ab-angle->ABCF B

Percentage Accurate: 53.9% → 57.3%

Time: 43.5s

Alternatives: 14

Speedup: 32.2×

• 33.7% 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: 57.3% accurate, 0.8× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \pi \cdot \left(0.005555555555555556 \cdot angle_m\right)\\ t_1 := \left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \sin t_0\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 2 \cdot 10^{-194}:\\ \;\;\;\;2 \cdot \left(t_1 \cdot \cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(t_0\right)\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \left(t_1 \cdot \cos \left(0.005555555555555556 \cdot \left(angle_m \cdot \pi\right)\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* PI (* 0.005555555555555556 angle_m)))
        (t_1 (* (* (- b a) (+ b a)) (sin t_0))))
   (*
    angle_s
    (if (<= (pow a 2.0) 2e-194)
      (* 2.0 (* t_1 (cos (expm1 (log1p t_0)))))
      (* 2.0 (* t_1 (cos (* 0.005555555555555556 (* angle_m PI)))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = ((double) M_PI) * (0.005555555555555556 * angle_m);
	double t_1 = ((b - a) * (b + a)) * sin(t_0);
	double tmp;
	if (pow(a, 2.0) <= 2e-194) {
		tmp = 2.0 * (t_1 * cos(expm1(log1p(t_0))));
	} else {
		tmp = 2.0 * (t_1 * cos((0.005555555555555556 * (angle_m * ((double) M_PI)))));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = Math.PI * (0.005555555555555556 * angle_m);
	double t_1 = ((b - a) * (b + a)) * Math.sin(t_0);
	double tmp;
	if (Math.pow(a, 2.0) <= 2e-194) {
		tmp = 2.0 * (t_1 * Math.cos(Math.expm1(Math.log1p(t_0))));
	} else {
		tmp = 2.0 * (t_1 * Math.cos((0.005555555555555556 * (angle_m * Math.PI))));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = math.pi * (0.005555555555555556 * angle_m)
	t_1 = ((b - a) * (b + a)) * math.sin(t_0)
	tmp = 0
	if math.pow(a, 2.0) <= 2e-194:
		tmp = 2.0 * (t_1 * math.cos(math.expm1(math.log1p(t_0))))
	else:
		tmp = 2.0 * (t_1 * math.cos((0.005555555555555556 * (angle_m * math.pi))))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(pi * Float64(0.005555555555555556 * angle_m))
	t_1 = Float64(Float64(Float64(b - a) * Float64(b + a)) * sin(t_0))
	tmp = 0.0
	if ((a ^ 2.0) <= 2e-194)
		tmp = Float64(2.0 * Float64(t_1 * cos(expm1(log1p(t_0)))));
	else
		tmp = Float64(2.0 * Float64(t_1 * cos(Float64(0.005555555555555556 * Float64(angle_m * 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[(0.005555555555555556 * angle$95$m), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision] * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 2e-194], N[(2.0 * N[(t$95$1 * N[Cos[N[(Exp[N[Log[1 + t$95$0], $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 * N[(t$95$1 * N[Cos[N[(0.005555555555555556 * N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a 2) < 2.00000000000000004e-194

   1. Initial program 56.1%

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

      1. associate-*l* 56.1%

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

      2. associate-*l* 56.1%

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

   3. Simplified 56.1%

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

      1. unpow2 56.1%

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

      2. unpow2 56.1%

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

      3. difference-of-squares 56.1%

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

   6. Applied egg-rr 56.1%

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

   7. Taylor expanded in angle around inf 55.0%

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

      1. *-commutative 55.0%

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

      2. +-commutative 55.0%

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

      3. associate-*r* 56.2%

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

      4. *-commutative 56.2%

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

      5. *-commutative 56.2%

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

      6. *-commutative 56.2%

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

      7. +-commutative 56.2%

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

      8. *-commutative 56.2%

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

   9. Simplified 56.2%

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

       1. div-inv 56.5%

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

       2. metadata-eval 56.5%

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

       3. expm1-log1p-u 44.6%

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

   11. Applied egg-rr 44.6%

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

   if 2.00000000000000004e-194 < (pow.f64 a 2)

   1. Initial program 48.1%

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

      1. associate-*l* 48.1%

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

      2. associate-*l* 48.1%

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

   3. Simplified 48.1%

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

      1. unpow2 48.1%

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

      2. unpow2 48.1%

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

      3. difference-of-squares 56.6%

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

   6. Applied egg-rr 56.6%

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

   7. Taylor expanded in angle around inf 56.3%

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

      1. *-commutative 56.3%

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

      2. +-commutative 56.3%

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

      3. associate-*r* 56.0%

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

      4. *-commutative 56.0%

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

      5. *-commutative 56.0%

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

      6. *-commutative 56.0%

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

      7. +-commutative 56.0%

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

      8. *-commutative 56.0%

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

   9. Simplified 56.0%

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

   10. Taylor expanded in angle around inf 59.9%

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

4. Final simplification 54.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 2 \cdot 10^{-194}:\\ \;\;\;\;2 \cdot \left(\left(\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \sin \left(\pi \cdot \left(0.005555555555555556 \cdot angle\right)\right)\right) \cdot \cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(0.005555555555555556 \cdot angle\right)\right)\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \left(\left(\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \sin \left(\pi \cdot \left(0.005555555555555556 \cdot angle\right)\right)\right) \cdot \cos \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 57.1% accurate, 0.6× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \pi \cdot \frac{angle_m}{180}\\ t_1 := \left(b - a\right) \cdot \left(b + a\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin t_0\right) \cdot \cos t_0 \leq 2 \cdot 10^{+93}:\\ \;\;\;\;2 \cdot \left(\left(t_1 \cdot \sin \left(\pi \cdot \left(0.005555555555555556 \cdot angle_m\right)\right)\right) \cdot \cos \left(\frac{\pi}{\frac{180}{angle_m}}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \left(t_1 \cdot \sin \left(\frac{angle_m}{180} \cdot {\left(\sqrt{\pi}\right)}^{2}\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* PI (/ angle_m 180.0))) (t_1 (* (- b a) (+ b a))))
   (*
    angle_s
    (if (<=
         (* (* (* 2.0 (- (pow b 2.0) (pow a 2.0))) (sin t_0)) (cos t_0))
         2e+93)
      (*
       2.0
       (*
        (* t_1 (sin (* PI (* 0.005555555555555556 angle_m))))
        (cos (/ PI (/ 180.0 angle_m)))))
      (* 2.0 (* t_1 (sin (* (/ angle_m 180.0) (pow (sqrt PI) 2.0)))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = ((double) M_PI) * (angle_m / 180.0);
	double t_1 = (b - a) * (b + a);
	double tmp;
	if ((((2.0 * (pow(b, 2.0) - pow(a, 2.0))) * sin(t_0)) * cos(t_0)) <= 2e+93) {
		tmp = 2.0 * ((t_1 * sin((((double) M_PI) * (0.005555555555555556 * angle_m)))) * cos((((double) M_PI) / (180.0 / angle_m))));
	} else {
		tmp = 2.0 * (t_1 * sin(((angle_m / 180.0) * pow(sqrt(((double) M_PI)), 2.0))));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = Math.PI * (angle_m / 180.0);
	double t_1 = (b - a) * (b + a);
	double tmp;
	if ((((2.0 * (Math.pow(b, 2.0) - Math.pow(a, 2.0))) * Math.sin(t_0)) * Math.cos(t_0)) <= 2e+93) {
		tmp = 2.0 * ((t_1 * Math.sin((Math.PI * (0.005555555555555556 * angle_m)))) * Math.cos((Math.PI / (180.0 / angle_m))));
	} else {
		tmp = 2.0 * (t_1 * Math.sin(((angle_m / 180.0) * Math.pow(Math.sqrt(Math.PI), 2.0))));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = math.pi * (angle_m / 180.0)
	t_1 = (b - a) * (b + a)
	tmp = 0
	if (((2.0 * (math.pow(b, 2.0) - math.pow(a, 2.0))) * math.sin(t_0)) * math.cos(t_0)) <= 2e+93:
		tmp = 2.0 * ((t_1 * math.sin((math.pi * (0.005555555555555556 * angle_m)))) * math.cos((math.pi / (180.0 / angle_m))))
	else:
		tmp = 2.0 * (t_1 * math.sin(((angle_m / 180.0) * math.pow(math.sqrt(math.pi), 2.0))))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(pi * Float64(angle_m / 180.0))
	t_1 = Float64(Float64(b - a) * Float64(b + a))
	tmp = 0.0
	if (Float64(Float64(Float64(2.0 * Float64((b ^ 2.0) - (a ^ 2.0))) * sin(t_0)) * cos(t_0)) <= 2e+93)
		tmp = Float64(2.0 * Float64(Float64(t_1 * sin(Float64(pi * Float64(0.005555555555555556 * angle_m)))) * cos(Float64(pi / Float64(180.0 / angle_m)))));
	else
		tmp = Float64(2.0 * Float64(t_1 * sin(Float64(Float64(angle_m / 180.0) * (sqrt(pi) ^ 2.0)))));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = pi * (angle_m / 180.0);
	t_1 = (b - a) * (b + a);
	tmp = 0.0;
	if ((((2.0 * ((b ^ 2.0) - (a ^ 2.0))) * sin(t_0)) * cos(t_0)) <= 2e+93)
		tmp = 2.0 * ((t_1 * sin((pi * (0.005555555555555556 * angle_m)))) * cos((pi / (180.0 / angle_m))));
	else
		tmp = 2.0 * (t_1 * sin(((angle_m / 180.0) * (sqrt(pi) ^ 2.0))));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[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], 2e+93], N[(2.0 * N[(N[(t$95$1 * N[Sin[N[(Pi * N[(0.005555555555555556 * angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(Pi / N[(180.0 / angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 * N[(t$95$1 * N[Sin[N[(N[(angle$95$m / 180.0), $MachinePrecision] * N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (*.f64 (*.f64 (*.f64 2 (-.f64 (pow.f64 b 2) (pow.f64 a 2))) (sin.f64 (*.f64 (PI.f64) (/.f64 angle 180)))) (cos.f64 (*.f64 (PI.f64) (/.f64 angle 180)))) < 2.00000000000000009e93

   1. Initial program 58.1%

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

      1. associate-*l* 58.1%

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

      2. associate-*l* 58.1%

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

   3. Simplified 58.1%

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

      1. unpow2 58.1%

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

      2. unpow2 58.1%

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

      3. difference-of-squares 58.1%

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

   6. Applied egg-rr 58.1%

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

   7. Taylor expanded in angle around inf 57.0%

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

      1. *-commutative 57.0%

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

      2. +-commutative 57.0%

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

      3. associate-*r* 57.6%

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

      4. *-commutative 57.6%

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

      5. *-commutative 57.6%

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

      6. *-commutative 57.6%

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

      7. +-commutative 57.6%

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

      8. *-commutative 57.6%

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

   9. Simplified 57.6%

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

       1. clear-num 55.6%

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

       2. un-div-inv 56.3%

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

   11. Applied egg-rr 58.8%

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

   if 2.00000000000000009e93 < (*.f64 (*.f64 (*.f64 2 (-.f64 (pow.f64 b 2) (pow.f64 a 2))) (sin.f64 (*.f64 (PI.f64) (/.f64 angle 180)))) (cos.f64 (*.f64 (PI.f64) (/.f64 angle 180))))

   1. Initial program 39.3%

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

      1. associate-*l* 39.3%

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

      2. associate-*l* 39.3%

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

   3. Simplified 39.3%

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

      1. unpow2 39.3%

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

      2. unpow2 39.3%

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

      3. difference-of-squares 53.8%

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

   6. Applied egg-rr 53.8%

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

   7. Taylor expanded in angle around 0 52.9%

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

      1. add-sqr-sqrt 50.4%

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

      2. pow2 50.4%

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

   9. Applied egg-rr 50.4%

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

4. Final simplification 55.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \leq 2 \cdot 10^{+93}:\\ \;\;\;\;2 \cdot \left(\left(\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \sin \left(\pi \cdot \left(0.005555555555555556 \cdot angle\right)\right)\right) \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \left(\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \sin \left(\frac{angle}{180} \cdot {\left(\sqrt{\pi}\right)}^{2}\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 57.5% accurate, 1.0× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ angle_s \cdot \left(2 \cdot \left(\left(\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(angle_m \cdot \pi\right)\right)\right) \cdot \cos \left({\left(\sqrt[3]{\pi \cdot \left(0.005555555555555556 \cdot angle_m\right)}\right)}^{3}\right)\right)\right) \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (*
  angle_s
  (*
   2.0
   (*
    (* (* (- b a) (+ b a)) (sin (* 0.005555555555555556 (* angle_m PI))))
    (cos (pow (cbrt (* PI (* 0.005555555555555556 angle_m))) 3.0))))))

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 * (2.0 * ((((b - a) * (b + a)) * sin((0.005555555555555556 * (angle_m * ((double) M_PI))))) * cos(pow(cbrt((((double) M_PI) * (0.005555555555555556 * angle_m))), 3.0))));
}

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 * (2.0 * ((((b - a) * (b + a)) * Math.sin((0.005555555555555556 * (angle_m * Math.PI)))) * Math.cos(Math.pow(Math.cbrt((Math.PI * (0.005555555555555556 * angle_m))), 3.0))));
}

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	return Float64(angle_s * Float64(2.0 * Float64(Float64(Float64(Float64(b - a) * Float64(b + a)) * sin(Float64(0.005555555555555556 * Float64(angle_m * pi)))) * cos((cbrt(Float64(pi * Float64(0.005555555555555556 * angle_m))) ^ 3.0)))))
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[(2.0 * N[(N[(N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(0.005555555555555556 * N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[Power[N[Power[N[(Pi * N[(0.005555555555555556 * angle$95$m), $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision], 3.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 50.8%

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

   1. associate-*l* 50.8%

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

   2. associate-*l* 50.8%

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

3. Simplified 50.8%

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

   1. unpow2 50.8%

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

   2. unpow2 50.8%

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

   3. difference-of-squares 56.4%

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

6. Applied egg-rr 56.4%

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

   1. add-cbrt-cube 57.4%

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

   2. pow3 57.4%

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

8. Applied egg-rr 57.4%

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

   1. add-cube-cbrt 53.4%

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

   2. pow3 54.0%

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

   3. div-inv 55.1%

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

   4. metadata-eval 55.1%

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

   5. *-commutative 55.1%

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

   6. *-commutative 55.1%

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

10. Applied egg-rr 55.1%

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

11. Taylor expanded in angle around inf 58.7%

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

12. Final simplification 58.7%

    \[\leadsto 2 \cdot \left(\left(\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right) \cdot \cos \left({\left(\sqrt[3]{\pi \cdot \left(0.005555555555555556 \cdot angle\right)}\right)}^{3}\right)\right) \]
13. Add Preprocessing

Alternative 4: 57.3% accurate, 1.3× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \left(b - a\right) \cdot \left(b + a\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{+262}:\\ \;\;\;\;2 \cdot \left(\left(t_0 \cdot \sin \left(0.005555555555555556 \cdot \left(angle_m \cdot \pi\right)\right)\right) \cdot \cos \left(\pi \cdot \frac{angle_m}{180}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(angle_m \cdot 0.011111111111111112\right) \cdot \left(t_0 \cdot \pi\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* (- b a) (+ b a))))
   (*
    angle_s
    (if (<= (pow a 2.0) 5e+262)
      (*
       2.0
       (*
        (* t_0 (sin (* 0.005555555555555556 (* angle_m PI))))
        (cos (* PI (/ angle_m 180.0)))))
      (* (* angle_m 0.011111111111111112) (* t_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 = (b - a) * (b + a);
	double tmp;
	if (pow(a, 2.0) <= 5e+262) {
		tmp = 2.0 * ((t_0 * sin((0.005555555555555556 * (angle_m * ((double) M_PI))))) * cos((((double) M_PI) * (angle_m / 180.0))));
	} else {
		tmp = (angle_m * 0.011111111111111112) * (t_0 * ((double) M_PI));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (b - a) * (b + a);
	double tmp;
	if (Math.pow(a, 2.0) <= 5e+262) {
		tmp = 2.0 * ((t_0 * Math.sin((0.005555555555555556 * (angle_m * Math.PI)))) * Math.cos((Math.PI * (angle_m / 180.0))));
	} else {
		tmp = (angle_m * 0.011111111111111112) * (t_0 * Math.PI);
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = (b - a) * (b + a)
	tmp = 0
	if math.pow(a, 2.0) <= 5e+262:
		tmp = 2.0 * ((t_0 * math.sin((0.005555555555555556 * (angle_m * math.pi)))) * math.cos((math.pi * (angle_m / 180.0))))
	else:
		tmp = (angle_m * 0.011111111111111112) * (t_0 * math.pi)
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(Float64(b - a) * Float64(b + a))
	tmp = 0.0
	if ((a ^ 2.0) <= 5e+262)
		tmp = Float64(2.0 * Float64(Float64(t_0 * sin(Float64(0.005555555555555556 * Float64(angle_m * pi)))) * cos(Float64(pi * Float64(angle_m / 180.0)))));
	else
		tmp = Float64(Float64(angle_m * 0.011111111111111112) * Float64(t_0 * pi));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = (b - a) * (b + a);
	tmp = 0.0;
	if ((a ^ 2.0) <= 5e+262)
		tmp = 2.0 * ((t_0 * sin((0.005555555555555556 * (angle_m * pi)))) * cos((pi * (angle_m / 180.0))));
	else
		tmp = (angle_m * 0.011111111111111112) * (t_0 * pi);
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 5e+262], N[(2.0 * N[(N[(t$95$0 * N[Sin[N[(0.005555555555555556 * N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(angle$95$m * 0.011111111111111112), $MachinePrecision] * N[(t$95$0 * Pi), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a 2) < 5.00000000000000008e262

   1. Initial program 56.4%

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

      1. associate-*l* 56.4%

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

      2. associate-*l* 56.4%

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

   3. Simplified 56.4%

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

      1. unpow2 56.4%

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

      2. unpow2 56.4%

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

      3. difference-of-squares 56.4%

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

   6. Applied egg-rr 56.4%

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

   7. Taylor expanded in angle around inf 56.2%

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

   if 5.00000000000000008e262 < (pow.f64 a 2)

   1. Initial program 37.8%

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

      1. associate-*l* 37.8%

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

      2. associate-*l* 37.8%

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

   3. Simplified 37.8%

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

      1. unpow2 37.8%

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

      2. unpow2 37.8%

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

      3. difference-of-squares 56.4%

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

   6. Applied egg-rr 56.4%

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

   7. Taylor expanded in angle around 0 59.0%

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

   8. Taylor expanded in angle around 0 61.4%

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

   9. Taylor expanded in angle around 0 61.4%

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

       1. associate-*r* 61.4%

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

   11. Simplified 61.4%

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

4. Final simplification 57.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{+262}:\\ \;\;\;\;2 \cdot \left(\left(\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(angle \cdot 0.011111111111111112\right) \cdot \left(\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \pi\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 57.3% accurate, 1.3× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \left(b - a\right) \cdot \left(b + a\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{+262}:\\ \;\;\;\;2 \cdot \left(\cos \left(\pi \cdot \frac{angle_m}{180}\right) \cdot \left(t_0 \cdot \sin \left(angle_m \cdot \left(0.005555555555555556 \cdot \pi\right)\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(angle_m \cdot 0.011111111111111112\right) \cdot \left(t_0 \cdot \pi\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* (- b a) (+ b a))))
   (*
    angle_s
    (if (<= (pow a 2.0) 5e+262)
      (*
       2.0
       (*
        (cos (* PI (/ angle_m 180.0)))
        (* t_0 (sin (* angle_m (* 0.005555555555555556 PI))))))
      (* (* angle_m 0.011111111111111112) (* t_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 = (b - a) * (b + a);
	double tmp;
	if (pow(a, 2.0) <= 5e+262) {
		tmp = 2.0 * (cos((((double) M_PI) * (angle_m / 180.0))) * (t_0 * sin((angle_m * (0.005555555555555556 * ((double) M_PI))))));
	} else {
		tmp = (angle_m * 0.011111111111111112) * (t_0 * ((double) M_PI));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (b - a) * (b + a);
	double tmp;
	if (Math.pow(a, 2.0) <= 5e+262) {
		tmp = 2.0 * (Math.cos((Math.PI * (angle_m / 180.0))) * (t_0 * Math.sin((angle_m * (0.005555555555555556 * Math.PI)))));
	} else {
		tmp = (angle_m * 0.011111111111111112) * (t_0 * Math.PI);
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = (b - a) * (b + a)
	tmp = 0
	if math.pow(a, 2.0) <= 5e+262:
		tmp = 2.0 * (math.cos((math.pi * (angle_m / 180.0))) * (t_0 * math.sin((angle_m * (0.005555555555555556 * math.pi)))))
	else:
		tmp = (angle_m * 0.011111111111111112) * (t_0 * math.pi)
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(Float64(b - a) * Float64(b + a))
	tmp = 0.0
	if ((a ^ 2.0) <= 5e+262)
		tmp = Float64(2.0 * Float64(cos(Float64(pi * Float64(angle_m / 180.0))) * Float64(t_0 * sin(Float64(angle_m * Float64(0.005555555555555556 * pi))))));
	else
		tmp = Float64(Float64(angle_m * 0.011111111111111112) * Float64(t_0 * pi));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = (b - a) * (b + a);
	tmp = 0.0;
	if ((a ^ 2.0) <= 5e+262)
		tmp = 2.0 * (cos((pi * (angle_m / 180.0))) * (t_0 * sin((angle_m * (0.005555555555555556 * pi)))));
	else
		tmp = (angle_m * 0.011111111111111112) * (t_0 * pi);
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 5e+262], N[(2.0 * N[(N[Cos[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(t$95$0 * N[Sin[N[(angle$95$m * N[(0.005555555555555556 * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(angle$95$m * 0.011111111111111112), $MachinePrecision] * N[(t$95$0 * Pi), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a 2) < 5.00000000000000008e262

   1. Initial program 56.4%

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

      1. associate-*l* 56.4%

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

      2. associate-*l* 56.4%

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

   3. Simplified 56.4%

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

      1. unpow2 56.4%

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

      2. unpow2 56.4%

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

      3. difference-of-squares 56.4%

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

   6. Applied egg-rr 56.4%

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

   7. Taylor expanded in angle around inf 56.2%

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

      1. *-commutative 56.2%

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

      2. associate-*l* 56.5%

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

   9. Simplified 56.5%

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

   if 5.00000000000000008e262 < (pow.f64 a 2)

   1. Initial program 37.8%

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

      1. associate-*l* 37.8%

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

      2. associate-*l* 37.8%

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

   3. Simplified 37.8%

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

      1. unpow2 37.8%

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

      2. unpow2 37.8%

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

      3. difference-of-squares 56.4%

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

   6. Applied egg-rr 56.4%

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

   7. Taylor expanded in angle around 0 59.0%

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

   8. Taylor expanded in angle around 0 61.4%

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

   9. Taylor expanded in angle around 0 61.4%

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

       1. associate-*r* 61.4%

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

   11. Simplified 61.4%

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

4. Final simplification 58.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{+262}:\\ \;\;\;\;2 \cdot \left(\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \sin \left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(angle \cdot 0.011111111111111112\right) \cdot \left(\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \pi\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 57.5% accurate, 1.9× speedup

Math

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

FPCore

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

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	return angle_s * (2.0 * (cos((0.005555555555555556 * (angle_m * ((double) M_PI)))) * (((b - a) * (b + a)) * sin((((double) M_PI) * (angle_m / 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 * (2.0 * (Math.cos((0.005555555555555556 * (angle_m * Math.PI))) * (((b - a) * (b + a)) * Math.sin((Math.PI * (angle_m / 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 * (2.0 * (math.cos((0.005555555555555556 * (angle_m * math.pi))) * (((b - a) * (b + a)) * math.sin((math.pi * (angle_m / 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(2.0 * Float64(cos(Float64(0.005555555555555556 * Float64(angle_m * pi))) * Float64(Float64(Float64(b - a) * Float64(b + a)) * sin(Float64(pi * Float64(angle_m / 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 * (2.0 * (cos((0.005555555555555556 * (angle_m * pi))) * (((b - a) * (b + a)) * sin((pi * (angle_m / 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[(2.0 * N[(N[Cos[N[(0.005555555555555556 * N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 50.8%

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

   1. associate-*l* 50.8%

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

   2. associate-*l* 50.8%

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

3. Simplified 50.8%

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

   1. unpow2 50.8%

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

   2. unpow2 50.8%

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

   3. difference-of-squares 56.4%

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

6. Applied egg-rr 56.4%

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

7. Taylor expanded in angle around inf 56.6%

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

8. Final simplification 56.6%

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

Alternative 7: 57.8% accurate, 1.9× speedup

Math

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

FPCore

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

C

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

Java

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

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	return angle_s * (2.0 * ((((b - a) * (b + a)) * math.sin((math.pi * (angle_m / 180.0)))) * math.cos((angle_m * (0.005555555555555556 * 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(2.0 * Float64(Float64(Float64(Float64(b - a) * Float64(b + a)) * sin(Float64(pi * Float64(angle_m / 180.0)))) * cos(Float64(angle_m * Float64(0.005555555555555556 * pi))))))
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp = code(angle_s, a, b, angle_m)
	tmp = angle_s * (2.0 * ((((b - a) * (b + a)) * sin((pi * (angle_m / 180.0)))) * cos((angle_m * (0.005555555555555556 * 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[(2.0 * N[(N[(N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(angle$95$m * N[(0.005555555555555556 * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 50.8%

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

   1. associate-*l* 50.8%

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

   2. associate-*l* 50.8%

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

3. Simplified 50.8%

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

   1. unpow2 50.8%

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

   2. unpow2 50.8%

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

   3. difference-of-squares 56.4%

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

6. Applied egg-rr 56.4%

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

7. Taylor expanded in angle around inf 56.6%

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

   1. *-commutative 56.6%

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

   2. associate-*l* 56.9%

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

9. Simplified 56.9%

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

10. Final simplification 56.9%

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

Alternative 8: 57.5% accurate, 1.9× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 0.005555555555555556 \cdot \left(angle_m \cdot \pi\right)\\ angle_s \cdot \left(2 \cdot \left(\left(\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \sin t_0\right) \cdot \cos t_0\right)\right) \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* 0.005555555555555556 (* angle_m PI))))
   (* angle_s (* 2.0 (* (* (* (- b a) (+ b a)) (sin t_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 = 0.005555555555555556 * (angle_m * ((double) M_PI));
	return angle_s * (2.0 * ((((b - a) * (b + a)) * sin(t_0)) * cos(t_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) {
	double t_0 = 0.005555555555555556 * (angle_m * Math.PI);
	return angle_s * (2.0 * ((((b - a) * (b + a)) * Math.sin(t_0)) * Math.cos(t_0)));
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = 0.005555555555555556 * (angle_m * math.pi)
	return angle_s * (2.0 * ((((b - a) * (b + a)) * math.sin(t_0)) * math.cos(t_0)))

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(0.005555555555555556 * Float64(angle_m * pi))
	return Float64(angle_s * Float64(2.0 * Float64(Float64(Float64(Float64(b - a) * Float64(b + a)) * sin(t_0)) * cos(t_0))))
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp = code(angle_s, a, b, angle_m)
	t_0 = 0.005555555555555556 * (angle_m * pi);
	tmp = angle_s * (2.0 * ((((b - a) * (b + a)) * sin(t_0)) * cos(t_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_] := Block[{t$95$0 = N[(0.005555555555555556 * N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * N[(2.0 * N[(N[(N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision] * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Initial program 50.8%

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

   1. associate-*l* 50.8%

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

   2. associate-*l* 50.8%

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

3. Simplified 50.8%

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

   1. unpow2 50.8%

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

   2. unpow2 50.8%

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

   3. difference-of-squares 56.4%

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

6. Applied egg-rr 56.4%

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

7. Taylor expanded in angle around inf 55.8%

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

   1. *-commutative 55.8%

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

   2. +-commutative 55.8%

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

   3. associate-*r* 56.1%

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

   4. *-commutative 56.1%

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

   5. *-commutative 56.1%

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

   6. *-commutative 56.1%

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

   7. +-commutative 56.1%

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

   8. *-commutative 56.1%

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

9. Simplified 56.1%

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

10. Taylor expanded in angle around inf 57.4%

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

11. Taylor expanded in angle around inf 57.5%

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

12. Final simplification 57.5%

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

Alternative 9: 55.1% accurate, 3.4× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \left(b - a\right) \cdot \left(b + a\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;\frac{angle_m}{180} \leq 4 \cdot 10^{+205}:\\ \;\;\;\;2 \cdot \left(t_0 \cdot \left(angle_m \cdot \left(0.005555555555555556 \cdot \pi\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \left(t_0 \cdot \sin \left(0.005555555555555556 \cdot \left(angle_m \cdot \pi\right)\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* (- b a) (+ b a))))
   (*
    angle_s
    (if (<= (/ angle_m 180.0) 4e+205)
      (* 2.0 (* t_0 (* angle_m (* 0.005555555555555556 PI))))
      (* 2.0 (* t_0 (sin (* 0.005555555555555556 (* angle_m PI)))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (b - a) * (b + a);
	double tmp;
	if ((angle_m / 180.0) <= 4e+205) {
		tmp = 2.0 * (t_0 * (angle_m * (0.005555555555555556 * ((double) M_PI))));
	} else {
		tmp = 2.0 * (t_0 * sin((0.005555555555555556 * (angle_m * ((double) M_PI)))));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (b - a) * (b + a);
	double tmp;
	if ((angle_m / 180.0) <= 4e+205) {
		tmp = 2.0 * (t_0 * (angle_m * (0.005555555555555556 * Math.PI)));
	} else {
		tmp = 2.0 * (t_0 * Math.sin((0.005555555555555556 * (angle_m * Math.PI))));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = (b - a) * (b + a)
	tmp = 0
	if (angle_m / 180.0) <= 4e+205:
		tmp = 2.0 * (t_0 * (angle_m * (0.005555555555555556 * math.pi)))
	else:
		tmp = 2.0 * (t_0 * math.sin((0.005555555555555556 * (angle_m * math.pi))))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(Float64(b - a) * Float64(b + a))
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 4e+205)
		tmp = Float64(2.0 * Float64(t_0 * Float64(angle_m * Float64(0.005555555555555556 * pi))));
	else
		tmp = Float64(2.0 * Float64(t_0 * sin(Float64(0.005555555555555556 * Float64(angle_m * pi)))));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = (b - a) * (b + a);
	tmp = 0.0;
	if ((angle_m / 180.0) <= 4e+205)
		tmp = 2.0 * (t_0 * (angle_m * (0.005555555555555556 * pi)));
	else
		tmp = 2.0 * (t_0 * sin((0.005555555555555556 * (angle_m * pi))));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 4e+205], N[(2.0 * N[(t$95$0 * N[(angle$95$m * N[(0.005555555555555556 * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 * N[(t$95$0 * N[Sin[N[(0.005555555555555556 * N[(angle$95$m * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 angle 180) < 4.00000000000000007e205

   1. Initial program 51.5%

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

      1. associate-*l* 51.5%

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

      2. associate-*l* 51.5%

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

   3. Simplified 51.5%

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

      1. unpow2 51.5%

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

      2. unpow2 51.5%

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

      3. difference-of-squares 56.9%

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

   6. Applied egg-rr 56.9%

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

   7. Taylor expanded in angle around 0 56.0%

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

   8. Taylor expanded in angle around 0 56.2%

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

      1. *-commutative 56.2%

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

      2. associate-*l* 56.2%

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

   10. Simplified 56.2%

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

   if 4.00000000000000007e205 < (/.f64 angle 180)

   1. Initial program 45.1%

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

      1. associate-*l* 45.1%

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

      2. associate-*l* 45.1%

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

   3. Simplified 45.1%

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

      1. unpow2 45.1%

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

      2. unpow2 45.1%

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

      3. difference-of-squares 52.2%

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

   6. Applied egg-rr 52.2%

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

   7. Taylor expanded in angle around 0 37.3%

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

   8. Taylor expanded in angle around inf 41.0%

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

4. Final simplification 54.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 4 \cdot 10^{+205}:\\ \;\;\;\;2 \cdot \left(\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \left(\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 10: 56.1% accurate, 3.6× speedup

Math

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

FPCore

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 50.8%

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

   1. associate-*l* 50.8%

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

   2. associate-*l* 50.8%

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

3. Simplified 50.8%

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

   1. unpow2 50.8%

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

   2. unpow2 50.8%

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

   3. difference-of-squares 56.4%

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

6. Applied egg-rr 56.4%

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

7. Taylor expanded in angle around 0 54.0%

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

   1. clear-num 55.1%

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

   2. un-div-inv 55.1%

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

9. Applied egg-rr 55.1%

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

10. Final simplification 55.1%

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

Alternative 11: 54.7% accurate, 27.9× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 50.8%

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

   1. associate-*l* 50.8%

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

   2. associate-*l* 50.8%

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

3. Simplified 50.8%

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

   1. unpow2 50.8%

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

   2. unpow2 50.8%

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

   3. difference-of-squares 56.4%

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

6. Applied egg-rr 56.4%

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

7. Taylor expanded in angle around 0 54.0%

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

8. Taylor expanded in angle around 0 53.4%

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

9. Final simplification 53.4%

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

Alternative 12: 54.7% accurate, 27.9× speedup

Math

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

FPCore

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

C

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

Java

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

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	return angle_s * (2.0 * (((b - a) * (b + a)) * (angle_m * (0.005555555555555556 * 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(2.0 * Float64(Float64(Float64(b - a) * Float64(b + a)) * Float64(angle_m * Float64(0.005555555555555556 * pi)))))
end

MATLAB

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

Derivation

1. Initial program 50.8%

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

   1. associate-*l* 50.8%

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

   2. associate-*l* 50.8%

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

3. Simplified 50.8%

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

   1. unpow2 50.8%

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

   2. unpow2 50.8%

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

   3. difference-of-squares 56.4%

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

6. Applied egg-rr 56.4%

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

7. Taylor expanded in angle around 0 54.0%

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

8. Taylor expanded in angle around 0 53.4%

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

   1. *-commutative 53.4%

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

   2. associate-*l* 53.4%

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

10. Simplified 53.4%

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

11. Final simplification 53.4%

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

Alternative 13: 54.6% accurate, 32.2× speedup

Math

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

FPCore

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

Derivation

1. Initial program 50.8%

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

   1. associate-*l* 50.8%

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

   2. associate-*l* 50.8%

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

3. Simplified 50.8%

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

   1. unpow2 50.8%

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

   2. unpow2 50.8%

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

   3. difference-of-squares 56.4%

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

6. Applied egg-rr 56.4%

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

7. Taylor expanded in angle around 0 54.0%

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

8. Taylor expanded in angle around 0 53.3%

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

9. Taylor expanded in angle around 0 53.3%

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

10. Final simplification 53.3%

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

Alternative 14: 54.6% accurate, 32.2× speedup

Math

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

FPCore

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

Derivation

1. Initial program 50.8%

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

   1. associate-*l* 50.8%

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

   2. associate-*l* 50.8%

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

3. Simplified 50.8%

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

   1. unpow2 50.8%

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

   2. unpow2 50.8%

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

   3. difference-of-squares 56.4%

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

6. Applied egg-rr 56.4%

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

7. Taylor expanded in angle around 0 54.0%

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

8. Taylor expanded in angle around 0 53.3%

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

9. Taylor expanded in angle around 0 53.3%

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

    1. associate-*r* 53.3%

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

11. Simplified 53.3%

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

12. Final simplification 53.3%

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

Reproduce

herbie shell --seed 2024017 
(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)))))