368541380886e-7 * Float64((angle_m ^ 2.0) * (pi ^ 3.0))) + Float64(pi * 0.011111111111111112))));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle\_m = abs(angle);
angle\_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = (b - a) * (b + a);
	tmp = 0.0;
	if ((a ^ 2.0) <= 1e+303)
		tmp = t_0 * (2.0 * (cos(((angle_m / 180.0) * pi)) * sin((((180.0 / angle_m) / pi) ^ -1.0))));
	else
		tmp = t_0 * (angle_m * ((-2.2862368541380886e-7 * ((angle_m ^ 2.0) * (pi ^ 3.0))) + (pi * 0.011111111111111112)));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle\_m = N[Abs[angle], $MachinePrecision]
angle\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 1e+303], N[(t$95$0 * N[(2.0 * N[(N[Cos[N[(N[(angle$95$m / 180.0), $MachinePrecision] * Pi), $MachinePrecision]], $MachinePrecision] * N[Sin[N[Power[N[(N[(180.0 / angle$95$m), $MachinePrecision] / Pi), $MachinePrecision], -1.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(t$95$0 * N[(angle$95$m * N[(N[(-2.2862368541380886e-7 * N[(N[Power[angle$95$m, 2.0], $MachinePrecision] * N[Power[Pi, 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(Pi * 0.011111111111111112), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a #s(literal 2 binary64)) < 1e303

   1. Initial program 58.5%

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

      1. associate-*l* 58.5%

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

      2. *-commutative 58.5%

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

      3. associate-*l* 58.5%

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

   3. Simplified 58.5%

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

      1. unpow2 58.5%

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

      2. unpow2 58.5%

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

      3. difference-of-squares 58.5%

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

   6. Applied egg-rr 58.5%

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

      1. add-sqr-sqrt 58.2%

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

      2. pow2 58.2%

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

   8. Applied egg-rr 60.8%

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

      1. unpow2 60.8%

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

      2. add-sqr-sqrt 58.5%

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

      3. clear-num 58.4%

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

      4. un-div-inv 57.9%

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

   10. Applied egg-rr 57.9%

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

       1. clear-num 60.5%

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

       2. inv-pow 60.5%

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

   12. Applied egg-rr 60.5%

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

   if 1e303 < (pow.f64 a #s(literal 2 binary64))

   1. Initial program 35.1%

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

      1. associate-*l* 35.1%

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

      2. *-commutative 35.1%

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

      3. associate-*l* 35.1%

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

   3. Simplified 35.1%

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

      1. unpow2 35.1%

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

      2. unpow2 35.1%

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

      3. difference-of-squares 54.7%

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

   6. Applied egg-rr 54.7%

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

      1. add-sqr-sqrt 63.6%

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

      2. pow2 63.6%

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

   8. Applied egg-rr 69.6%

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

      1. log1p-expm1-u 69.6%

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

      2. log1p-undefine 38.8%

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

      3. sin-cos-mult 29.9%

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

      4. div-inv 29.9%

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

   10. Applied egg-rr 22.4%

       \[\leadsto \left(\left(b + a\right) \cdot \left(b - a\right)\right) \cdot \left(2 \cdot \color{blue}{\log \left(1 + \mathsf{expm1}\left(\left(\sin \left(\left(\pi \cdot angle\right) \cdot 0.011111111111111112\right) + \sin 0\right) \cdot 0.5\right)\right)}\right) \]

   11. Taylor expanded in angle around 0 69.6%

       \[\leadsto \left(\left(b + a\right) \cdot \left(b - a\right)\right) \cdot \color{blue}{\left(angle \cdot \left(-2.2862368541380886 \cdot 10^{-7} \cdot \left({angle}^{2} \cdot {\pi}^{3}\right) + 0.011111111111111112 \cdot \pi\right)\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 62.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 10^{+303}:\\ \;\;\;\;\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \left(2 \cdot \left(\cos \left(\frac{angle}{180} \cdot \pi\right) \cdot \sin \left({\left(\frac{\frac{180}{angle}}{\pi}\right)}^{-1}\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \left(angle \cdot \left(-2.2862368541380886 \cdot 10^{-7} \cdot \left({angle}^{2} \cdot {\pi}^{3}\right) + \pi \cdot 0.011111111111111112\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 57.1% accurate, 1.0× speedup

Math

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

FPCore

angle\_m = (fabs.f64 angle)
angle\_s = (copysign.f64 #s(literal 1 binary64) angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (*
  angle_s
  (*
   (* (- b a) (+ b a))
   (*
    2.0
    (*
     (sin (* (/ angle_m 180.0) (pow (sqrt PI) 2.0)))
     (cos (* (/ angle_m 180.0) PI)))))))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 52.3%

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

   1. associate-*l* 52.3%

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

   2. *-commutative 52.3%

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

   3. associate-*l* 52.3%

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

3. Simplified 52.3%

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

   1. unpow2 52.3%

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

   2. unpow2 52.3%

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

   3. difference-of-squares 57.5%

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

6. Applied egg-rr 57.5%

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

   1. add-sqr-sqrt 59.6%

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

   2. pow2 59.6%

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

8. Applied egg-rr 63.1%

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

9. Final simplification 63.1%

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

Alternative 6: 57.4% accurate, 1.3× speedup

Math

\[\begin{array}{l} angle\_m = \left|angle\right| \\ angle\_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \left(b - a\right) \cdot \left(b + a\right)\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 10^{+281}:\\ \;\;\;\;t\_0 \cdot \left(2 \cdot \left(\sin \left(\frac{angle\_m}{180} \cdot \pi\right) \cdot \cos \left(\frac{angle\_m \cdot \pi}{180}\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;t\_0 \cdot \left(angle\_m \cdot \left(-2.2862368541380886 \cdot 10^{-7} \cdot \left({angle\_m}^{2} \cdot {\pi}^{3}\right) + \pi \cdot 0.011111111111111112\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle\_m = (fabs.f64 angle)
angle\_s = (copysign.f64 #s(literal 1 binary64) angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* (- b a) (+ b a))))
   (*
    angle_s
    (if (<= (pow a 2.0) 1e+281)
      (*
       t_0
       (*
        2.0
        (* (sin (* (/ angle_m 180.0) PI)) (cos (/ (* angle_m PI) 180.0)))))
      (*
       t_0
       (*
        angle_m
        (+
         (* -2.2862368541380886e-7 (* (pow angle_m 2.0) (pow PI 3.0)))
         (* PI 0.011111111111111112))))))))

C

angle\_m = fabs(angle);
angle\_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (b - a) * (b + a);
	double tmp;
	if (pow(a, 2.0) <= 1e+281) {
		tmp = t_0 * (2.0 * (sin(((angle_m / 180.0) * ((double) M_PI))) * cos(((angle_m * ((double) M_PI)) / 180.0))));
	} else {
		tmp = t_0 * (angle_m * ((-2.2862368541380886e-7 * (pow(angle_m, 2.0) * pow(((double) M_PI), 3.0))) + (((double) M_PI) * 0.011111111111111112)));
	}
	return angle_s * tmp;
}

Java

angle\_m = Math.abs(angle);
angle\_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (b - a) * (b + a);
	double tmp;
	if (Math.pow(a, 2.0) <= 1e+281) {
		tmp = t_0 * (2.0 * (Math.sin(((angle_m / 180.0) * Math.PI)) * Math.cos(((angle_m * Math.PI) / 180.0))));
	} else {
		tmp = t_0 * (angle_m * ((-2.2862368541380886e-7 * (Math.pow(angle_m, 2.0) * Math.pow(Math.PI, 3.0))) + (Math.PI * 0.011111111111111112)));
	}
	return angle_s * tmp;
}

Python

angle\_m = math.fabs(angle)
angle\_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = (b - a) * (b + a)
	tmp = 0
	if math.pow(a, 2.0) <= 1e+281:
		tmp = t_0 * (2.0 * (math.sin(((angle_m / 180.0) * math.pi)) * math.cos(((angle_m * math.pi) / 180.0))))
	else:
		tmp = t_0 * (angle_m * ((-2.2862368541380886e-7 * (math.pow(angle_m, 2.0) * math.pow(math.pi, 3.0))) + (math.pi * 0.011111111111111112)))
	return angle_s * tmp

Julia

angle\_m = abs(angle)
angle\_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(Float64(b - a) * Float64(b + a))
	tmp = 0.0
	if ((a ^ 2.0) <= 1e+281)
		tmp = Float64(t_0 * Float64(2.0 * Float64(sin(Float64(Float64(angle_m / 180.0) * pi)) * cos(Float64(Float64(angle_m * pi) / 180.0)))));
	else
		tmp = Float64(t_0 * Float64(angle_m * Float64(Float64(-2.2862368541380886e-7 * Float64((angle_m ^ 2.0) * (pi ^ 3.0))) + Float64(pi * 0.011111111111111112))));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle\_m = abs(angle);
angle\_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = (b - a) * (b + a);
	tmp = 0.0;
	if ((a ^ 2.0) <= 1e+281)
		tmp = t_0 * (2.0 * (sin(((angle_m / 180.0) * pi)) * cos(((angle_m * pi) / 180.0))));
	else
		tmp = t_0 * (angle_m * ((-2.2862368541380886e-7 * ((angle_m ^ 2.0) * (pi ^ 3.0))) + (pi * 0.011111111111111112)));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle\_m = N[Abs[angle], $MachinePrecision]
angle\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 1e+281], N[(t$95$0 * N[(2.0 * N[(N[Sin[N[(N[(angle$95$m / 180.0), $MachinePrecision] * Pi), $MachinePrecision]], $MachinePrecision] * N[Cos[N[(N[(angle$95$m * Pi), $MachinePrecision] / 180.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(t$95$0 * N[(angle$95$m * N[(N[(-2.2862368541380886e-7 * N[(N[Power[angle$95$m, 2.0], $MachinePrecision] * N[Power[Pi, 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(Pi * 0.011111111111111112), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a #s(literal 2 binary64)) < 1e281

   1. Initial program 59.2%

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

      1. associate-*l* 59.2%

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

      2. *-commutative 59.2%

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

      3. associate-*l* 59.2%

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

   3. Simplified 59.2%

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

      1. unpow2 59.2%

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

      2. unpow2 59.2%

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

      3. difference-of-squares 59.2%

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

   6. Applied egg-rr 59.2%

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

      1. add-sqr-sqrt 58.8%

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

      2. pow2 58.8%

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

   8. Applied egg-rr 58.8%

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

      1. unpow2 58.8%

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

      2. add-sqr-sqrt 59.2%

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

      3. associate-*r/ 60.5%

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

   10. Applied egg-rr 60.5%

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

   if 1e281 < (pow.f64 a #s(literal 2 binary64))

   1. Initial program 34.5%

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

      1. associate-*l* 34.5%

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

      2. *-commutative 34.5%

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

      3. associate-*l* 34.5%

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

   3. Simplified 34.5%

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

      1. unpow2 34.5%

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

      2. unpow2 34.5%

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

      3. difference-of-squares 53.0%

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

   6. Applied egg-rr 53.0%

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

      1. add-sqr-sqrt 61.8%

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

      2. pow2 61.8%

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

   8. Applied egg-rr 68.5%

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

      1. log1p-expm1-u 68.5%

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

      2. log1p-undefine 38.1%

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

      3. sin-cos-mult 28.2%

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

      4. div-inv 28.2%

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

   10. Applied egg-rr 22.6%

       \[\leadsto \left(\left(b + a\right) \cdot \left(b - a\right)\right) \cdot \left(2 \cdot \color{blue}{\log \left(1 + \mathsf{expm1}\left(\left(\sin \left(\left(\pi \cdot angle\right) \cdot 0.011111111111111112\right) + \sin 0\right) \cdot 0.5\right)\right)}\right) \]

   11. Taylor expanded in angle around 0 68.5%

       \[\leadsto \left(\left(b + a\right) \cdot \left(b - a\right)\right) \cdot \color{blue}{\left(angle \cdot \left(-2.2862368541380886 \cdot 10^{-7} \cdot \left({angle}^{2} \cdot {\pi}^{3}\right) + 0.011111111111111112 \cdot \pi\right)\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 62.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 10^{+281}:\\ \;\;\;\;\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \left(2 \cdot \left(\sin \left(\frac{angle}{180} \cdot \pi\right) \cdot \cos \left(\frac{angle \cdot \pi}{180}\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \left(angle \cdot \left(-2.2862368541380886 \cdot 10^{-7} \cdot \left({angle}^{2} \cdot {\pi}^{3}\right) + \pi \cdot 0.011111111111111112\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 57.6% accurate, 1.9× speedup

Math

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

FPCore

angle\_m = (fabs.f64 angle)
angle\_s = (copysign.f64 #s(literal 1 binary64) angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (*
  angle_s
  (*
   (* (- b a) (+ b a))
   (* 2.0 (* (cos (* (/ angle_m 180.0) PI)) (sin (/ (* angle_m PI) 180.0)))))))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 52.3%

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

   1. associate-*l* 52.3%

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

   2. *-commutative 52.3%

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

   3. associate-*l* 52.3%

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

3. Simplified 52.3%

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

   1. unpow2 52.3%

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

   2. unpow2 52.3%

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

   3. difference-of-squares 57.5%

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

6. Applied egg-rr 57.5%

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

   1. add-sqr-sqrt 59.6%

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

   2. pow2 59.6%

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

8. Applied egg-rr 63.1%

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

   1. unpow2 59.6%

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

   2. add-sqr-sqrt 57.5%

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

   3. associate-*r/ 58.8%

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

10. Applied egg-rr 60.2%

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

11. Final simplification 60.2%

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

Alternative 8: 57.5% accurate, 1.9× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 52.3%

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

   1. associate-*l* 52.3%

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

   2. *-commutative 52.3%

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

   3. associate-*l* 52.3%

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

3. Simplified 52.3%

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

   1. unpow2 52.3%

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

   2. unpow2 52.3%

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

   3. difference-of-squares 57.5%

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

6. Applied egg-rr 57.5%

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

7. Taylor expanded in angle around inf 59.4%

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

8. Final simplification 59.4%

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

Alternative 9: 57.6% accurate, 3.7× speedup

Math

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

FPCore

angle\_m = (fabs.f64 angle)
angle\_s = (copysign.f64 #s(literal 1 binary64) angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (*
  angle_s
  (* (* (- b a) (+ b a)) (sin (* PI (* angle_m 0.011111111111111112))))))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

angle\_m = N[Abs[angle], $MachinePrecision]
angle\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := N[(angle$95$s * N[(N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(Pi * N[(angle$95$m * 0.011111111111111112), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 52.3%

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

   1. associate-*l* 52.3%

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

   2. *-commutative 52.3%

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

   3. associate-*l* 52.3%

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

3. Simplified 52.3%

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

   1. unpow2 52.3%

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

   2. unpow2 52.3%

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

   3. difference-of-squares 57.5%

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

6. Applied egg-rr 57.5%

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

   1. add-sqr-sqrt 59.6%

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

   2. pow2 59.6%

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

8. Applied egg-rr 63.1%

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

10. Applied egg-rr 56.7%

    \[\leadsto \left(\left(b + a\right) \cdot \left(b - a\right)\right) \cdot \color{blue}{\left(\left(\sin \left(\left(\pi \cdot angle\right) \cdot 0.011111111111111112\right) + \sin 0\right) \cdot 0.5 + \left(\sin \left(\left(\pi \cdot angle\right) \cdot 0.011111111111111112\right) + \sin 0\right) \cdot 0.5\right)} \]
11. Step-by-step derivation

    1. distribute-lft-out 56.7%

       \[\leadsto \left(\left(b + a\right) \cdot \left(b - a\right)\right) \cdot \color{blue}{\left(\left(\sin \left(\left(\pi \cdot angle\right) \cdot 0.011111111111111112\right) + \sin 0\right) \cdot \left(0.5 + 0.5\right)\right)} \]

    2. sin-0 56.7%

       \[\leadsto \left(\left(b + a\right) \cdot \left(b - a\right)\right) \cdot \left(\left(\sin \left(\left(\pi \cdot angle\right) \cdot 0.011111111111111112\right) + \color{blue}{0}\right) \cdot \left(0.5 + 0.5\right)\right) \]

    3. +-rgt-identity 56.7%

       \[\leadsto \left(\left(b + a\right) \cdot \left(b - a\right)\right) \cdot \left(\color{blue}{\sin \left(\left(\pi \cdot angle\right) \cdot 0.011111111111111112\right)} \cdot \left(0.5 + 0.5\right)\right) \]

    4. metadata-eval 56.7%

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

    5. *-rgt-identity 56.7%

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

    6. associate-*l* 58.6%

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

12. Simplified 58.6%

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

13. Final simplification 58.6%

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

Alternative 10: 57.6% accurate, 3.7× speedup

Math

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

FPCore

angle\_m = (fabs.f64 angle)
angle\_s = (copysign.f64 #s(literal 1 binary64) angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (*
  angle_s
  (* (* (- b a) (+ b a)) (sin (* angle_m (* PI 0.011111111111111112))))))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

angle\_m = N[Abs[angle], $MachinePrecision]
angle\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := N[(angle$95$s * N[(N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(angle$95$m * N[(Pi * 0.011111111111111112), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 52.3%

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

   1. associate-*l* 52.3%

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

   2. *-commutative 52.3%

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

   3. associate-*l* 52.3%

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

3. Simplified 52.3%

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

   1. unpow2 52.3%

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

   2. unpow2 52.3%

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

   3. difference-of-squares 57.5%

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

6. Applied egg-rr 57.5%

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

   1. add-sqr-sqrt 59.6%

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

   2. pow2 59.6%

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

8. Applied egg-rr 63.1%

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

   1. log1p-expm1-u 63.1%

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

   2. log1p-undefine 31.8%

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

   3. sin-cos-mult 27.8%

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

   4. div-inv 27.8%

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

10. Applied egg-rr 25.2%

    \[\leadsto \left(\left(b + a\right) \cdot \left(b - a\right)\right) \cdot \left(2 \cdot \color{blue}{\log \left(1 + \mathsf{expm1}\left(\left(\sin \left(\left(\pi \cdot angle\right) \cdot 0.011111111111111112\right) + \sin 0\right) \cdot 0.5\right)\right)}\right) \]

11. Taylor expanded in angle around inf 56.7%

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

    1. associate-*r* 58.6%

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

    2. *-commutative 58.6%

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

    3. *-commutative 58.6%

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

    4. associate-*r* 56.7%

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

    5. *-commutative 56.7%

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

    6. associate-*r* 57.8%

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

13. Simplified 57.8%

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

14. Final simplification 57.8%

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

Alternative 11: 54.0% accurate, 32.2× speedup

Math

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

FPCore

angle\_m = (fabs.f64 angle)
angle\_s = (copysign.f64 #s(literal 1 binary64) angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (* angle_s (* (* (- b a) (+ b a)) (* 0.011111111111111112 (* angle_m PI)))))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

angle\_m = N[Abs[angle], $MachinePrecision]
angle\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := N[(angle$95$s * N[(N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision] * N[(0.011111111111111112 * N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 52.3%

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

   1. associate-*l* 52.3%

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

   2. *-commutative 52.3%

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

   3. associate-*l* 52.3%

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

3. Simplified 52.3%

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

   1. unpow2 52.3%

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

   2. unpow2 52.3%

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

   3. difference-of-squares 57.5%

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

6. Applied egg-rr 57.5%

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

   1. add-sqr-sqrt 59.6%

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

   2. pow2 59.6%

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

8. Applied egg-rr 63.1%

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

9. Taylor expanded in angle around 0 54.8%

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

10. Final simplification 54.8%

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

Alternative 12: 54.0% accurate, 32.2× speedup

Math

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

FPCore

angle\_m = (fabs.f64 angle)
angle\_s = (copysign.f64 #s(literal 1 binary64) angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (* angle_s (* 0.011111111111111112 (* angle_m (* PI (* (- b a) (+ b a)))))))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

angle\_m = N[Abs[angle], $MachinePrecision]
angle\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := N[(angle$95$s * N[(0.011111111111111112 * N[(angle$95$m * N[(Pi * N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 52.3%

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

   1. associate-*l* 52.3%

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

   2. *-commutative 52.3%

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

   3. associate-*l* 52.3%

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

3. Simplified 52.3%

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

   1. unpow2 52.3%

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

   2. unpow2 52.3%

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

   3. difference-of-squares 57.5%

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

6. Applied egg-rr 57.5%

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

   1. add-sqr-sqrt 59.6%

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

   2. pow2 59.6%

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

8. Applied egg-rr 59.6%

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

9. Taylor expanded in angle around 0 54.8%

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

10. Final simplification 54.8%

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

Reproduce

herbie shell --seed 2024111 
(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)))))