ab-angle->ABCF B

Percentage Accurate: 53.8% → 58.4%

Time: 29.4s

Alternatives: 12

Speedup: 27.9×

• 32.0% 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: 53.8% 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: 58.4% accurate, 0.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 0.005555555555555556 \cdot \left(angle \cdot \pi\right)\\ \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{-137}:\\ \;\;\;\;2 \cdot \left(\cos \left(angle \cdot \frac{\pi}{-180}\right) \cdot {\left(\sqrt[3]{\sin t\_0} \cdot {\left(\sqrt[3]{b}\right)}^{2}\right)}^{3}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\right) \cdot \cos t\_0\\ \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* 0.005555555555555556 (* angle PI))))
   (if (<= (pow a 2.0) 5e-137)
     (*
      2.0
      (*
       (cos (* angle (/ PI -180.0)))
       (pow (* (cbrt (sin t_0)) (pow (cbrt b) 2.0)) 3.0)))
     (*
      (*
       (* 2.0 (* (- b a) (+ a b)))
       (sin (* (pow (sqrt PI) 2.0) (/ angle 180.0))))
      (cos t_0)))))

C

double code(double a, double b, double angle) {
	double t_0 = 0.005555555555555556 * (angle * ((double) M_PI));
	double tmp;
	if (pow(a, 2.0) <= 5e-137) {
		tmp = 2.0 * (cos((angle * (((double) M_PI) / -180.0))) * pow((cbrt(sin(t_0)) * pow(cbrt(b), 2.0)), 3.0));
	} else {
		tmp = ((2.0 * ((b - a) * (a + b))) * sin((pow(sqrt(((double) M_PI)), 2.0) * (angle / 180.0)))) * cos(t_0);
	}
	return tmp;
}

Java

public static double code(double a, double b, double angle) {
	double t_0 = 0.005555555555555556 * (angle * Math.PI);
	double tmp;
	if (Math.pow(a, 2.0) <= 5e-137) {
		tmp = 2.0 * (Math.cos((angle * (Math.PI / -180.0))) * Math.pow((Math.cbrt(Math.sin(t_0)) * Math.pow(Math.cbrt(b), 2.0)), 3.0));
	} else {
		tmp = ((2.0 * ((b - a) * (a + b))) * Math.sin((Math.pow(Math.sqrt(Math.PI), 2.0) * (angle / 180.0)))) * Math.cos(t_0);
	}
	return tmp;
}

Julia

function code(a, b, angle)
	t_0 = Float64(0.005555555555555556 * Float64(angle * pi))
	tmp = 0.0
	if ((a ^ 2.0) <= 5e-137)
		tmp = Float64(2.0 * Float64(cos(Float64(angle * Float64(pi / -180.0))) * (Float64(cbrt(sin(t_0)) * (cbrt(b) ^ 2.0)) ^ 3.0)));
	else
		tmp = Float64(Float64(Float64(2.0 * Float64(Float64(b - a) * Float64(a + b))) * sin(Float64((sqrt(pi) ^ 2.0) * Float64(angle / 180.0)))) * cos(t_0));
	end
	return tmp
end

Wolfram

code[a_, b_, angle_] := Block[{t$95$0 = N[(0.005555555555555556 * N[(angle * Pi), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 5e-137], N[(2.0 * N[(N[Cos[N[(angle * N[(Pi / -180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[Power[N[(N[Power[N[Sin[t$95$0], $MachinePrecision], 1/3], $MachinePrecision] * N[Power[N[Power[b, 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(a + b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision] * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a 2) < 5.00000000000000001e-137

   1. Initial program 55.2%

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

   3. Simplified 54.4%

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

   5. Taylor expanded in a around 0 54.5%

      \[\leadsto 2 \cdot \left(\cos \left(angle \cdot \frac{\pi}{-180}\right) \cdot \color{blue}{\left(-1 \cdot \left({b}^{2} \cdot \sin \left(-0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\right)}\right) \]
   6. Step-by-step derivation

      1. mul-1-neg 54.5%

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

      2. *-commutative 54.5%

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

      3. distribute-rgt-neg-in 54.5%

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

      4. *-commutative 54.5%

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

      5. *-commutative 54.5%

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

      6. associate-*l* 52.6%

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

      7. *-commutative 52.6%

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

   7. Simplified 52.6%

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

      1. add-cube-cbrt 52.4%

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

      2. pow3 52.4%

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

   9. Applied egg-rr 52.4%

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

       1. cbrt-prod 52.4%

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

       2. unpow2 52.4%

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

       3. cbrt-prod 58.4%

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

       4. pow2 58.4%

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

   11. Applied egg-rr 58.4%

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

       1. *-commutative 58.4%

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

       2. *-commutative 58.4%

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

       3. associate-*r* 60.4%

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

   13. Simplified 60.4%

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

   if 5.00000000000000001e-137 < (pow.f64 a 2)

   1. Initial program 48.6%

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

      1. unpow2 48.6%

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

      2. unpow2 48.6%

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

      3. difference-of-squares 53.2%

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

   4. Applied egg-rr 53.2%

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

      1. add-sqr-sqrt 56.5%

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

      2. pow2 56.5%

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

   6. Applied egg-rr 56.5%

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

   7. Taylor expanded in angle around inf 57.3%

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

4. Final simplification 58.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{-137}:\\ \;\;\;\;2 \cdot \left(\cos \left(angle \cdot \frac{\pi}{-180}\right) \cdot {\left(\sqrt[3]{\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)} \cdot {\left(\sqrt[3]{b}\right)}^{2}\right)}^{3}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 53.1% accurate, 0.8× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \cos \left(\pi \cdot \frac{angle}{180}\right)\\ \mathbf{if}\;{a}^{2} \leq 10^{+65}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(\frac{angle}{180} \cdot \sqrt[3]{{\pi}^{3}}\right)\right) \cdot t\_0\\ \mathbf{else}:\\ \;\;\;\;t\_0 \cdot \left(\sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right) \cdot \left(2 \cdot \left(b \cdot \left(b - a\right) + a \cdot \left(b - a\right)\right)\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (cos (* PI (/ angle 180.0)))))
   (if (<= (pow a 2.0) 1e+65)
     (*
      (*
       (* 2.0 (* (- b a) (+ a b)))
       (sin (* (/ angle 180.0) (cbrt (pow PI 3.0)))))
      t_0)
     (*
      t_0
      (*
       (sin (* (pow (sqrt PI) 2.0) (/ angle 180.0)))
       (* 2.0 (+ (* b (- b a)) (* a (- b a)))))))))

C

double code(double a, double b, double angle) {
	double t_0 = cos((((double) M_PI) * (angle / 180.0)));
	double tmp;
	if (pow(a, 2.0) <= 1e+65) {
		tmp = ((2.0 * ((b - a) * (a + b))) * sin(((angle / 180.0) * cbrt(pow(((double) M_PI), 3.0))))) * t_0;
	} else {
		tmp = t_0 * (sin((pow(sqrt(((double) M_PI)), 2.0) * (angle / 180.0))) * (2.0 * ((b * (b - a)) + (a * (b - a)))));
	}
	return tmp;
}

Java

public static double code(double a, double b, double angle) {
	double t_0 = Math.cos((Math.PI * (angle / 180.0)));
	double tmp;
	if (Math.pow(a, 2.0) <= 1e+65) {
		tmp = ((2.0 * ((b - a) * (a + b))) * Math.sin(((angle / 180.0) * Math.cbrt(Math.pow(Math.PI, 3.0))))) * t_0;
	} else {
		tmp = t_0 * (Math.sin((Math.pow(Math.sqrt(Math.PI), 2.0) * (angle / 180.0))) * (2.0 * ((b * (b - a)) + (a * (b - a)))));
	}
	return tmp;
}

Julia

function code(a, b, angle)
	t_0 = cos(Float64(pi * Float64(angle / 180.0)))
	tmp = 0.0
	if ((a ^ 2.0) <= 1e+65)
		tmp = Float64(Float64(Float64(2.0 * Float64(Float64(b - a) * Float64(a + b))) * sin(Float64(Float64(angle / 180.0) * cbrt((pi ^ 3.0))))) * t_0);
	else
		tmp = Float64(t_0 * Float64(sin(Float64((sqrt(pi) ^ 2.0) * Float64(angle / 180.0))) * Float64(2.0 * Float64(Float64(b * Float64(b - a)) + Float64(a * Float64(b - a))))));
	end
	return tmp
end

Wolfram

code[a_, b_, angle_] := Block[{t$95$0 = N[Cos[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 1e+65], N[(N[(N[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(a + b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(N[(angle / 180.0), $MachinePrecision] * N[Power[N[Power[Pi, 3.0], $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * t$95$0), $MachinePrecision], N[(t$95$0 * N[(N[Sin[N[(N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision] * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(2.0 * N[(N[(b * N[(b - a), $MachinePrecision]), $MachinePrecision] + N[(a * N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a 2) < 9.9999999999999999e64

   1. Initial program 52.7%

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

      1. unpow2 52.7%

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

      2. unpow2 52.7%

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

      3. difference-of-squares 52.7%

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

   4. Applied egg-rr 52.7%

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

      1. add-cbrt-cube 52.9%

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

      2. pow3 52.9%

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

   6. Applied egg-rr 55.2%

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

   if 9.9999999999999999e64 < (pow.f64 a 2)

   1. Initial program 49.2%

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

      1. unpow2 49.2%

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

      2. unpow2 49.2%

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

      3. difference-of-squares 55.4%

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

   4. Applied egg-rr 55.4%

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

      1. add-sqr-sqrt 60.2%

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

      2. pow2 60.2%

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

   6. Applied egg-rr 60.2%

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

      1. *-commutative 60.2%

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

      2. distribute-lft-in 50.4%

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

   8. Applied egg-rr 50.4%

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

4. Final simplification 53.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 10^{+65}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(\frac{angle}{180} \cdot \sqrt[3]{{\pi}^{3}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right) \cdot \left(2 \cdot \left(b \cdot \left(b - a\right) + a \cdot \left(b - a\right)\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 57.4% accurate, 0.8× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\\ t_1 := \pi \cdot \frac{angle}{180}\\ \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{-168}:\\ \;\;\;\;\cos t\_1 \cdot \left(t\_0 \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(t\_0 \cdot \sin t\_1\right) \cdot \cos \left(\frac{angle}{180} \cdot \sqrt[3]{{\pi}^{3}}\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* 2.0 (* (- b a) (+ a b)))) (t_1 (* PI (/ angle 180.0))))
   (if (<= (pow a 2.0) 5e-168)
     (* (cos t_1) (* t_0 (sin (* PI (* angle 0.005555555555555556)))))
     (* (* t_0 (sin t_1)) (cos (* (/ angle 180.0) (cbrt (pow PI 3.0))))))))

C

double code(double a, double b, double angle) {
	double t_0 = 2.0 * ((b - a) * (a + b));
	double t_1 = ((double) M_PI) * (angle / 180.0);
	double tmp;
	if (pow(a, 2.0) <= 5e-168) {
		tmp = cos(t_1) * (t_0 * sin((((double) M_PI) * (angle * 0.005555555555555556))));
	} else {
		tmp = (t_0 * sin(t_1)) * cos(((angle / 180.0) * cbrt(pow(((double) M_PI), 3.0))));
	}
	return tmp;
}

Java

public static double code(double a, double b, double angle) {
	double t_0 = 2.0 * ((b - a) * (a + b));
	double t_1 = Math.PI * (angle / 180.0);
	double tmp;
	if (Math.pow(a, 2.0) <= 5e-168) {
		tmp = Math.cos(t_1) * (t_0 * Math.sin((Math.PI * (angle * 0.005555555555555556))));
	} else {
		tmp = (t_0 * Math.sin(t_1)) * Math.cos(((angle / 180.0) * Math.cbrt(Math.pow(Math.PI, 3.0))));
	}
	return tmp;
}

Julia

function code(a, b, angle)
	t_0 = Float64(2.0 * Float64(Float64(b - a) * Float64(a + b)))
	t_1 = Float64(pi * Float64(angle / 180.0))
	tmp = 0.0
	if ((a ^ 2.0) <= 5e-168)
		tmp = Float64(cos(t_1) * Float64(t_0 * sin(Float64(pi * Float64(angle * 0.005555555555555556)))));
	else
		tmp = Float64(Float64(t_0 * sin(t_1)) * cos(Float64(Float64(angle / 180.0) * cbrt((pi ^ 3.0)))));
	end
	return tmp
end

Wolfram

code[a_, b_, angle_] := Block[{t$95$0 = N[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(a + b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 5e-168], N[(N[Cos[t$95$1], $MachinePrecision] * N[(t$95$0 * N[Sin[N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(t$95$0 * N[Sin[t$95$1], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(N[(angle / 180.0), $MachinePrecision] * N[Power[N[Power[Pi, 3.0], $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a 2) < 5.00000000000000001e-168

   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. Add Preprocessing
   3. 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 57.3%

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

   4. Applied egg-rr 57.3%

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

   5. Taylor expanded in angle around inf 58.2%

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

      1. associate-*r* 59.3%

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

      2. *-commutative 59.3%

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

   7. Simplified 59.3%

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

   if 5.00000000000000001e-168 < (pow.f64 a 2)

   1. Initial program 47.8%

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

      1. unpow2 47.8%

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

      2. unpow2 47.8%

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

      3. difference-of-squares 52.1%

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

   4. Applied egg-rr 52.1%

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

      1. add-cbrt-cube 55.4%

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

      2. pow3 55.4%

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

   6. Applied egg-rr 55.4%

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

4. Final simplification 56.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{-168}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\frac{angle}{180} \cdot \sqrt[3]{{\pi}^{3}}\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 57.0% accurate, 0.8× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\\ \mathbf{if}\;{a}^{2} \leq 2 \cdot 10^{-116}:\\ \;\;\;\;\left(t\_0 \cdot \sin \left(\frac{angle}{180} \cdot \sqrt[3]{{\pi}^{3}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(t\_0 \cdot \sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* 2.0 (* (- b a) (+ a b)))))
   (if (<= (pow a 2.0) 2e-116)
     (*
      (* t_0 (sin (* (/ angle 180.0) (cbrt (pow PI 3.0)))))
      (cos (* PI (/ angle 180.0))))
     (*
      (* t_0 (sin (* (pow (sqrt PI) 2.0) (/ angle 180.0))))
      (cos (* 0.005555555555555556 (* angle PI)))))))

C

double code(double a, double b, double angle) {
	double t_0 = 2.0 * ((b - a) * (a + b));
	double tmp;
	if (pow(a, 2.0) <= 2e-116) {
		tmp = (t_0 * sin(((angle / 180.0) * cbrt(pow(((double) M_PI), 3.0))))) * cos((((double) M_PI) * (angle / 180.0)));
	} else {
		tmp = (t_0 * sin((pow(sqrt(((double) M_PI)), 2.0) * (angle / 180.0)))) * cos((0.005555555555555556 * (angle * ((double) M_PI))));
	}
	return tmp;
}

Java

public static double code(double a, double b, double angle) {
	double t_0 = 2.0 * ((b - a) * (a + b));
	double tmp;
	if (Math.pow(a, 2.0) <= 2e-116) {
		tmp = (t_0 * Math.sin(((angle / 180.0) * Math.cbrt(Math.pow(Math.PI, 3.0))))) * Math.cos((Math.PI * (angle / 180.0)));
	} else {
		tmp = (t_0 * Math.sin((Math.pow(Math.sqrt(Math.PI), 2.0) * (angle / 180.0)))) * Math.cos((0.005555555555555556 * (angle * Math.PI)));
	}
	return tmp;
}

Julia

function code(a, b, angle)
	t_0 = Float64(2.0 * Float64(Float64(b - a) * Float64(a + b)))
	tmp = 0.0
	if ((a ^ 2.0) <= 2e-116)
		tmp = Float64(Float64(t_0 * sin(Float64(Float64(angle / 180.0) * cbrt((pi ^ 3.0))))) * cos(Float64(pi * Float64(angle / 180.0))));
	else
		tmp = Float64(Float64(t_0 * sin(Float64((sqrt(pi) ^ 2.0) * Float64(angle / 180.0)))) * cos(Float64(0.005555555555555556 * Float64(angle * pi))));
	end
	return tmp
end

Wolfram

code[a_, b_, angle_] := Block[{t$95$0 = N[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(a + b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 2e-116], N[(N[(t$95$0 * N[Sin[N[(N[(angle / 180.0), $MachinePrecision] * N[Power[N[Power[Pi, 3.0], $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[(t$95$0 * N[Sin[N[(N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision] * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(0.005555555555555556 * N[(angle * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a 2) < 2e-116

   1. Initial program 54.7%

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

      1. unpow2 54.7%

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

      2. unpow2 54.7%

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

      3. difference-of-squares 54.7%

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

   4. Applied egg-rr 54.7%

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

      1. add-cbrt-cube 53.2%

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

      2. pow3 53.2%

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

   6. Applied egg-rr 56.4%

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

   if 2e-116 < (pow.f64 a 2)

   1. Initial program 48.7%

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

      1. unpow2 48.7%

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

      2. unpow2 48.7%

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

      3. difference-of-squares 53.4%

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

   4. Applied egg-rr 53.4%

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

      1. add-sqr-sqrt 57.1%

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

      2. pow2 57.1%

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

   6. Applied egg-rr 57.1%

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

   7. Taylor expanded in angle around inf 57.7%

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

4. Final simplification 57.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 2 \cdot 10^{-116}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(\frac{angle}{180} \cdot \sqrt[3]{{\pi}^{3}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 56.9% accurate, 0.8× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \cos \left(\pi \cdot \frac{angle}{180}\right)\\ t_1 := 2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\\ \mathbf{if}\;{a}^{2} \leq 10^{+65}:\\ \;\;\;\;\left(t\_1 \cdot \sin \left(\frac{angle}{180} \cdot \sqrt[3]{{\pi}^{3}}\right)\right) \cdot t\_0\\ \mathbf{else}:\\ \;\;\;\;\left(t\_1 \cdot \sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\right) \cdot t\_0\\ \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (cos (* PI (/ angle 180.0)))) (t_1 (* 2.0 (* (- b a) (+ a b)))))
   (if (<= (pow a 2.0) 1e+65)
     (* (* t_1 (sin (* (/ angle 180.0) (cbrt (pow PI 3.0))))) t_0)
     (* (* t_1 (sin (* (pow (sqrt PI) 2.0) (/ angle 180.0)))) t_0))))

C

double code(double a, double b, double angle) {
	double t_0 = cos((((double) M_PI) * (angle / 180.0)));
	double t_1 = 2.0 * ((b - a) * (a + b));
	double tmp;
	if (pow(a, 2.0) <= 1e+65) {
		tmp = (t_1 * sin(((angle / 180.0) * cbrt(pow(((double) M_PI), 3.0))))) * t_0;
	} else {
		tmp = (t_1 * sin((pow(sqrt(((double) M_PI)), 2.0) * (angle / 180.0)))) * t_0;
	}
	return tmp;
}

Java

public static double code(double a, double b, double angle) {
	double t_0 = Math.cos((Math.PI * (angle / 180.0)));
	double t_1 = 2.0 * ((b - a) * (a + b));
	double tmp;
	if (Math.pow(a, 2.0) <= 1e+65) {
		tmp = (t_1 * Math.sin(((angle / 180.0) * Math.cbrt(Math.pow(Math.PI, 3.0))))) * t_0;
	} else {
		tmp = (t_1 * Math.sin((Math.pow(Math.sqrt(Math.PI), 2.0) * (angle / 180.0)))) * t_0;
	}
	return tmp;
}

Julia

function code(a, b, angle)
	t_0 = cos(Float64(pi * Float64(angle / 180.0)))
	t_1 = Float64(2.0 * Float64(Float64(b - a) * Float64(a + b)))
	tmp = 0.0
	if ((a ^ 2.0) <= 1e+65)
		tmp = Float64(Float64(t_1 * sin(Float64(Float64(angle / 180.0) * cbrt((pi ^ 3.0))))) * t_0);
	else
		tmp = Float64(Float64(t_1 * sin(Float64((sqrt(pi) ^ 2.0) * Float64(angle / 180.0)))) * t_0);
	end
	return tmp
end

Wolfram

code[a_, b_, angle_] := Block[{t$95$0 = N[Cos[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]}, Block[{t$95$1 = N[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(a + b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 1e+65], N[(N[(t$95$1 * N[Sin[N[(N[(angle / 180.0), $MachinePrecision] * N[Power[N[Power[Pi, 3.0], $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * t$95$0), $MachinePrecision], N[(N[(t$95$1 * N[Sin[N[(N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision] * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * t$95$0), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a 2) < 9.9999999999999999e64

   1. Initial program 52.7%

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

      1. unpow2 52.7%

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

      2. unpow2 52.7%

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

      3. difference-of-squares 52.7%

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

   4. Applied egg-rr 52.7%

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

      1. add-cbrt-cube 52.9%

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

      2. pow3 52.9%

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

   6. Applied egg-rr 55.2%

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

   if 9.9999999999999999e64 < (pow.f64 a 2)

   1. Initial program 49.2%

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

      1. unpow2 49.2%

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

      2. unpow2 49.2%

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

      3. difference-of-squares 55.4%

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

   4. Applied egg-rr 55.4%

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

      1. add-sqr-sqrt 60.2%

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

      2. pow2 60.2%

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

   6. Applied egg-rr 60.2%

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

4. Final simplification 57.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 10^{+65}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(\frac{angle}{180} \cdot \sqrt[3]{{\pi}^{3}}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi}\right)}^{2} \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 57.4% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := angle \cdot \frac{\pi}{-180}\\ t_1 := \sin t\_0 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\\ \mathbf{if}\;{b}^{2} - {a}^{2} \leq -4 \cdot 10^{+302}:\\ \;\;\;\;2 \cdot t\_1\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \left(\cos t\_0 \cdot t\_1\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* angle (/ PI -180.0))) (t_1 (* (sin t_0) (* (+ a b) (- a b)))))
   (if (<= (- (pow b 2.0) (pow a 2.0)) -4e+302)
     (* 2.0 t_1)
     (* 2.0 (* (cos t_0) t_1)))))

C

double code(double a, double b, double angle) {
	double t_0 = angle * (((double) M_PI) / -180.0);
	double t_1 = sin(t_0) * ((a + b) * (a - b));
	double tmp;
	if ((pow(b, 2.0) - pow(a, 2.0)) <= -4e+302) {
		tmp = 2.0 * t_1;
	} else {
		tmp = 2.0 * (cos(t_0) * t_1);
	}
	return tmp;
}

Java

public static double code(double a, double b, double angle) {
	double t_0 = angle * (Math.PI / -180.0);
	double t_1 = Math.sin(t_0) * ((a + b) * (a - b));
	double tmp;
	if ((Math.pow(b, 2.0) - Math.pow(a, 2.0)) <= -4e+302) {
		tmp = 2.0 * t_1;
	} else {
		tmp = 2.0 * (Math.cos(t_0) * t_1);
	}
	return tmp;
}

Python

def code(a, b, angle):
	t_0 = angle * (math.pi / -180.0)
	t_1 = math.sin(t_0) * ((a + b) * (a - b))
	tmp = 0
	if (math.pow(b, 2.0) - math.pow(a, 2.0)) <= -4e+302:
		tmp = 2.0 * t_1
	else:
		tmp = 2.0 * (math.cos(t_0) * t_1)
	return tmp

Julia

function code(a, b, angle)
	t_0 = Float64(angle * Float64(pi / -180.0))
	t_1 = Float64(sin(t_0) * Float64(Float64(a + b) * Float64(a - b)))
	tmp = 0.0
	if (Float64((b ^ 2.0) - (a ^ 2.0)) <= -4e+302)
		tmp = Float64(2.0 * t_1);
	else
		tmp = Float64(2.0 * Float64(cos(t_0) * t_1));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, b, angle)
	t_0 = angle * (pi / -180.0);
	t_1 = sin(t_0) * ((a + b) * (a - b));
	tmp = 0.0;
	if (((b ^ 2.0) - (a ^ 2.0)) <= -4e+302)
		tmp = 2.0 * t_1;
	else
		tmp = 2.0 * (cos(t_0) * t_1);
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, b_, angle_] := Block[{t$95$0 = N[(angle * N[(Pi / -180.0), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[Sin[t$95$0], $MachinePrecision] * N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision], -4e+302], N[(2.0 * t$95$1), $MachinePrecision], N[(2.0 * N[(N[Cos[t$95$0], $MachinePrecision] * t$95$1), $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if (-.f64 (pow.f64 b 2) (pow.f64 a 2)) < -4.0000000000000003e302

   1. Initial program 52.9%

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

   3. Simplified 47.8%

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

   5. Taylor expanded in angle around 0 54.9%

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

      1. unpow2 54.9%

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

      2. unpow2 54.9%

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

      3. difference-of-squares 54.9%

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

   7. Applied egg-rr 54.9%

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

   if -4.0000000000000003e302 < (-.f64 (pow.f64 b 2) (pow.f64 a 2))

   1. Initial program 50.6%

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

   3. Simplified 50.6%

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

      1. unpow2 48.2%

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

      2. unpow2 48.2%

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

      3. difference-of-squares 51.4%

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

   6. Applied egg-rr 54.8%

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

4. Final simplification 54.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq -4 \cdot 10^{+302}:\\ \;\;\;\;2 \cdot \left(\sin \left(angle \cdot \frac{\pi}{-180}\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \left(\cos \left(angle \cdot \frac{\pi}{-180}\right) \cdot \left(\sin \left(angle \cdot \frac{\pi}{-180}\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 57.1% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq -1 \cdot 10^{+267}:\\ \;\;\;\;2 \cdot \left(\sin \left(angle \cdot \frac{\pi}{-180}\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(2 \cdot \left(\left(\left(b - a\right) \cdot \left(a + b\right)\right) \cdot \sin \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (if (<= (- (pow b 2.0) (pow a 2.0)) -1e+267)
   (* 2.0 (* (sin (* angle (/ PI -180.0))) (* (+ a b) (- a b))))
   (*
    (cos (* PI (/ angle 180.0)))
    (*
     2.0
     (* (* (- b a) (+ a b)) (sin (* angle (* PI 0.005555555555555556))))))))

C

double code(double a, double b, double angle) {
	double tmp;
	if ((pow(b, 2.0) - pow(a, 2.0)) <= -1e+267) {
		tmp = 2.0 * (sin((angle * (((double) M_PI) / -180.0))) * ((a + b) * (a - b)));
	} else {
		tmp = cos((((double) M_PI) * (angle / 180.0))) * (2.0 * (((b - a) * (a + b)) * sin((angle * (((double) M_PI) * 0.005555555555555556)))));
	}
	return tmp;
}

Java

public static double code(double a, double b, double angle) {
	double tmp;
	if ((Math.pow(b, 2.0) - Math.pow(a, 2.0)) <= -1e+267) {
		tmp = 2.0 * (Math.sin((angle * (Math.PI / -180.0))) * ((a + b) * (a - b)));
	} else {
		tmp = Math.cos((Math.PI * (angle / 180.0))) * (2.0 * (((b - a) * (a + b)) * Math.sin((angle * (Math.PI * 0.005555555555555556)))));
	}
	return tmp;
}

Python

def code(a, b, angle):
	tmp = 0
	if (math.pow(b, 2.0) - math.pow(a, 2.0)) <= -1e+267:
		tmp = 2.0 * (math.sin((angle * (math.pi / -180.0))) * ((a + b) * (a - b)))
	else:
		tmp = math.cos((math.pi * (angle / 180.0))) * (2.0 * (((b - a) * (a + b)) * math.sin((angle * (math.pi * 0.005555555555555556)))))
	return tmp

Julia

function code(a, b, angle)
	tmp = 0.0
	if (Float64((b ^ 2.0) - (a ^ 2.0)) <= -1e+267)
		tmp = Float64(2.0 * Float64(sin(Float64(angle * Float64(pi / -180.0))) * Float64(Float64(a + b) * Float64(a - b))));
	else
		tmp = Float64(cos(Float64(pi * Float64(angle / 180.0))) * Float64(2.0 * Float64(Float64(Float64(b - a) * Float64(a + b)) * sin(Float64(angle * Float64(pi * 0.005555555555555556))))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, b, angle)
	tmp = 0.0;
	if (((b ^ 2.0) - (a ^ 2.0)) <= -1e+267)
		tmp = 2.0 * (sin((angle * (pi / -180.0))) * ((a + b) * (a - b)));
	else
		tmp = cos((pi * (angle / 180.0))) * (2.0 * (((b - a) * (a + b)) * sin((angle * (pi * 0.005555555555555556)))));
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, b_, angle_] := If[LessEqual[N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision], -1e+267], N[(2.0 * N[(N[Sin[N[(angle * N[(Pi / -180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Cos[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(2.0 * N[(N[(N[(b - a), $MachinePrecision] * N[(a + b), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(angle * N[(Pi * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (-.f64 (pow.f64 b 2) (pow.f64 a 2)) < -9.9999999999999997e266

   1. Initial program 51.3%

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

   3. Simplified 47.5%

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

   5. Taylor expanded in angle around 0 53.5%

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

      1. unpow2 53.5%

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

      2. unpow2 53.5%

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

      3. difference-of-squares 53.5%

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

   7. Applied egg-rr 53.5%

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

   if -9.9999999999999997e266 < (-.f64 (pow.f64 b 2) (pow.f64 a 2))

   1. Initial program 51.1%

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

      1. unpow2 51.1%

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

      2. unpow2 51.1%

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

      3. difference-of-squares 54.9%

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

   4. Applied egg-rr 54.9%

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

      1. add-sqr-sqrt 56.7%

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

      2. pow2 56.7%

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

   6. Applied egg-rr 56.7%

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

      1. *-commutative 56.7%

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

      2. distribute-lft-in 52.4%

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

   8. Applied egg-rr 52.4%

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

   9. Taylor expanded in angle around inf 53.5%

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

       1. associate-*r* 53.2%

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

       2. *-commutative 53.2%

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

       3. associate-*l* 54.0%

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

       4. distribute-rgt-out 56.1%

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

   11. Simplified 56.1%

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

4. Final simplification 55.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq -1 \cdot 10^{+267}:\\ \;\;\;\;2 \cdot \left(\sin \left(angle \cdot \frac{\pi}{-180}\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(2 \cdot \left(\left(\left(b - a\right) \cdot \left(a + b\right)\right) \cdot \sin \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 8: 57.1% accurate, 1.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \left(b - a\right) \cdot \left(a + b\right)\\ \mathbf{if}\;{b}^{2} \leq 5 \cdot 10^{+166}:\\ \;\;\;\;2 \cdot \left(t\_0 \cdot \left(\sin \left(\pi \cdot \frac{angle}{180}\right) \cdot \cos \left(\frac{angle \cdot \pi}{-180}\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \left(0.005555555555555556 \cdot \left(angle \cdot \left(\pi \cdot t\_0\right)\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* (- b a) (+ a b))))
   (if (<= (pow b 2.0) 5e+166)
     (*
      2.0
      (* t_0 (* (sin (* PI (/ angle 180.0))) (cos (/ (* angle PI) -180.0)))))
     (* 2.0 (* 0.005555555555555556 (* angle (* PI t_0)))))))

C

double code(double a, double b, double angle) {
	double t_0 = (b - a) * (a + b);
	double tmp;
	if (pow(b, 2.0) <= 5e+166) {
		tmp = 2.0 * (t_0 * (sin((((double) M_PI) * (angle / 180.0))) * cos(((angle * ((double) M_PI)) / -180.0))));
	} else {
		tmp = 2.0 * (0.005555555555555556 * (angle * (((double) M_PI) * t_0)));
	}
	return tmp;
}

Java

public static double code(double a, double b, double angle) {
	double t_0 = (b - a) * (a + b);
	double tmp;
	if (Math.pow(b, 2.0) <= 5e+166) {
		tmp = 2.0 * (t_0 * (Math.sin((Math.PI * (angle / 180.0))) * Math.cos(((angle * Math.PI) / -180.0))));
	} else {
		tmp = 2.0 * (0.005555555555555556 * (angle * (Math.PI * t_0)));
	}
	return tmp;
}

Python

def code(a, b, angle):
	t_0 = (b - a) * (a + b)
	tmp = 0
	if math.pow(b, 2.0) <= 5e+166:
		tmp = 2.0 * (t_0 * (math.sin((math.pi * (angle / 180.0))) * math.cos(((angle * math.pi) / -180.0))))
	else:
		tmp = 2.0 * (0.005555555555555556 * (angle * (math.pi * t_0)))
	return tmp

Julia

function code(a, b, angle)
	t_0 = Float64(Float64(b - a) * Float64(a + b))
	tmp = 0.0
	if ((b ^ 2.0) <= 5e+166)
		tmp = Float64(2.0 * Float64(t_0 * Float64(sin(Float64(pi * Float64(angle / 180.0))) * cos(Float64(Float64(angle * pi) / -180.0)))));
	else
		tmp = Float64(2.0 * Float64(0.005555555555555556 * Float64(angle * Float64(pi * t_0))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, b, angle)
	t_0 = (b - a) * (a + b);
	tmp = 0.0;
	if ((b ^ 2.0) <= 5e+166)
		tmp = 2.0 * (t_0 * (sin((pi * (angle / 180.0))) * cos(((angle * pi) / -180.0))));
	else
		tmp = 2.0 * (0.005555555555555556 * (angle * (pi * t_0)));
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, b_, angle_] := Block[{t$95$0 = N[(N[(b - a), $MachinePrecision] * N[(a + b), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[b, 2.0], $MachinePrecision], 5e+166], N[(2.0 * N[(t$95$0 * N[(N[Sin[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[Cos[N[(N[(angle * Pi), $MachinePrecision] / -180.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 * N[(0.005555555555555556 * N[(angle * N[(Pi * t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 b 2) < 5.0000000000000002e166

   1. Initial program 56.8%

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

   3. Simplified 57.3%

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

      1. unpow2 56.8%

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

      2. unpow2 56.8%

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

      3. difference-of-squares 56.8%

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

   6. Applied egg-rr 57.3%

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

   if 5.0000000000000002e166 < (pow.f64 b 2)

   1. Initial program 40.0%

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

   3. Simplified 36.6%

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

   5. Taylor expanded in angle around 0 41.5%

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

      1. unpow2 40.0%

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

      2. unpow2 40.0%

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

      3. difference-of-squares 48.4%

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

   7. Applied egg-rr 51.0%

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

4. Final simplification 55.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} \leq 5 \cdot 10^{+166}:\\ \;\;\;\;2 \cdot \left(\left(\left(b - a\right) \cdot \left(a + b\right)\right) \cdot \left(\sin \left(\pi \cdot \frac{angle}{180}\right) \cdot \cos \left(\frac{angle \cdot \pi}{-180}\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \left(0.005555555555555556 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 9: 57.6% accurate, 1.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \pi \cdot \frac{angle}{180}\\ t_1 := 2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\\ \mathbf{if}\;{b}^{2} \leq 10^{+47}:\\ \;\;\;\;\left(t\_1 \cdot \sin t\_0\right) \cdot \cos \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\cos t\_0 \cdot \left(\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right) \cdot t\_1\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* PI (/ angle 180.0))) (t_1 (* 2.0 (* (- b a) (+ a b)))))
   (if (<= (pow b 2.0) 1e+47)
     (* (* t_1 (sin t_0)) (cos (* angle (* PI 0.005555555555555556))))
     (* (cos t_0) (* (sin (* 0.005555555555555556 (* angle PI))) t_1)))))

C

double code(double a, double b, double angle) {
	double t_0 = ((double) M_PI) * (angle / 180.0);
	double t_1 = 2.0 * ((b - a) * (a + b));
	double tmp;
	if (pow(b, 2.0) <= 1e+47) {
		tmp = (t_1 * sin(t_0)) * cos((angle * (((double) M_PI) * 0.005555555555555556)));
	} else {
		tmp = cos(t_0) * (sin((0.005555555555555556 * (angle * ((double) M_PI)))) * t_1);
	}
	return tmp;
}

Java

public static double code(double a, double b, double angle) {
	double t_0 = Math.PI * (angle / 180.0);
	double t_1 = 2.0 * ((b - a) * (a + b));
	double tmp;
	if (Math.pow(b, 2.0) <= 1e+47) {
		tmp = (t_1 * Math.sin(t_0)) * Math.cos((angle * (Math.PI * 0.005555555555555556)));
	} else {
		tmp = Math.cos(t_0) * (Math.sin((0.005555555555555556 * (angle * Math.PI))) * t_1);
	}
	return tmp;
}

Python

def code(a, b, angle):
	t_0 = math.pi * (angle / 180.0)
	t_1 = 2.0 * ((b - a) * (a + b))
	tmp = 0
	if math.pow(b, 2.0) <= 1e+47:
		tmp = (t_1 * math.sin(t_0)) * math.cos((angle * (math.pi * 0.005555555555555556)))
	else:
		tmp = math.cos(t_0) * (math.sin((0.005555555555555556 * (angle * math.pi))) * t_1)
	return tmp

Julia

function code(a, b, angle)
	t_0 = Float64(pi * Float64(angle / 180.0))
	t_1 = Float64(2.0 * Float64(Float64(b - a) * Float64(a + b)))
	tmp = 0.0
	if ((b ^ 2.0) <= 1e+47)
		tmp = Float64(Float64(t_1 * sin(t_0)) * cos(Float64(angle * Float64(pi * 0.005555555555555556))));
	else
		tmp = Float64(cos(t_0) * Float64(sin(Float64(0.005555555555555556 * Float64(angle * pi))) * t_1));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, b, angle)
	t_0 = pi * (angle / 180.0);
	t_1 = 2.0 * ((b - a) * (a + b));
	tmp = 0.0;
	if ((b ^ 2.0) <= 1e+47)
		tmp = (t_1 * sin(t_0)) * cos((angle * (pi * 0.005555555555555556)));
	else
		tmp = cos(t_0) * (sin((0.005555555555555556 * (angle * pi))) * t_1);
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, b_, angle_] := Block[{t$95$0 = N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(a + b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[b, 2.0], $MachinePrecision], 1e+47], N[(N[(t$95$1 * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(angle * N[(Pi * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[Cos[t$95$0], $MachinePrecision] * N[(N[Sin[N[(0.005555555555555556 * N[(angle * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * t$95$1), $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 b 2) < 1e47

   1. Initial program 55.4%

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

      1. unpow2 55.4%

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

      2. unpow2 55.4%

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

      3. difference-of-squares 55.4%

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

   4. Applied egg-rr 55.4%

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

   5. Taylor expanded in angle around inf 56.0%

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

      1. *-commutative 56.0%

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

      2. associate-*l* 56.4%

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

   7. Simplified 56.4%

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

   if 1e47 < (pow.f64 b 2)

   1. Initial program 45.5%

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

      1. unpow2 45.5%

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

      2. unpow2 45.5%

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

      3. difference-of-squares 52.1%

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

   4. Applied egg-rr 52.1%

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

   5. Taylor expanded in angle around inf 55.7%

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

4. Final simplification 56.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} \leq 10^{+47}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right) \cdot \left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 10: 57.6% accurate, 1.9× speedup

Math

\[\begin{array}{l} \\ \cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right) \cdot \left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right)\right) \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (*
  (cos (* PI (/ angle 180.0)))
  (* (sin (* 0.005555555555555556 (* angle PI))) (* 2.0 (* (- b a) (+ a b))))))

C

double code(double a, double b, double angle) {
	return cos((((double) M_PI) * (angle / 180.0))) * (sin((0.005555555555555556 * (angle * ((double) M_PI)))) * (2.0 * ((b - a) * (a + b))));
}

Java

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

Python

def code(a, b, angle):
	return math.cos((math.pi * (angle / 180.0))) * (math.sin((0.005555555555555556 * (angle * math.pi))) * (2.0 * ((b - a) * (a + b))))

Julia

function code(a, b, angle)
	return Float64(cos(Float64(pi * Float64(angle / 180.0))) * Float64(sin(Float64(0.005555555555555556 * Float64(angle * pi))) * Float64(2.0 * Float64(Float64(b - a) * Float64(a + b)))))
end

MATLAB

function tmp = code(a, b, angle)
	tmp = cos((pi * (angle / 180.0))) * (sin((0.005555555555555556 * (angle * pi))) * (2.0 * ((b - a) * (a + b))));
end

Wolfram

code[a_, b_, angle_] := N[(N[Cos[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sin[N[(0.005555555555555556 * N[(angle * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(a + b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 51.1%

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

   1. unpow2 51.1%

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

   2. unpow2 51.1%

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

   3. difference-of-squares 53.9%

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

4. Applied egg-rr 53.9%

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

5. Taylor expanded in angle around inf 54.5%

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

6. Final simplification 54.5%

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

Alternative 11: 56.1% accurate, 3.6× speedup

Math

\[\begin{array}{l} \\ 2 \cdot \left(\sin \left(angle \cdot \frac{\pi}{-180}\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (* 2.0 (* (sin (* angle (/ PI -180.0))) (* (+ a b) (- a b)))))

C

double code(double a, double b, double angle) {
	return 2.0 * (sin((angle * (((double) M_PI) / -180.0))) * ((a + b) * (a - b)));
}

Java

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

Python

def code(a, b, angle):
	return 2.0 * (math.sin((angle * (math.pi / -180.0))) * ((a + b) * (a - b)))

Julia

function code(a, b, angle)
	return Float64(2.0 * Float64(sin(Float64(angle * Float64(pi / -180.0))) * Float64(Float64(a + b) * Float64(a - b))))
end

MATLAB

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

Wolfram

code[a_, b_, angle_] := N[(2.0 * N[(N[Sin[N[(angle * N[(Pi / -180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 51.1%

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

3. Simplified 50.0%

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

5. Taylor expanded in angle around 0 49.8%

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

   1. unpow2 49.8%

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

   2. unpow2 49.8%

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

   3. difference-of-squares 52.2%

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

7. Applied egg-rr 52.2%

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

8. Final simplification 52.2%

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

Alternative 12: 54.6% accurate, 27.9× speedup

Math

\[\begin{array}{l} \\ 2 \cdot \left(0.005555555555555556 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right)\right)\right) \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (* 2.0 (* 0.005555555555555556 (* angle (* PI (* (- b a) (+ a b)))))))

C

double code(double a, double b, double angle) {
	return 2.0 * (0.005555555555555556 * (angle * (((double) M_PI) * ((b - a) * (a + b)))));
}

Java

public static double code(double a, double b, double angle) {
	return 2.0 * (0.005555555555555556 * (angle * (Math.PI * ((b - a) * (a + b)))));
}

Python

def code(a, b, angle):
	return 2.0 * (0.005555555555555556 * (angle * (math.pi * ((b - a) * (a + b)))))

Julia

function code(a, b, angle)
	return Float64(2.0 * Float64(0.005555555555555556 * Float64(angle * Float64(pi * Float64(Float64(b - a) * Float64(a + b))))))
end

MATLAB

function tmp = code(a, b, angle)
	tmp = 2.0 * (0.005555555555555556 * (angle * (pi * ((b - a) * (a + b)))));
end

Wolfram

code[a_, b_, angle_] := N[(2.0 * N[(0.005555555555555556 * N[(angle * N[(Pi * N[(N[(b - a), $MachinePrecision] * N[(a + b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 51.1%

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

3. Simplified 50.3%

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

5. Taylor expanded in angle around 0 48.0%

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

   1. unpow2 51.1%

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

   2. unpow2 51.1%

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

   3. difference-of-squares 53.9%

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

7. Applied egg-rr 51.3%

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

8. Final simplification 51.3%

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

Reproduce

herbie shell --seed 2024040 
(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)))))