ab-angle->ABCF B

Percentage Accurate: 53.7% → 56.8%

Time: 23.9s

Alternatives: 8

Speedup: 5.5×

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

Specification

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Initial Program: 53.7% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Alternative 1: 56.8% accurate, 1.0× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ angle_s \cdot \left(\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle_m \cdot 0.005555555555555556\right)\right)\right)\right)\right) \cdot \cos \left(\pi \cdot \frac{angle_m}{180}\right)\right) \end{array} \]

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 (expm1 (log1p (* PI (* angle_m 0.005555555555555556))))))
   (cos (* PI (/ angle_m 180.0))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	return angle_s * (((2.0 * ((b + a) * (b - a))) * sin(expm1(log1p((((double) M_PI) * (angle_m * 0.005555555555555556)))))) * cos((((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 * ((b + a) * (b - a))) * Math.sin(Math.expm1(Math.log1p((Math.PI * (angle_m * 0.005555555555555556)))))) * Math.cos((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 * ((b + a) * (b - a))) * math.sin(math.expm1(math.log1p((math.pi * (angle_m * 0.005555555555555556)))))) * math.cos((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(Float64(Float64(2.0 * Float64(Float64(b + a) * Float64(b - a))) * sin(expm1(log1p(Float64(pi * Float64(angle_m * 0.005555555555555556)))))) * cos(Float64(pi * Float64(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[(N[(N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(Exp[N[Log[1 + N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 57.0%

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

   1. unpow2 57.0%

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

   2. unpow2 57.0%

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

   3. difference-of-squares 60.6%

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

3. Applied egg-rr 60.6%

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

   1. div-inv 62.2%

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

   2. metadata-eval 62.2%

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

   3. expm1-log1p-u 54.2%

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

5. Applied egg-rr 54.2%

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

6. Final simplification 54.2%

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

Alternative 2: 56.7% accurate, 1.0× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\\ t_1 := 0.005555555555555556 \cdot \left(\pi \cdot angle_m\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq 2 \cdot 10^{+305}:\\ \;\;\;\;\left(t_0 \cdot \sin t_1\right) \cdot \cos t_1\\ \mathbf{else}:\\ \;\;\;\;t_0 \cdot \sin \left({\left(\sqrt[3]{\pi \cdot \left(angle_m \cdot 0.005555555555555556\right)}\right)}^{3}\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* 2.0 (* (+ b a) (- b a))))
        (t_1 (* 0.005555555555555556 (* PI angle_m))))
   (*
    angle_s
    (if (<= (- (pow b 2.0) (pow a 2.0)) 2e+305)
      (* (* t_0 (sin t_1)) (cos t_1))
      (*
       t_0
       (sin (pow (cbrt (* PI (* angle_m 0.005555555555555556))) 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) {
	double t_0 = 2.0 * ((b + a) * (b - a));
	double t_1 = 0.005555555555555556 * (((double) M_PI) * angle_m);
	double tmp;
	if ((pow(b, 2.0) - pow(a, 2.0)) <= 2e+305) {
		tmp = (t_0 * sin(t_1)) * cos(t_1);
	} else {
		tmp = t_0 * sin(pow(cbrt((((double) M_PI) * (angle_m * 0.005555555555555556))), 3.0));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b + a) * (b - a));
	double t_1 = 0.005555555555555556 * (Math.PI * angle_m);
	double tmp;
	if ((Math.pow(b, 2.0) - Math.pow(a, 2.0)) <= 2e+305) {
		tmp = (t_0 * Math.sin(t_1)) * Math.cos(t_1);
	} else {
		tmp = t_0 * Math.sin(Math.pow(Math.cbrt((Math.PI * (angle_m * 0.005555555555555556))), 3.0));
	}
	return angle_s * tmp;
}

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(2.0 * Float64(Float64(b + a) * Float64(b - a)))
	t_1 = Float64(0.005555555555555556 * Float64(pi * angle_m))
	tmp = 0.0
	if (Float64((b ^ 2.0) - (a ^ 2.0)) <= 2e+305)
		tmp = Float64(Float64(t_0 * sin(t_1)) * cos(t_1));
	else
		tmp = Float64(t_0 * sin((cbrt(Float64(pi * Float64(angle_m * 0.005555555555555556))) ^ 3.0)));
	end
	return Float64(angle_s * tmp)
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(0.005555555555555556 * N[(Pi * angle$95$m), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision], 2e+305], N[(N[(t$95$0 * N[Sin[t$95$1], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$1], $MachinePrecision]), $MachinePrecision], N[(t$95$0 * N[Sin[N[Power[N[Power[N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision], 3.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (-.f64 (pow.f64 b 2) (pow.f64 a 2)) < 1.9999999999999999e305

   1. Initial program 64.5%

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

      1. unpow2 64.5%

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

      2. unpow2 64.5%

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

      3. difference-of-squares 64.5%

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

   3. Applied egg-rr 64.5%

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

   4. Taylor expanded in angle around 0 64.5%

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

   5. Taylor expanded in angle around inf 65.6%

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

   if 1.9999999999999999e305 < (-.f64 (pow.f64 b 2) (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. unpow2 37.8%

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

      2. unpow2 37.8%

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

      3. difference-of-squares 50.4%

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

   3. Applied egg-rr 50.4%

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

   4. Taylor expanded in angle around 0 52.1%

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

      1. add-cube-cbrt 56.3%

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

      2. pow3 61.8%

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

      3. div-inv 61.8%

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

      4. metadata-eval 61.8%

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

   6. Applied egg-rr 61.8%

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

4. Final simplification 64.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq 2 \cdot 10^{+305}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right) \cdot \cos \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left({\left(\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right)}^{3}\right)\\ \end{array} \]

Alternative 3: 55.3% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 57.0%

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

   1. unpow2 57.0%

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

   2. unpow2 57.0%

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

   3. difference-of-squares 60.6%

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

3. Applied egg-rr 60.6%

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

   1. div-inv 62.2%

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

   2. metadata-eval 62.2%

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

   3. add-exp-log 30.2%

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

5. Applied egg-rr 30.2%

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

6. Final simplification 30.2%

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

Alternative 4: 56.6% accurate, 1.0× speedup

Math

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

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* PI (* angle_m 0.005555555555555556))))
   (*
    angle_s
    (* (* (* 2.0 (* (+ b a) (- b a))) (sin (expm1 (log1p 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 = ((double) M_PI) * (angle_m * 0.005555555555555556);
	return angle_s * (((2.0 * ((b + a) * (b - a))) * sin(expm1(log1p(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 = Math.PI * (angle_m * 0.005555555555555556);
	return angle_s * (((2.0 * ((b + a) * (b - a))) * Math.sin(Math.expm1(Math.log1p(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 = math.pi * (angle_m * 0.005555555555555556)
	return angle_s * (((2.0 * ((b + a) * (b - a))) * math.sin(math.expm1(math.log1p(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(pi * Float64(angle_m * 0.005555555555555556))
	return Float64(angle_s * Float64(Float64(Float64(2.0 * Float64(Float64(b + a) * Float64(b - a))) * sin(expm1(log1p(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[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * N[(N[(N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(Exp[N[Log[1 + t$95$0], $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Initial program 57.0%

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

   1. unpow2 57.0%

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

   2. unpow2 57.0%

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

   3. difference-of-squares 60.6%

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

3. Applied egg-rr 60.6%

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

   1. div-inv 62.2%

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

   2. metadata-eval 62.2%

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

   3. expm1-log1p-u 54.2%

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

5. Applied egg-rr 54.2%

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

6. Taylor expanded in angle around inf 54.2%

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

   1. associate-*r* 54.1%

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

   2. *-commutative 54.1%

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

   3. *-commutative 54.1%

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

   4. *-commutative 54.1%

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

8. Simplified 54.1%

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

9. Final simplification 54.1%

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

Alternative 5: 55.8% accurate, 1.5× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \pi \cdot \left(angle_m \cdot 0.005555555555555556\right)\\ t_1 := 2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;b \leq 1.05 \cdot 10^{+188}:\\ \;\;\;\;t_1 \cdot \sin t_0\\ \mathbf{else}:\\ \;\;\;\;t_1 \cdot \sin \left({\left(\sqrt[3]{t_0}\right)}^{3}\right)\\ \end{array} \end{array} \end{array} \]

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 0.005555555555555556)))
        (t_1 (* 2.0 (* (+ b a) (- b a)))))
   (*
    angle_s
    (if (<= b 1.05e+188)
      (* t_1 (sin t_0))
      (* t_1 (sin (pow (cbrt t_0) 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) {
	double t_0 = ((double) M_PI) * (angle_m * 0.005555555555555556);
	double t_1 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if (b <= 1.05e+188) {
		tmp = t_1 * sin(t_0);
	} else {
		tmp = t_1 * sin(pow(cbrt(t_0), 3.0));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = Math.PI * (angle_m * 0.005555555555555556);
	double t_1 = 2.0 * ((b + a) * (b - a));
	double tmp;
	if (b <= 1.05e+188) {
		tmp = t_1 * Math.sin(t_0);
	} else {
		tmp = t_1 * Math.sin(Math.pow(Math.cbrt(t_0), 3.0));
	}
	return angle_s * tmp;
}

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(pi * Float64(angle_m * 0.005555555555555556))
	t_1 = Float64(2.0 * Float64(Float64(b + a) * Float64(b - a)))
	tmp = 0.0
	if (b <= 1.05e+188)
		tmp = Float64(t_1 * sin(t_0));
	else
		tmp = Float64(t_1 * sin((cbrt(t_0) ^ 3.0)));
	end
	return Float64(angle_s * tmp)
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[b, 1.05e+188], N[(t$95$1 * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision], N[(t$95$1 * N[Sin[N[Power[N[Power[t$95$0, 1/3], $MachinePrecision], 3.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if b < 1.04999999999999993e188

   1. Initial program 57.3%

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

      1. unpow2 57.3%

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

      2. unpow2 57.3%

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

      3. difference-of-squares 60.0%

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

   3. Applied egg-rr 60.0%

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

   4. Taylor expanded in angle around 0 60.8%

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

   5. Taylor expanded in angle around inf 60.1%

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

      1. associate-*r* 60.9%

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

      2. *-commutative 60.9%

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

      3. *-commutative 60.9%

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

      4. *-commutative 60.9%

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

   7. Simplified 60.9%

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

   if 1.04999999999999993e188 < b

   1. Initial program 54.3%

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

      1. unpow2 54.3%

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

      2. unpow2 54.3%

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

      3. difference-of-squares 65.1%

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

   3. Applied egg-rr 65.1%

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

   4. Taylor expanded in angle around 0 65.1%

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

      1. add-cube-cbrt 72.2%

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

      2. pow3 82.9%

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

      3. div-inv 79.3%

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

      4. metadata-eval 79.3%

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

   6. Applied egg-rr 79.3%

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

4. Final simplification 62.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq 1.05 \cdot 10^{+188}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left({\left(\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right)}^{3}\right)\\ \end{array} \]

Alternative 6: 55.3% accurate, 2.9× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ angle_s \cdot \begin{array}{l} \mathbf{if}\;b \leq 2.7 \cdot 10^{+187}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle_m\right)\right)\\ \mathbf{else}:\\ \;\;\;\;0.011111111111111112 \cdot \left({b}^{2} \cdot \left(\pi \cdot angle_m\right)\right)\\ \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (*
  angle_s
  (if (<= b 2.7e+187)
    (*
     (* 2.0 (* (+ b a) (- b a)))
     (sin (* 0.005555555555555556 (* PI angle_m))))
    (* 0.011111111111111112 (* (pow b 2.0) (* PI angle_m))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double tmp;
	if (b <= 2.7e+187) {
		tmp = (2.0 * ((b + a) * (b - a))) * sin((0.005555555555555556 * (((double) M_PI) * angle_m)));
	} else {
		tmp = 0.011111111111111112 * (pow(b, 2.0) * (((double) M_PI) * angle_m));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double tmp;
	if (b <= 2.7e+187) {
		tmp = (2.0 * ((b + a) * (b - a))) * Math.sin((0.005555555555555556 * (Math.PI * angle_m)));
	} else {
		tmp = 0.011111111111111112 * (Math.pow(b, 2.0) * (Math.PI * angle_m));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	tmp = 0
	if b <= 2.7e+187:
		tmp = (2.0 * ((b + a) * (b - a))) * math.sin((0.005555555555555556 * (math.pi * angle_m)))
	else:
		tmp = 0.011111111111111112 * (math.pow(b, 2.0) * (math.pi * angle_m))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	tmp = 0.0
	if (b <= 2.7e+187)
		tmp = Float64(Float64(2.0 * Float64(Float64(b + a) * Float64(b - a))) * sin(Float64(0.005555555555555556 * Float64(pi * angle_m))));
	else
		tmp = Float64(0.011111111111111112 * Float64((b ^ 2.0) * Float64(pi * angle_m)));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	tmp = 0.0;
	if (b <= 2.7e+187)
		tmp = (2.0 * ((b + a) * (b - a))) * sin((0.005555555555555556 * (pi * angle_m)));
	else
		tmp = 0.011111111111111112 * ((b ^ 2.0) * (pi * angle_m));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := N[(angle$95$s * If[LessEqual[b, 2.7e+187], N[(N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(0.005555555555555556 * N[(Pi * angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(0.011111111111111112 * N[(N[Power[b, 2.0], $MachinePrecision] * N[(Pi * angle$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if b < 2.70000000000000008e187

   1. Initial program 57.6%

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

      1. unpow2 57.6%

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

      2. unpow2 57.6%

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

      3. difference-of-squares 60.3%

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

   3. Applied egg-rr 60.3%

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

   4. Taylor expanded in angle around 0 61.2%

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

   5. Taylor expanded in angle around 0 60.3%

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

   if 2.70000000000000008e187 < b

   1. Initial program 52.5%

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

      1. unpow2 52.5%

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

      2. unpow2 52.5%

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

      3. difference-of-squares 62.8%

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

   3. Applied egg-rr 62.8%

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

   4. Taylor expanded in angle around 0 66.3%

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

   5. Taylor expanded in angle around 0 76.6%

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

   6. Taylor expanded in a around 0 80.0%

      \[\leadsto \left(0.011111111111111112 \cdot \color{blue}{\left(angle \cdot \left({b}^{2} \cdot \pi\right)\right)}\right) \cdot 1 \]
   7. Step-by-step derivation

      1. *-commutative 80.0%

         \[\leadsto \left(0.011111111111111112 \cdot \color{blue}{\left(\left({b}^{2} \cdot \pi\right) \cdot angle\right)}\right) \cdot 1 \]

      2. associate-*l* 80.0%

         \[\leadsto \left(0.011111111111111112 \cdot \color{blue}{\left({b}^{2} \cdot \left(\pi \cdot angle\right)\right)}\right) \cdot 1 \]

      3. *-commutative 80.0%

         \[\leadsto \left(0.011111111111111112 \cdot \left({b}^{2} \cdot \color{blue}{\left(angle \cdot \pi\right)}\right)\right) \cdot 1 \]

   8. Simplified 80.0%

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

4. Final simplification 62.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq 2.7 \cdot 10^{+187}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\\ \mathbf{else}:\\ \;\;\;\;0.011111111111111112 \cdot \left({b}^{2} \cdot \left(\pi \cdot angle\right)\right)\\ \end{array} \]

Alternative 7: 55.4% accurate, 2.9× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ angle_s \cdot \begin{array}{l} \mathbf{if}\;b \leq 5 \cdot 10^{+183}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle_m \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;0.011111111111111112 \cdot \left({b}^{2} \cdot \left(\pi \cdot angle_m\right)\right)\\ \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (*
  angle_s
  (if (<= b 5e+183)
    (*
     (* 2.0 (* (+ b a) (- b a)))
     (sin (* PI (* angle_m 0.005555555555555556))))
    (* 0.011111111111111112 (* (pow b 2.0) (* PI angle_m))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double tmp;
	if (b <= 5e+183) {
		tmp = (2.0 * ((b + a) * (b - a))) * sin((((double) M_PI) * (angle_m * 0.005555555555555556)));
	} else {
		tmp = 0.011111111111111112 * (pow(b, 2.0) * (((double) M_PI) * angle_m));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double tmp;
	if (b <= 5e+183) {
		tmp = (2.0 * ((b + a) * (b - a))) * Math.sin((Math.PI * (angle_m * 0.005555555555555556)));
	} else {
		tmp = 0.011111111111111112 * (Math.pow(b, 2.0) * (Math.PI * angle_m));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	tmp = 0
	if b <= 5e+183:
		tmp = (2.0 * ((b + a) * (b - a))) * math.sin((math.pi * (angle_m * 0.005555555555555556)))
	else:
		tmp = 0.011111111111111112 * (math.pow(b, 2.0) * (math.pi * angle_m))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	tmp = 0.0
	if (b <= 5e+183)
		tmp = Float64(Float64(2.0 * Float64(Float64(b + a) * Float64(b - a))) * sin(Float64(pi * Float64(angle_m * 0.005555555555555556))));
	else
		tmp = Float64(0.011111111111111112 * Float64((b ^ 2.0) * Float64(pi * angle_m)));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	tmp = 0.0;
	if (b <= 5e+183)
		tmp = (2.0 * ((b + a) * (b - a))) * sin((pi * (angle_m * 0.005555555555555556)));
	else
		tmp = 0.011111111111111112 * ((b ^ 2.0) * (pi * angle_m));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := N[(angle$95$s * If[LessEqual[b, 5e+183], N[(N[(2.0 * N[(N[(b + a), $MachinePrecision] * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(0.011111111111111112 * N[(N[Power[b, 2.0], $MachinePrecision] * N[(Pi * angle$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if b < 5.00000000000000009e183

   1. Initial program 57.6%

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

      1. unpow2 57.6%

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

      2. unpow2 57.6%

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

      3. difference-of-squares 60.3%

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

   3. Applied egg-rr 60.3%

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

   4. Taylor expanded in angle around 0 60.6%

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

   5. Taylor expanded in angle around inf 60.3%

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

      1. associate-*r* 60.7%

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

      2. *-commutative 60.7%

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

      3. *-commutative 60.7%

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

      4. *-commutative 60.7%

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

   7. Simplified 60.7%

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

   if 5.00000000000000009e183 < b

   1. Initial program 52.5%

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

      1. unpow2 52.5%

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

      2. unpow2 52.5%

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

      3. difference-of-squares 62.8%

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

   3. Applied egg-rr 62.8%

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

   4. Taylor expanded in angle around 0 66.3%

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

   5. Taylor expanded in angle around 0 76.6%

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

   6. Taylor expanded in a around 0 80.0%

      \[\leadsto \left(0.011111111111111112 \cdot \color{blue}{\left(angle \cdot \left({b}^{2} \cdot \pi\right)\right)}\right) \cdot 1 \]
   7. Step-by-step derivation

      1. *-commutative 80.0%

         \[\leadsto \left(0.011111111111111112 \cdot \color{blue}{\left(\left({b}^{2} \cdot \pi\right) \cdot angle\right)}\right) \cdot 1 \]

      2. associate-*l* 80.0%

         \[\leadsto \left(0.011111111111111112 \cdot \color{blue}{\left({b}^{2} \cdot \left(\pi \cdot angle\right)\right)}\right) \cdot 1 \]

      3. *-commutative 80.0%

         \[\leadsto \left(0.011111111111111112 \cdot \left({b}^{2} \cdot \color{blue}{\left(angle \cdot \pi\right)}\right)\right) \cdot 1 \]

   8. Simplified 80.0%

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

4. Final simplification 62.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq 5 \cdot 10^{+183}:\\ \;\;\;\;\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;0.011111111111111112 \cdot \left({b}^{2} \cdot \left(\pi \cdot angle\right)\right)\\ \end{array} \]

Alternative 8: 53.9% accurate, 5.5× speedup

Math

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

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 57.0%

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

   1. unpow2 57.0%

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

   2. unpow2 57.0%

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

   3. difference-of-squares 60.6%

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

3. Applied egg-rr 60.6%

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

4. Taylor expanded in angle around 0 61.2%

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

5. Taylor expanded in angle around 0 60.0%

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

6. Final simplification 60.0%

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

Reproduce

herbie shell --seed 2023331 
(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)))))