ab-angle->ABCF B

Percentage Accurate: 54.2% → 58.1%

Time: 57.0s

Alternatives: 13

Speedup: 5.5×

• 32.3% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Initial Program: 54.2% 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: 58.1% accurate, 1.0× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \sqrt{\pi \cdot angle_m}\\ t_1 := \pi \cdot \left(angle_m \cdot 0.005555555555555556\right)\\ t_2 := 2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\\ t_3 := t_2 \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle_m\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;\frac{angle_m}{180} \leq 2 \cdot 10^{+81}:\\ \;\;\;\;t_3 \cdot \cos t_1\\ \mathbf{elif}\;\frac{angle_m}{180} \leq 2 \cdot 10^{+153}:\\ \;\;\;\;t_2 \cdot \sin \left(\frac{t_0}{\frac{180}{t_0}}\right)\\ \mathbf{elif}\;\frac{angle_m}{180} \leq 2 \cdot 10^{+181}:\\ \;\;\;\;\left(t_2 \cdot \sin \left(\pi \cdot \frac{angle_m}{180}\right)\right) \cdot \mathsf{expm1}\left(\mathsf{log1p}\left(\cos \left(angle_m \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\right)\right)\\ \mathbf{elif}\;\frac{angle_m}{180} \leq 10^{+230}:\\ \;\;\;\;t_3\\ \mathbf{elif}\;\frac{angle_m}{180} \leq 4 \cdot 10^{+237}:\\ \;\;\;\;t_3 \cdot \cos \left(\frac{\pi \cdot angle_m}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;t_2 \cdot \sin \left({\left(\sqrt{t_1}\right)}^{2}\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (sqrt (* PI angle_m)))
        (t_1 (* PI (* angle_m 0.005555555555555556)))
        (t_2 (* 2.0 (* (- b a) (+ a b))))
        (t_3 (* t_2 (sin (* 0.005555555555555556 (* PI angle_m))))))
   (*
    angle_s
    (if (<= (/ angle_m 180.0) 2e+81)
      (* t_3 (cos t_1))
      (if (<= (/ angle_m 180.0) 2e+153)
        (* t_2 (sin (/ t_0 (/ 180.0 t_0))))
        (if (<= (/ angle_m 180.0) 2e+181)
          (*
           (* t_2 (sin (* PI (/ angle_m 180.0))))
           (expm1 (log1p (cos (* angle_m (* PI 0.005555555555555556))))))
          (if (<= (/ angle_m 180.0) 1e+230)
            t_3
            (if (<= (/ angle_m 180.0) 4e+237)
              (* t_3 (cos (/ (* PI angle_m) 180.0)))
              (* t_2 (sin (pow (sqrt t_1) 2.0)))))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = sqrt((((double) M_PI) * angle_m));
	double t_1 = ((double) M_PI) * (angle_m * 0.005555555555555556);
	double t_2 = 2.0 * ((b - a) * (a + b));
	double t_3 = t_2 * sin((0.005555555555555556 * (((double) M_PI) * angle_m)));
	double tmp;
	if ((angle_m / 180.0) <= 2e+81) {
		tmp = t_3 * cos(t_1);
	} else if ((angle_m / 180.0) <= 2e+153) {
		tmp = t_2 * sin((t_0 / (180.0 / t_0)));
	} else if ((angle_m / 180.0) <= 2e+181) {
		tmp = (t_2 * sin((((double) M_PI) * (angle_m / 180.0)))) * expm1(log1p(cos((angle_m * (((double) M_PI) * 0.005555555555555556)))));
	} else if ((angle_m / 180.0) <= 1e+230) {
		tmp = t_3;
	} else if ((angle_m / 180.0) <= 4e+237) {
		tmp = t_3 * cos(((((double) M_PI) * angle_m) / 180.0));
	} else {
		tmp = t_2 * sin(pow(sqrt(t_1), 2.0));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = Math.sqrt((Math.PI * angle_m));
	double t_1 = Math.PI * (angle_m * 0.005555555555555556);
	double t_2 = 2.0 * ((b - a) * (a + b));
	double t_3 = t_2 * Math.sin((0.005555555555555556 * (Math.PI * angle_m)));
	double tmp;
	if ((angle_m / 180.0) <= 2e+81) {
		tmp = t_3 * Math.cos(t_1);
	} else if ((angle_m / 180.0) <= 2e+153) {
		tmp = t_2 * Math.sin((t_0 / (180.0 / t_0)));
	} else if ((angle_m / 180.0) <= 2e+181) {
		tmp = (t_2 * Math.sin((Math.PI * (angle_m / 180.0)))) * Math.expm1(Math.log1p(Math.cos((angle_m * (Math.PI * 0.005555555555555556)))));
	} else if ((angle_m / 180.0) <= 1e+230) {
		tmp = t_3;
	} else if ((angle_m / 180.0) <= 4e+237) {
		tmp = t_3 * Math.cos(((Math.PI * angle_m) / 180.0));
	} else {
		tmp = t_2 * Math.sin(Math.pow(Math.sqrt(t_1), 2.0));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = math.sqrt((math.pi * angle_m))
	t_1 = math.pi * (angle_m * 0.005555555555555556)
	t_2 = 2.0 * ((b - a) * (a + b))
	t_3 = t_2 * math.sin((0.005555555555555556 * (math.pi * angle_m)))
	tmp = 0
	if (angle_m / 180.0) <= 2e+81:
		tmp = t_3 * math.cos(t_1)
	elif (angle_m / 180.0) <= 2e+153:
		tmp = t_2 * math.sin((t_0 / (180.0 / t_0)))
	elif (angle_m / 180.0) <= 2e+181:
		tmp = (t_2 * math.sin((math.pi * (angle_m / 180.0)))) * math.expm1(math.log1p(math.cos((angle_m * (math.pi * 0.005555555555555556)))))
	elif (angle_m / 180.0) <= 1e+230:
		tmp = t_3
	elif (angle_m / 180.0) <= 4e+237:
		tmp = t_3 * math.cos(((math.pi * angle_m) / 180.0))
	else:
		tmp = t_2 * math.sin(math.pow(math.sqrt(t_1), 2.0))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = sqrt(Float64(pi * angle_m))
	t_1 = Float64(pi * Float64(angle_m * 0.005555555555555556))
	t_2 = Float64(2.0 * Float64(Float64(b - a) * Float64(a + b)))
	t_3 = Float64(t_2 * sin(Float64(0.005555555555555556 * Float64(pi * angle_m))))
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 2e+81)
		tmp = Float64(t_3 * cos(t_1));
	elseif (Float64(angle_m / 180.0) <= 2e+153)
		tmp = Float64(t_2 * sin(Float64(t_0 / Float64(180.0 / t_0))));
	elseif (Float64(angle_m / 180.0) <= 2e+181)
		tmp = Float64(Float64(t_2 * sin(Float64(pi * Float64(angle_m / 180.0)))) * expm1(log1p(cos(Float64(angle_m * Float64(pi * 0.005555555555555556))))));
	elseif (Float64(angle_m / 180.0) <= 1e+230)
		tmp = t_3;
	elseif (Float64(angle_m / 180.0) <= 4e+237)
		tmp = Float64(t_3 * cos(Float64(Float64(pi * angle_m) / 180.0)));
	else
		tmp = Float64(t_2 * sin((sqrt(t_1) ^ 2.0)));
	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[Sqrt[N[(Pi * angle$95$m), $MachinePrecision]], $MachinePrecision]}, Block[{t$95$1 = N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(a + b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$3 = N[(t$95$2 * N[Sin[N[(0.005555555555555556 * N[(Pi * angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 2e+81], N[(t$95$3 * N[Cos[t$95$1], $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 2e+153], N[(t$95$2 * N[Sin[N[(t$95$0 / N[(180.0 / t$95$0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 2e+181], N[(N[(t$95$2 * N[Sin[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[(Exp[N[Log[1 + N[Cos[N[(angle$95$m * N[(Pi * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 1e+230], t$95$3, If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 4e+237], N[(t$95$3 * N[Cos[N[(N[(Pi * angle$95$m), $MachinePrecision] / 180.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(t$95$2 * N[Sin[N[Power[N[Sqrt[t$95$1], $MachinePrecision], 2.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]]]]), $MachinePrecision]]]]]

Derivation

1. Split input into 6 regimes

2. if (/.f64 angle 180) < 1.99999999999999984e81

   1. Initial program 60.6%

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

      1. unpow2 60.6%

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

      2. unpow2 60.6%

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

      3. difference-of-squares 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 64.9%

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

   5. Taylor expanded in angle around inf 63.8%

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

      1. *-commutative 63.8%

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

      2. *-commutative 63.8%

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

      3. associate-*r* 62.9%

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

   7. Simplified 62.9%

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

   if 1.99999999999999984e81 < (/.f64 angle 180) < 2e153

   1. Initial program 42.6%

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

      1. unpow2 42.6%

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

      2. unpow2 42.6%

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

      3. difference-of-squares 42.6%

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

   3. Applied egg-rr 42.6%

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

   4. Taylor expanded in angle around 0 46.8%

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

      1. associate-*r/ 36.8%

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

      2. *-commutative 36.8%

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

      3. add-sqr-sqrt 23.9%

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

      4. associate-/l* 53.9%

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

      5. *-commutative 53.9%

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

      6. *-commutative 53.9%

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

   6. Applied egg-rr 53.9%

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

   if 2e153 < (/.f64 angle 180) < 1.9999999999999998e181

   1. Initial program 40.0%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. 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 40.0%

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

   3. Applied egg-rr 40.0%

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

      1. associate-*r/ 52.4%

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

   5. Applied egg-rr 52.4%

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

      1. div-inv 67.0%

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

      2. metadata-eval 67.0%

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

      3. associate-*r* 54.5%

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

      4. rem-cbrt-cube 0.0%

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

      5. unpow1/3 0.0%

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

      6. expm1-log1p-u 0.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\cos \left({\left({\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{3}\right)}^{0.3333333333333333}\right)\right)\right)} \]

      7. unpow1/3 0.0%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \mathsf{expm1}\left(\mathsf{log1p}\left(\cos \color{blue}{\left(\sqrt[3]{{\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{3}}\right)}\right)\right) \]

      8. rem-cbrt-cube 54.5%

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

      9. *-commutative 54.5%

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

      10. associate-*l* 67.0%

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

   7. Applied egg-rr 67.0%

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

   if 1.9999999999999998e181 < (/.f64 angle 180) < 1.0000000000000001e230

   1. Initial program 30.1%

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

      1. unpow2 30.1%

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

      2. unpow2 30.1%

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

      3. difference-of-squares 30.1%

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

   3. Applied egg-rr 30.1%

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

   4. Taylor expanded in angle around 0 40.1%

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

   5. Taylor expanded in angle around 0 81.9%

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

   if 1.0000000000000001e230 < (/.f64 angle 180) < 3.99999999999999976e237

   1. Initial program 50.0%

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

      1. unpow2 50.0%

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

      2. unpow2 50.0%

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

      3. difference-of-squares 50.0%

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

   3. Applied egg-rr 50.0%

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

   4. Taylor expanded in angle around 0 50.0%

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

      1. associate-*r/ 100.0%

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

   6. Applied egg-rr 100.0%

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

   if 3.99999999999999976e237 < (/.f64 angle 180)

   1. Initial program 24.3%

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

      1. unpow2 24.3%

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

      2. unpow2 24.3%

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

      3. difference-of-squares 24.3%

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

   3. Applied egg-rr 24.3%

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

   4. Taylor expanded in angle around 0 49.5%

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

      1. div-inv 44.3%

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

      2. metadata-eval 44.3%

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

      3. add-sqr-sqrt 65.3%

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

      4. pow2 65.3%

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

   6. Applied egg-rr 65.3%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left({\left(\sqrt{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right)}^{2}\right)}\right) \cdot 1 \]
3. Recombined 6 regimes into one program.

4. Final simplification 63.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 2 \cdot 10^{+81}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \mathbf{elif}\;\frac{angle}{180} \leq 2 \cdot 10^{+153}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(\frac{\sqrt{\pi \cdot angle}}{\frac{180}{\sqrt{\pi \cdot angle}}}\right)\\ \mathbf{elif}\;\frac{angle}{180} \leq 2 \cdot 10^{+181}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \mathsf{expm1}\left(\mathsf{log1p}\left(\cos \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\right)\right)\\ \mathbf{elif}\;\frac{angle}{180} \leq 10^{+230}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\\ \mathbf{elif}\;\frac{angle}{180} \leq 4 \cdot 10^{+237}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right) \cdot \cos \left(\frac{\pi \cdot angle}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right)}^{2}\right)\\ \end{array} \]

Alternative 2: 57.3% accurate, 0.9× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{-6}:\\ \;\;\;\;\left(t_0 \cdot \sin \left(\sqrt[3]{{\pi}^{3}} \cdot \frac{angle_m}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle_m}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;t_0 \cdot \sin \left({\left(\sqrt{\pi \cdot \left(angle_m \cdot 0.005555555555555556\right)}\right)}^{2}\right)\\ \end{array} \end{array} \end{array} \]

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) (+ a b)))))
   (*
    angle_s
    (if (<= (pow a 2.0) 5e-6)
      (*
       (* t_0 (sin (* (cbrt (pow PI 3.0)) (/ angle_m 180.0))))
       (cos (* PI (/ angle_m 180.0))))
      (*
       t_0
       (sin (pow (sqrt (* PI (* angle_m 0.005555555555555556))) 2.0)))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b - a) * (a + b));
	double tmp;
	if (pow(a, 2.0) <= 5e-6) {
		tmp = (t_0 * sin((cbrt(pow(((double) M_PI), 3.0)) * (angle_m / 180.0)))) * cos((((double) M_PI) * (angle_m / 180.0)));
	} else {
		tmp = t_0 * sin(pow(sqrt((((double) M_PI) * (angle_m * 0.005555555555555556))), 2.0));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b - a) * (a + b));
	double tmp;
	if (Math.pow(a, 2.0) <= 5e-6) {
		tmp = (t_0 * Math.sin((Math.cbrt(Math.pow(Math.PI, 3.0)) * (angle_m / 180.0)))) * Math.cos((Math.PI * (angle_m / 180.0)));
	} else {
		tmp = t_0 * Math.sin(Math.pow(Math.sqrt((Math.PI * (angle_m * 0.005555555555555556))), 2.0));
	}
	return angle_s * tmp;
}

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(2.0 * Float64(Float64(b - a) * Float64(a + b)))
	tmp = 0.0
	if ((a ^ 2.0) <= 5e-6)
		tmp = Float64(Float64(t_0 * sin(Float64(cbrt((pi ^ 3.0)) * Float64(angle_m / 180.0)))) * cos(Float64(pi * Float64(angle_m / 180.0))));
	else
		tmp = Float64(t_0 * sin((sqrt(Float64(pi * Float64(angle_m * 0.005555555555555556))) ^ 2.0)));
	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[(a + b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 5e-6], N[(N[(t$95$0 * N[Sin[N[(N[Power[N[Power[Pi, 3.0], $MachinePrecision], 1/3], $MachinePrecision] * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(t$95$0 * N[Sin[N[Power[N[Sqrt[N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision], 2.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a 2) < 5.00000000000000041e-6

   1. Initial program 60.0%

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

      1. unpow2 60.0%

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

      2. unpow2 60.0%

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

      3. difference-of-squares 60.0%

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

   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. Step-by-step derivation

      1. add-cbrt-cube 63.8%

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

      2. pow3 63.8%

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

   5. Applied egg-rr 63.8%

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

   if 5.00000000000000041e-6 < (pow.f64 a 2)

   1. Initial program 51.4%

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

      1. unpow2 51.4%

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

      2. unpow2 51.4%

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

      3. difference-of-squares 57.9%

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

   3. Applied egg-rr 57.9%

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

   4. Taylor expanded in angle around 0 59.2%

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

      1. div-inv 60.9%

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

      2. metadata-eval 60.9%

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

      3. add-sqr-sqrt 28.8%

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

      4. pow2 28.8%

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

   6. Applied egg-rr 28.8%

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

4. Final simplification 44.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{-6}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(\sqrt[3]{{\pi}^{3}} \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right)}^{2}\right)\\ \end{array} \]

Alternative 3: 57.5% accurate, 0.9× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{-6}:\\ \;\;\;\;\left(t_0 \cdot \sin \left(\pi \cdot \frac{angle_m}{180}\right)\right) \cdot \cos \left(\frac{angle_m}{180} \cdot {\left(\sqrt{\pi}\right)}^{2}\right)\\ \mathbf{else}:\\ \;\;\;\;t_0 \cdot \sin \left({\left(\sqrt{\pi \cdot \left(angle_m \cdot 0.005555555555555556\right)}\right)}^{2}\right)\\ \end{array} \end{array} \end{array} \]

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) (+ a b)))))
   (*
    angle_s
    (if (<= (pow a 2.0) 5e-6)
      (*
       (* t_0 (sin (* PI (/ angle_m 180.0))))
       (cos (* (/ angle_m 180.0) (pow (sqrt PI) 2.0))))
      (*
       t_0
       (sin (pow (sqrt (* PI (* angle_m 0.005555555555555556))) 2.0)))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b - a) * (a + b));
	double tmp;
	if (pow(a, 2.0) <= 5e-6) {
		tmp = (t_0 * sin((((double) M_PI) * (angle_m / 180.0)))) * cos(((angle_m / 180.0) * pow(sqrt(((double) M_PI)), 2.0)));
	} else {
		tmp = t_0 * sin(pow(sqrt((((double) M_PI) * (angle_m * 0.005555555555555556))), 2.0));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b - a) * (a + b));
	double tmp;
	if (Math.pow(a, 2.0) <= 5e-6) {
		tmp = (t_0 * Math.sin((Math.PI * (angle_m / 180.0)))) * Math.cos(((angle_m / 180.0) * Math.pow(Math.sqrt(Math.PI), 2.0)));
	} else {
		tmp = t_0 * Math.sin(Math.pow(Math.sqrt((Math.PI * (angle_m * 0.005555555555555556))), 2.0));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = 2.0 * ((b - a) * (a + b))
	tmp = 0
	if math.pow(a, 2.0) <= 5e-6:
		tmp = (t_0 * math.sin((math.pi * (angle_m / 180.0)))) * math.cos(((angle_m / 180.0) * math.pow(math.sqrt(math.pi), 2.0)))
	else:
		tmp = t_0 * math.sin(math.pow(math.sqrt((math.pi * (angle_m * 0.005555555555555556))), 2.0))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(2.0 * Float64(Float64(b - a) * Float64(a + b)))
	tmp = 0.0
	if ((a ^ 2.0) <= 5e-6)
		tmp = Float64(Float64(t_0 * sin(Float64(pi * Float64(angle_m / 180.0)))) * cos(Float64(Float64(angle_m / 180.0) * (sqrt(pi) ^ 2.0))));
	else
		tmp = Float64(t_0 * sin((sqrt(Float64(pi * Float64(angle_m * 0.005555555555555556))) ^ 2.0)));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = 2.0 * ((b - a) * (a + b));
	tmp = 0.0;
	if ((a ^ 2.0) <= 5e-6)
		tmp = (t_0 * sin((pi * (angle_m / 180.0)))) * cos(((angle_m / 180.0) * (sqrt(pi) ^ 2.0)));
	else
		tmp = t_0 * sin((sqrt((pi * (angle_m * 0.005555555555555556))) ^ 2.0));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(a + b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 5e-6], N[(N[(t$95$0 * N[Sin[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(N[(angle$95$m / 180.0), $MachinePrecision] * N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(t$95$0 * N[Sin[N[Power[N[Sqrt[N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision], 2.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a 2) < 5.00000000000000041e-6

   1. Initial program 60.0%

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

      1. unpow2 60.0%

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

      2. unpow2 60.0%

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

      3. difference-of-squares 60.0%

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

   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. Step-by-step derivation

      1. add-sqr-sqrt 63.5%

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

      2. pow2 63.5%

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

   5. Applied egg-rr 63.5%

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

   if 5.00000000000000041e-6 < (pow.f64 a 2)

   1. Initial program 51.4%

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

      1. unpow2 51.4%

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

      2. unpow2 51.4%

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

      3. difference-of-squares 57.9%

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

   3. Applied egg-rr 57.9%

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

   4. Taylor expanded in angle around 0 59.2%

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

      1. div-inv 60.9%

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

      2. metadata-eval 60.9%

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

      3. add-sqr-sqrt 28.8%

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

      4. pow2 28.8%

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

   6. Applied egg-rr 28.8%

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

4. Final simplification 44.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 5 \cdot 10^{-6}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\frac{angle}{180} \cdot {\left(\sqrt{\pi}\right)}^{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right)}^{2}\right)\\ \end{array} \]

Alternative 4: 57.1% accurate, 1.0× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle_m\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq 5 \cdot 10^{+281}:\\ \;\;\;\;t_0 \cdot \cos \left(\frac{\pi}{\frac{180}{angle_m}}\right)\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \end{array} \end{array} \]

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) (+ a b)))
          (sin (* 0.005555555555555556 (* PI angle_m))))))
   (*
    angle_s
    (if (<= (- (pow b 2.0) (pow a 2.0)) 5e+281)
      (* t_0 (cos (/ PI (/ 180.0 angle_m))))
      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 = (2.0 * ((b - a) * (a + b))) * sin((0.005555555555555556 * (((double) M_PI) * angle_m)));
	double tmp;
	if ((pow(b, 2.0) - pow(a, 2.0)) <= 5e+281) {
		tmp = t_0 * cos((((double) M_PI) / (180.0 / angle_m)));
	} else {
		tmp = t_0;
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (2.0 * ((b - a) * (a + b))) * Math.sin((0.005555555555555556 * (Math.PI * angle_m)));
	double tmp;
	if ((Math.pow(b, 2.0) - Math.pow(a, 2.0)) <= 5e+281) {
		tmp = t_0 * Math.cos((Math.PI / (180.0 / angle_m)));
	} else {
		tmp = t_0;
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = (2.0 * ((b - a) * (a + b))) * math.sin((0.005555555555555556 * (math.pi * angle_m)))
	tmp = 0
	if (math.pow(b, 2.0) - math.pow(a, 2.0)) <= 5e+281:
		tmp = t_0 * math.cos((math.pi / (180.0 / angle_m)))
	else:
		tmp = t_0
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(Float64(2.0 * Float64(Float64(b - a) * Float64(a + b))) * sin(Float64(0.005555555555555556 * Float64(pi * angle_m))))
	tmp = 0.0
	if (Float64((b ^ 2.0) - (a ^ 2.0)) <= 5e+281)
		tmp = Float64(t_0 * cos(Float64(pi / Float64(180.0 / angle_m))));
	else
		tmp = t_0;
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = (2.0 * ((b - a) * (a + b))) * sin((0.005555555555555556 * (pi * angle_m)));
	tmp = 0.0;
	if (((b ^ 2.0) - (a ^ 2.0)) <= 5e+281)
		tmp = t_0 * cos((pi / (180.0 / angle_m)));
	else
		tmp = t_0;
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(N[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(a + b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(0.005555555555555556 * N[(Pi * angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision], 5e+281], N[(t$95$0 * N[Cos[N[(Pi / N[(180.0 / angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], t$95$0]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (-.f64 (pow.f64 b 2) (pow.f64 a 2)) < 5.00000000000000016e281

   1. Initial program 58.6%

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

      1. unpow2 58.6%

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

      2. unpow2 58.6%

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

      3. difference-of-squares 58.6%

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

   3. Applied egg-rr 58.6%

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

   4. Taylor expanded in angle around 0 60.9%

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

      1. clear-num 59.9%

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

      2. un-div-inv 61.0%

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

   6. Applied egg-rr 61.0%

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

   if 5.00000000000000016e281 < (-.f64 (pow.f64 b 2) (pow.f64 a 2))

   1. Initial program 47.9%

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

      1. unpow2 47.9%

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

      2. unpow2 47.9%

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

      3. difference-of-squares 59.5%

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

   3. Applied egg-rr 59.5%

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

   4. Taylor expanded in angle around 0 60.8%

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

   5. Taylor expanded in angle around 0 65.8%

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

4. Final simplification 62.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq 5 \cdot 10^{+281}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right) \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\\ \end{array} \]

Alternative 5: 58.1% accurate, 1.2× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \sqrt{\pi \cdot angle_m}\\ t_1 := 0.005555555555555556 \cdot \left(\pi \cdot angle_m\right)\\ t_2 := \pi \cdot \left(angle_m \cdot 0.005555555555555556\right)\\ t_3 := 2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\\ t_4 := t_3 \cdot \sin t_1\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;\frac{angle_m}{180} \leq 2 \cdot 10^{+81}:\\ \;\;\;\;t_4 \cdot \cos t_2\\ \mathbf{elif}\;\frac{angle_m}{180} \leq 2 \cdot 10^{+153}:\\ \;\;\;\;t_3 \cdot \sin \left(\frac{t_0}{\frac{180}{t_0}}\right)\\ \mathbf{elif}\;\frac{angle_m}{180} \leq 2 \cdot 10^{+181}:\\ \;\;\;\;\left(t_3 \cdot \sin \left(\pi \cdot \frac{angle_m}{180}\right)\right) \cdot \cos t_1\\ \mathbf{elif}\;\frac{angle_m}{180} \leq 10^{+230}:\\ \;\;\;\;t_4\\ \mathbf{elif}\;\frac{angle_m}{180} \leq 4 \cdot 10^{+237}:\\ \;\;\;\;t_4 \cdot \cos \left(\frac{\pi \cdot angle_m}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;t_3 \cdot \sin \left({\left(\sqrt{t_2}\right)}^{2}\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (sqrt (* PI angle_m)))
        (t_1 (* 0.005555555555555556 (* PI angle_m)))
        (t_2 (* PI (* angle_m 0.005555555555555556)))
        (t_3 (* 2.0 (* (- b a) (+ a b))))
        (t_4 (* t_3 (sin t_1))))
   (*
    angle_s
    (if (<= (/ angle_m 180.0) 2e+81)
      (* t_4 (cos t_2))
      (if (<= (/ angle_m 180.0) 2e+153)
        (* t_3 (sin (/ t_0 (/ 180.0 t_0))))
        (if (<= (/ angle_m 180.0) 2e+181)
          (* (* t_3 (sin (* PI (/ angle_m 180.0)))) (cos t_1))
          (if (<= (/ angle_m 180.0) 1e+230)
            t_4
            (if (<= (/ angle_m 180.0) 4e+237)
              (* t_4 (cos (/ (* PI angle_m) 180.0)))
              (* t_3 (sin (pow (sqrt t_2) 2.0)))))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = sqrt((((double) M_PI) * angle_m));
	double t_1 = 0.005555555555555556 * (((double) M_PI) * angle_m);
	double t_2 = ((double) M_PI) * (angle_m * 0.005555555555555556);
	double t_3 = 2.0 * ((b - a) * (a + b));
	double t_4 = t_3 * sin(t_1);
	double tmp;
	if ((angle_m / 180.0) <= 2e+81) {
		tmp = t_4 * cos(t_2);
	} else if ((angle_m / 180.0) <= 2e+153) {
		tmp = t_3 * sin((t_0 / (180.0 / t_0)));
	} else if ((angle_m / 180.0) <= 2e+181) {
		tmp = (t_3 * sin((((double) M_PI) * (angle_m / 180.0)))) * cos(t_1);
	} else if ((angle_m / 180.0) <= 1e+230) {
		tmp = t_4;
	} else if ((angle_m / 180.0) <= 4e+237) {
		tmp = t_4 * cos(((((double) M_PI) * angle_m) / 180.0));
	} else {
		tmp = t_3 * sin(pow(sqrt(t_2), 2.0));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = Math.sqrt((Math.PI * angle_m));
	double t_1 = 0.005555555555555556 * (Math.PI * angle_m);
	double t_2 = Math.PI * (angle_m * 0.005555555555555556);
	double t_3 = 2.0 * ((b - a) * (a + b));
	double t_4 = t_3 * Math.sin(t_1);
	double tmp;
	if ((angle_m / 180.0) <= 2e+81) {
		tmp = t_4 * Math.cos(t_2);
	} else if ((angle_m / 180.0) <= 2e+153) {
		tmp = t_3 * Math.sin((t_0 / (180.0 / t_0)));
	} else if ((angle_m / 180.0) <= 2e+181) {
		tmp = (t_3 * Math.sin((Math.PI * (angle_m / 180.0)))) * Math.cos(t_1);
	} else if ((angle_m / 180.0) <= 1e+230) {
		tmp = t_4;
	} else if ((angle_m / 180.0) <= 4e+237) {
		tmp = t_4 * Math.cos(((Math.PI * angle_m) / 180.0));
	} else {
		tmp = t_3 * Math.sin(Math.pow(Math.sqrt(t_2), 2.0));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = math.sqrt((math.pi * angle_m))
	t_1 = 0.005555555555555556 * (math.pi * angle_m)
	t_2 = math.pi * (angle_m * 0.005555555555555556)
	t_3 = 2.0 * ((b - a) * (a + b))
	t_4 = t_3 * math.sin(t_1)
	tmp = 0
	if (angle_m / 180.0) <= 2e+81:
		tmp = t_4 * math.cos(t_2)
	elif (angle_m / 180.0) <= 2e+153:
		tmp = t_3 * math.sin((t_0 / (180.0 / t_0)))
	elif (angle_m / 180.0) <= 2e+181:
		tmp = (t_3 * math.sin((math.pi * (angle_m / 180.0)))) * math.cos(t_1)
	elif (angle_m / 180.0) <= 1e+230:
		tmp = t_4
	elif (angle_m / 180.0) <= 4e+237:
		tmp = t_4 * math.cos(((math.pi * angle_m) / 180.0))
	else:
		tmp = t_3 * math.sin(math.pow(math.sqrt(t_2), 2.0))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = sqrt(Float64(pi * angle_m))
	t_1 = Float64(0.005555555555555556 * Float64(pi * angle_m))
	t_2 = Float64(pi * Float64(angle_m * 0.005555555555555556))
	t_3 = Float64(2.0 * Float64(Float64(b - a) * Float64(a + b)))
	t_4 = Float64(t_3 * sin(t_1))
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 2e+81)
		tmp = Float64(t_4 * cos(t_2));
	elseif (Float64(angle_m / 180.0) <= 2e+153)
		tmp = Float64(t_3 * sin(Float64(t_0 / Float64(180.0 / t_0))));
	elseif (Float64(angle_m / 180.0) <= 2e+181)
		tmp = Float64(Float64(t_3 * sin(Float64(pi * Float64(angle_m / 180.0)))) * cos(t_1));
	elseif (Float64(angle_m / 180.0) <= 1e+230)
		tmp = t_4;
	elseif (Float64(angle_m / 180.0) <= 4e+237)
		tmp = Float64(t_4 * cos(Float64(Float64(pi * angle_m) / 180.0)));
	else
		tmp = Float64(t_3 * sin((sqrt(t_2) ^ 2.0)));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = sqrt((pi * angle_m));
	t_1 = 0.005555555555555556 * (pi * angle_m);
	t_2 = pi * (angle_m * 0.005555555555555556);
	t_3 = 2.0 * ((b - a) * (a + b));
	t_4 = t_3 * sin(t_1);
	tmp = 0.0;
	if ((angle_m / 180.0) <= 2e+81)
		tmp = t_4 * cos(t_2);
	elseif ((angle_m / 180.0) <= 2e+153)
		tmp = t_3 * sin((t_0 / (180.0 / t_0)));
	elseif ((angle_m / 180.0) <= 2e+181)
		tmp = (t_3 * sin((pi * (angle_m / 180.0)))) * cos(t_1);
	elseif ((angle_m / 180.0) <= 1e+230)
		tmp = t_4;
	elseif ((angle_m / 180.0) <= 4e+237)
		tmp = t_4 * cos(((pi * angle_m) / 180.0));
	else
		tmp = t_3 * sin((sqrt(t_2) ^ 2.0));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[Sqrt[N[(Pi * angle$95$m), $MachinePrecision]], $MachinePrecision]}, Block[{t$95$1 = N[(0.005555555555555556 * N[(Pi * angle$95$m), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$3 = N[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(a + b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$4 = N[(t$95$3 * N[Sin[t$95$1], $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 2e+81], N[(t$95$4 * N[Cos[t$95$2], $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 2e+153], N[(t$95$3 * N[Sin[N[(t$95$0 / N[(180.0 / t$95$0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 2e+181], N[(N[(t$95$3 * N[Sin[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$1], $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 1e+230], t$95$4, If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 4e+237], N[(t$95$4 * N[Cos[N[(N[(Pi * angle$95$m), $MachinePrecision] / 180.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(t$95$3 * N[Sin[N[Power[N[Sqrt[t$95$2], $MachinePrecision], 2.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]]]]), $MachinePrecision]]]]]]

Derivation

1. Split input into 6 regimes

2. if (/.f64 angle 180) < 1.99999999999999984e81

   1. Initial program 60.6%

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

      1. unpow2 60.6%

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

      2. unpow2 60.6%

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

      3. difference-of-squares 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 64.9%

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

   5. Taylor expanded in angle around inf 63.8%

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

      1. *-commutative 63.8%

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

      2. *-commutative 63.8%

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

      3. associate-*r* 62.9%

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

   7. Simplified 62.9%

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

   if 1.99999999999999984e81 < (/.f64 angle 180) < 2e153

   1. Initial program 42.6%

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

      1. unpow2 42.6%

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

      2. unpow2 42.6%

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

      3. difference-of-squares 42.6%

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

   3. Applied egg-rr 42.6%

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

   4. Taylor expanded in angle around 0 46.8%

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

      1. associate-*r/ 36.8%

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

      2. *-commutative 36.8%

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

      3. add-sqr-sqrt 23.9%

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

      4. associate-/l* 53.9%

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

      5. *-commutative 53.9%

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

      6. *-commutative 53.9%

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

   6. Applied egg-rr 53.9%

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

   if 2e153 < (/.f64 angle 180) < 1.9999999999999998e181

   1. Initial program 40.0%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. 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 40.0%

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

   3. Applied egg-rr 40.0%

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

   4. Taylor expanded in angle around 0 67.0%

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

   if 1.9999999999999998e181 < (/.f64 angle 180) < 1.0000000000000001e230

   1. Initial program 30.1%

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

      1. unpow2 30.1%

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

      2. unpow2 30.1%

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

      3. difference-of-squares 30.1%

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

   3. Applied egg-rr 30.1%

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

   4. Taylor expanded in angle around 0 40.1%

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

   5. Taylor expanded in angle around 0 81.9%

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

   if 1.0000000000000001e230 < (/.f64 angle 180) < 3.99999999999999976e237

   1. Initial program 50.0%

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

      1. unpow2 50.0%

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

      2. unpow2 50.0%

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

      3. difference-of-squares 50.0%

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

   3. Applied egg-rr 50.0%

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

   4. Taylor expanded in angle around 0 50.0%

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

      1. associate-*r/ 100.0%

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

   6. Applied egg-rr 100.0%

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

   if 3.99999999999999976e237 < (/.f64 angle 180)

   1. Initial program 24.3%

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

      1. unpow2 24.3%

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

      2. unpow2 24.3%

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

      3. difference-of-squares 24.3%

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

   3. Applied egg-rr 24.3%

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

   4. Taylor expanded in angle around 0 49.5%

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

      1. div-inv 44.3%

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

      2. metadata-eval 44.3%

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

      3. add-sqr-sqrt 65.3%

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

      4. pow2 65.3%

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

   6. Applied egg-rr 65.3%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \color{blue}{\left({\left(\sqrt{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right)}^{2}\right)}\right) \cdot 1 \]
3. Recombined 6 regimes into one program.

4. Final simplification 63.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 2 \cdot 10^{+81}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \mathbf{elif}\;\frac{angle}{180} \leq 2 \cdot 10^{+153}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(\frac{\sqrt{\pi \cdot angle}}{\frac{180}{\sqrt{\pi \cdot angle}}}\right)\\ \mathbf{elif}\;\frac{angle}{180} \leq 2 \cdot 10^{+181}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\\ \mathbf{elif}\;\frac{angle}{180} \leq 10^{+230}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\\ \mathbf{elif}\;\frac{angle}{180} \leq 4 \cdot 10^{+237}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right) \cdot \cos \left(\frac{\pi \cdot angle}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right)}^{2}\right)\\ \end{array} \]

Alternative 6: 57.6% accurate, 1.5× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \left(b - a\right) \cdot \left(a + b\right)\\ t_1 := \pi \cdot \left(angle_m \cdot 0.005555555555555556\right)\\ t_2 := 2 \cdot t_0\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;\frac{angle_m}{180} \leq 10^{+80}:\\ \;\;\;\;\left(t_2 \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle_m\right)\right)\right) \cdot \cos t_1\\ \mathbf{elif}\;\frac{angle_m}{180} \leq 10^{+203}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle_m}{180}\right) \cdot \left(0.011111111111111112 \cdot \left(angle_m \cdot \left(\pi \cdot t_0\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;t_2 \cdot \sin \left({\left(\sqrt{t_1}\right)}^{2}\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* (- b a) (+ a b)))
        (t_1 (* PI (* angle_m 0.005555555555555556)))
        (t_2 (* 2.0 t_0)))
   (*
    angle_s
    (if (<= (/ angle_m 180.0) 1e+80)
      (* (* t_2 (sin (* 0.005555555555555556 (* PI angle_m)))) (cos t_1))
      (if (<= (/ angle_m 180.0) 1e+203)
        (*
         (cos (* PI (/ angle_m 180.0)))
         (* 0.011111111111111112 (* angle_m (* PI t_0))))
        (* t_2 (sin (pow (sqrt t_1) 2.0))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (b - a) * (a + b);
	double t_1 = ((double) M_PI) * (angle_m * 0.005555555555555556);
	double t_2 = 2.0 * t_0;
	double tmp;
	if ((angle_m / 180.0) <= 1e+80) {
		tmp = (t_2 * sin((0.005555555555555556 * (((double) M_PI) * angle_m)))) * cos(t_1);
	} else if ((angle_m / 180.0) <= 1e+203) {
		tmp = cos((((double) M_PI) * (angle_m / 180.0))) * (0.011111111111111112 * (angle_m * (((double) M_PI) * t_0)));
	} else {
		tmp = t_2 * sin(pow(sqrt(t_1), 2.0));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (b - a) * (a + b);
	double t_1 = Math.PI * (angle_m * 0.005555555555555556);
	double t_2 = 2.0 * t_0;
	double tmp;
	if ((angle_m / 180.0) <= 1e+80) {
		tmp = (t_2 * Math.sin((0.005555555555555556 * (Math.PI * angle_m)))) * Math.cos(t_1);
	} else if ((angle_m / 180.0) <= 1e+203) {
		tmp = Math.cos((Math.PI * (angle_m / 180.0))) * (0.011111111111111112 * (angle_m * (Math.PI * t_0)));
	} else {
		tmp = t_2 * Math.sin(Math.pow(Math.sqrt(t_1), 2.0));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = (b - a) * (a + b)
	t_1 = math.pi * (angle_m * 0.005555555555555556)
	t_2 = 2.0 * t_0
	tmp = 0
	if (angle_m / 180.0) <= 1e+80:
		tmp = (t_2 * math.sin((0.005555555555555556 * (math.pi * angle_m)))) * math.cos(t_1)
	elif (angle_m / 180.0) <= 1e+203:
		tmp = math.cos((math.pi * (angle_m / 180.0))) * (0.011111111111111112 * (angle_m * (math.pi * t_0)))
	else:
		tmp = t_2 * math.sin(math.pow(math.sqrt(t_1), 2.0))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(Float64(b - a) * Float64(a + b))
	t_1 = Float64(pi * Float64(angle_m * 0.005555555555555556))
	t_2 = Float64(2.0 * t_0)
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 1e+80)
		tmp = Float64(Float64(t_2 * sin(Float64(0.005555555555555556 * Float64(pi * angle_m)))) * cos(t_1));
	elseif (Float64(angle_m / 180.0) <= 1e+203)
		tmp = Float64(cos(Float64(pi * Float64(angle_m / 180.0))) * Float64(0.011111111111111112 * Float64(angle_m * Float64(pi * t_0))));
	else
		tmp = Float64(t_2 * sin((sqrt(t_1) ^ 2.0)));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = (b - a) * (a + b);
	t_1 = pi * (angle_m * 0.005555555555555556);
	t_2 = 2.0 * t_0;
	tmp = 0.0;
	if ((angle_m / 180.0) <= 1e+80)
		tmp = (t_2 * sin((0.005555555555555556 * (pi * angle_m)))) * cos(t_1);
	elseif ((angle_m / 180.0) <= 1e+203)
		tmp = cos((pi * (angle_m / 180.0))) * (0.011111111111111112 * (angle_m * (pi * t_0)));
	else
		tmp = t_2 * sin((sqrt(t_1) ^ 2.0));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(N[(b - a), $MachinePrecision] * N[(a + b), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(2.0 * t$95$0), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 1e+80], N[(N[(t$95$2 * N[Sin[N[(0.005555555555555556 * N[(Pi * angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$1], $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 1e+203], N[(N[Cos[N[(Pi * N[(angle$95$m / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(0.011111111111111112 * N[(angle$95$m * N[(Pi * t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(t$95$2 * N[Sin[N[Power[N[Sqrt[t$95$1], $MachinePrecision], 2.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 angle 180) < 1e80

   1. Initial program 60.6%

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

      1. unpow2 60.6%

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

      2. unpow2 60.6%

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

      3. difference-of-squares 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 64.9%

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

   5. Taylor expanded in angle around inf 63.8%

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

      1. *-commutative 63.8%

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

      2. *-commutative 63.8%

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

      3. associate-*r* 62.9%

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

   7. Simplified 62.9%

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

   if 1e80 < (/.f64 angle 180) < 9.9999999999999999e202

   1. Initial program 38.4%

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

      1. unpow2 38.4%

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

      2. unpow2 38.4%

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

      3. difference-of-squares 38.4%

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

   3. Applied egg-rr 38.4%

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

   4. Taylor expanded in angle around 0 43.0%

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

   5. Taylor expanded in angle around 0 41.9%

      \[\leadsto \color{blue}{\left(0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right)\right)\right)} \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   if 9.9999999999999999e202 < (/.f64 angle 180)

   1. Initial program 28.3%

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

      1. unpow2 28.3%

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

      2. unpow2 28.3%

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

      3. difference-of-squares 28.3%

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

   3. Applied egg-rr 28.3%

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

   4. Taylor expanded in angle around 0 50.4%

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

      1. div-inv 42.9%

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

      2. metadata-eval 42.9%

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

      3. add-sqr-sqrt 61.4%

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

      4. pow2 61.4%

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

   6. Applied egg-rr 61.4%

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

4. Final simplification 60.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 10^{+80}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \mathbf{elif}\;\frac{angle}{180} \leq 10^{+203}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left({\left(\sqrt{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right)}^{2}\right)\\ \end{array} \]

Alternative 7: 56.3% accurate, 1.5× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\\ angle_s \cdot \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq -2 \cdot 10^{-141}:\\ \;\;\;\;t_0 \cdot \sin \left(\pi \cdot \left(angle_m \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;t_0 \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle_m\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* 2.0 (* (- b a) (+ a b)))))
   (*
    angle_s
    (if (<= (- (pow b 2.0) (pow a 2.0)) -2e-141)
      (* t_0 (sin (* PI (* angle_m 0.005555555555555556))))
      (* t_0 (sin (* 0.005555555555555556 (* 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 t_0 = 2.0 * ((b - a) * (a + b));
	double tmp;
	if ((pow(b, 2.0) - pow(a, 2.0)) <= -2e-141) {
		tmp = t_0 * sin((((double) M_PI) * (angle_m * 0.005555555555555556)));
	} else {
		tmp = t_0 * sin((0.005555555555555556 * (((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 t_0 = 2.0 * ((b - a) * (a + b));
	double tmp;
	if ((Math.pow(b, 2.0) - Math.pow(a, 2.0)) <= -2e-141) {
		tmp = t_0 * Math.sin((Math.PI * (angle_m * 0.005555555555555556)));
	} else {
		tmp = t_0 * Math.sin((0.005555555555555556 * (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):
	t_0 = 2.0 * ((b - a) * (a + b))
	tmp = 0
	if (math.pow(b, 2.0) - math.pow(a, 2.0)) <= -2e-141:
		tmp = t_0 * math.sin((math.pi * (angle_m * 0.005555555555555556)))
	else:
		tmp = t_0 * math.sin((0.005555555555555556 * (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)
	t_0 = Float64(2.0 * Float64(Float64(b - a) * Float64(a + b)))
	tmp = 0.0
	if (Float64((b ^ 2.0) - (a ^ 2.0)) <= -2e-141)
		tmp = Float64(t_0 * sin(Float64(pi * Float64(angle_m * 0.005555555555555556))));
	else
		tmp = Float64(t_0 * sin(Float64(0.005555555555555556 * 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)
	t_0 = 2.0 * ((b - a) * (a + b));
	tmp = 0.0;
	if (((b ^ 2.0) - (a ^ 2.0)) <= -2e-141)
		tmp = t_0 * sin((pi * (angle_m * 0.005555555555555556)));
	else
		tmp = t_0 * sin((0.005555555555555556 * (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_] := Block[{t$95$0 = N[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(a + b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision], -2e-141], N[(t$95$0 * N[Sin[N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(t$95$0 * N[Sin[N[(0.005555555555555556 * N[(Pi * angle$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (-.f64 (pow.f64 b 2) (pow.f64 a 2)) < -2.0000000000000001e-141

   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 57.6%

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

   3. Applied egg-rr 57.6%

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

   4. Taylor expanded in angle around 0 57.4%

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

   5. Taylor expanded in angle around inf 55.2%

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

      1. *-commutative 55.2%

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

      2. *-commutative 55.2%

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

      3. associate-*r* 60.5%

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

   7. Simplified 60.5%

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

   if -2.0000000000000001e-141 < (-.f64 (pow.f64 b 2) (pow.f64 a 2))

   1. Initial program 53.7%

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

      1. unpow2 53.7%

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

      2. unpow2 53.7%

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

      3. difference-of-squares 59.8%

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

   3. Applied egg-rr 59.8%

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

   4. Taylor expanded in angle around 0 60.4%

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

   5. Taylor expanded in angle around 0 62.3%

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

4. Final simplification 61.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq -2 \cdot 10^{-141}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\\ \end{array} \]

Alternative 8: 57.9% accurate, 1.5× speedup

Math

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

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) (+ a b)))
    (sin (* 0.005555555555555556 (* 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) {
	return angle_s * (cos((((double) M_PI) * (angle_m / 180.0))) * ((2.0 * ((b - a) * (a + b))) * sin((0.005555555555555556 * (((double) M_PI) * angle_m)))));
}

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 55.3%

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

   1. unpow2 55.3%

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

   2. unpow2 55.3%

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

   3. difference-of-squares 58.9%

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

3. Applied egg-rr 58.9%

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

4. Taylor expanded in angle around 0 60.9%

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

5. Final simplification 60.9%

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

Alternative 9: 56.2% accurate, 2.9× speedup

Math

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

FPCore

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 55.3%

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

   1. unpow2 55.3%

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

   2. unpow2 55.3%

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

   3. difference-of-squares 58.9%

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

3. Applied egg-rr 58.9%

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

4. Taylor expanded in angle around 0 60.9%

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

5. Taylor expanded in angle around 0 59.3%

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

6. Final simplification 59.3%

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

Alternative 10: 53.7% accurate, 3.0× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ angle_s \cdot \begin{array}{l} \mathbf{if}\;angle_m \leq 1.06 \cdot 10^{+232}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \left(angle_m \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\pi \cdot angle_m\right) \cdot \left({a}^{2} \cdot -0.011111111111111112\right)\\ \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (*
  angle_s
  (if (<= angle_m 1.06e+232)
    (* (* 2.0 (* (- b a) (+ a b))) (* angle_m (* PI 0.005555555555555556)))
    (* (* PI angle_m) (* (pow a 2.0) -0.011111111111111112)))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double tmp;
	if (angle_m <= 1.06e+232) {
		tmp = (2.0 * ((b - a) * (a + b))) * (angle_m * (((double) M_PI) * 0.005555555555555556));
	} else {
		tmp = (((double) M_PI) * angle_m) * (pow(a, 2.0) * -0.011111111111111112);
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double tmp;
	if (angle_m <= 1.06e+232) {
		tmp = (2.0 * ((b - a) * (a + b))) * (angle_m * (Math.PI * 0.005555555555555556));
	} else {
		tmp = (Math.PI * angle_m) * (Math.pow(a, 2.0) * -0.011111111111111112);
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	tmp = 0
	if angle_m <= 1.06e+232:
		tmp = (2.0 * ((b - a) * (a + b))) * (angle_m * (math.pi * 0.005555555555555556))
	else:
		tmp = (math.pi * angle_m) * (math.pow(a, 2.0) * -0.011111111111111112)
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	tmp = 0.0
	if (angle_m <= 1.06e+232)
		tmp = Float64(Float64(2.0 * Float64(Float64(b - a) * Float64(a + b))) * Float64(angle_m * Float64(pi * 0.005555555555555556)));
	else
		tmp = Float64(Float64(pi * angle_m) * Float64((a ^ 2.0) * -0.011111111111111112));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	tmp = 0.0;
	if (angle_m <= 1.06e+232)
		tmp = (2.0 * ((b - a) * (a + b))) * (angle_m * (pi * 0.005555555555555556));
	else
		tmp = (pi * angle_m) * ((a ^ 2.0) * -0.011111111111111112);
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := N[(angle$95$s * If[LessEqual[angle$95$m, 1.06e+232], N[(N[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(a + b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(angle$95$m * N[(Pi * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(Pi * angle$95$m), $MachinePrecision] * N[(N[Power[a, 2.0], $MachinePrecision] * -0.011111111111111112), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if angle < 1.06e232

   1. Initial program 57.7%

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

      1. unpow2 57.7%

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

      2. unpow2 57.7%

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

      3. difference-of-squares 61.6%

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

   3. Applied egg-rr 61.6%

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

   4. Taylor expanded in angle around 0 59.4%

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

   5. Taylor expanded in angle around 0 57.1%

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

      1. *-commutative 57.1%

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

      2. associate-*l* 57.2%

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

   7. Simplified 57.2%

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

   if 1.06e232 < angle

   1. Initial program 30.1%

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

      1. unpow2 30.1%

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

      2. unpow2 30.1%

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

      3. difference-of-squares 30.1%

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

   3. Applied egg-rr 30.1%

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

   4. Taylor expanded in angle around 0 42.8%

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

   5. Taylor expanded in angle around 0 18.8%

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

   6. Taylor expanded in b around 0 33.4%

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

      1. *-commutative 33.4%

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

      2. *-commutative 33.4%

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

      3. *-commutative 33.4%

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

      4. associate-*l* 33.4%

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

      5. *-commutative 33.4%

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

   8. Simplified 33.4%

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

4. Final simplification 55.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;angle \leq 1.06 \cdot 10^{+232}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(a + b\right)\right)\right) \cdot \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\pi \cdot angle\right) \cdot \left({a}^{2} \cdot -0.011111111111111112\right)\\ \end{array} \]

Alternative 11: 54.3% accurate, 5.4× speedup

Math

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

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (*
  angle_s
  (* (* 2.0 (* (- b a) (+ a b))) (* angle_m (* PI 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 * ((2.0 * ((b - a) * (a + b))) * (angle_m * (((double) M_PI) * 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 * ((2.0 * ((b - a) * (a + b))) * (angle_m * (Math.PI * 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 * ((2.0 * ((b - a) * (a + b))) * (angle_m * (math.pi * 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(Float64(2.0 * Float64(Float64(b - a) * Float64(a + b))) * Float64(angle_m * Float64(pi * 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 * ((2.0 * ((b - a) * (a + b))) * (angle_m * (pi * 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[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(a + b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(angle$95$m * N[(Pi * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 55.3%

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

   1. unpow2 55.3%

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

   2. unpow2 55.3%

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

   3. difference-of-squares 58.9%

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

3. Applied egg-rr 58.9%

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

4. Taylor expanded in angle around 0 58.0%

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

5. Taylor expanded in angle around 0 53.9%

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

   1. *-commutative 53.9%

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

   2. associate-*l* 53.9%

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

7. Simplified 53.9%

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

8. Final simplification 53.9%

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

Alternative 12: 54.2% accurate, 5.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 55.3%

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

   1. unpow2 55.3%

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

   2. unpow2 55.3%

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

   3. difference-of-squares 58.9%

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

3. Applied egg-rr 58.9%

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

4. Taylor expanded in angle around 0 58.0%

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

5. Taylor expanded in angle around 0 53.9%

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

6. Taylor expanded in angle around 0 53.8%

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

7. Final simplification 53.8%

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

Alternative 13: 54.3% accurate, 5.5× speedup

Math

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

FPCore

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

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 * ((((double) M_PI) * angle_m) * ((b - a) * (a + b))));
}

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 * ((Math.PI * angle_m) * ((b - a) * (a + b))));
}

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 * ((math.pi * angle_m) * ((b - a) * (a + b))))

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(Float64(pi * angle_m) * Float64(Float64(b - a) * Float64(a + b)))))
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 * ((pi * angle_m) * ((b - a) * (a + b))));
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[(N[(Pi * angle$95$m), $MachinePrecision] * N[(N[(b - a), $MachinePrecision] * N[(a + b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 55.3%

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

   1. unpow2 55.3%

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

   2. unpow2 55.3%

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

   3. difference-of-squares 58.9%

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

3. Applied egg-rr 58.9%

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

4. Taylor expanded in angle around 0 58.0%

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

5. Taylor expanded in angle around 0 53.9%

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

6. Taylor expanded in angle around 0 53.8%

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

   1. associate-*r* 53.9%

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

8. Simplified 53.9%

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

9. Final simplification 53.9%

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

Reproduce

herbie shell --seed 2023334 
(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)))))