ab-angle->ABCF B

Percentage Accurate: 54.2% → 57.3%

Time: 48.3s

Alternatives: 15

Speedup: 5.5×

• 32.6% 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: 57.3% accurate, 0.7× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} t_0 := \left(a + b\right) \cdot \left(a - b\right)\\ t_1 := \sqrt[3]{{\pi}^{3}}\\ t_2 := angle \cdot \frac{t_1}{-180}\\ \mathbf{if}\;a \leq 3.8 \cdot 10^{+148}:\\ \;\;\;\;\left(2 \cdot \sin \left(\left(angle \cdot t_1\right) \cdot -0.005555555555555556\right)\right) \cdot \left(\cos \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right) \cdot t_0\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \sin t_2\right) \cdot \left(t_0 \cdot \cos t_2\right)\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* (+ a b) (- a b)))
        (t_1 (cbrt (pow PI 3.0)))
        (t_2 (* angle (/ t_1 -180.0))))
   (if (<= a 3.8e+148)
     (*
      (* 2.0 (sin (* (* angle t_1) -0.005555555555555556)))
      (* (cos (* PI (* angle -0.005555555555555556))) t_0))
     (* (* 2.0 (sin t_2)) (* t_0 (cos t_2))))))

C

a = abs(a);
double code(double a, double b, double angle) {
	double t_0 = (a + b) * (a - b);
	double t_1 = cbrt(pow(((double) M_PI), 3.0));
	double t_2 = angle * (t_1 / -180.0);
	double tmp;
	if (a <= 3.8e+148) {
		tmp = (2.0 * sin(((angle * t_1) * -0.005555555555555556))) * (cos((((double) M_PI) * (angle * -0.005555555555555556))) * t_0);
	} else {
		tmp = (2.0 * sin(t_2)) * (t_0 * cos(t_2));
	}
	return tmp;
}

Java

a = Math.abs(a);
public static double code(double a, double b, double angle) {
	double t_0 = (a + b) * (a - b);
	double t_1 = Math.cbrt(Math.pow(Math.PI, 3.0));
	double t_2 = angle * (t_1 / -180.0);
	double tmp;
	if (a <= 3.8e+148) {
		tmp = (2.0 * Math.sin(((angle * t_1) * -0.005555555555555556))) * (Math.cos((Math.PI * (angle * -0.005555555555555556))) * t_0);
	} else {
		tmp = (2.0 * Math.sin(t_2)) * (t_0 * Math.cos(t_2));
	}
	return tmp;
}

Julia

a = abs(a)
function code(a, b, angle)
	t_0 = Float64(Float64(a + b) * Float64(a - b))
	t_1 = cbrt((pi ^ 3.0))
	t_2 = Float64(angle * Float64(t_1 / -180.0))
	tmp = 0.0
	if (a <= 3.8e+148)
		tmp = Float64(Float64(2.0 * sin(Float64(Float64(angle * t_1) * -0.005555555555555556))) * Float64(cos(Float64(pi * Float64(angle * -0.005555555555555556))) * t_0));
	else
		tmp = Float64(Float64(2.0 * sin(t_2)) * Float64(t_0 * cos(t_2)));
	end
	return tmp
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := Block[{t$95$0 = N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[Power[N[Power[Pi, 3.0], $MachinePrecision], 1/3], $MachinePrecision]}, Block[{t$95$2 = N[(angle * N[(t$95$1 / -180.0), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[a, 3.8e+148], N[(N[(2.0 * N[Sin[N[(N[(angle * t$95$1), $MachinePrecision] * -0.005555555555555556), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[(N[Cos[N[(Pi * N[(angle * -0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * t$95$0), $MachinePrecision]), $MachinePrecision], N[(N[(2.0 * N[Sin[t$95$2], $MachinePrecision]), $MachinePrecision] * N[(t$95$0 * N[Cos[t$95$2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]

Derivation

1. Split input into 2 regimes

2. if a < 3.7999999999999998e148

   1. Initial program 54.4%

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

   3. Simplified 53.5%

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

      1. unpow2 53.5%

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

      2. unpow2 53.5%

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

      3. difference-of-squares 55.8%

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

   5. Applied egg-rr 55.8%

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

   6. Taylor expanded in angle around inf 55.2%

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

      1. *-commutative 55.2%

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

   8. Simplified 55.2%

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

   9. Taylor expanded in angle around inf 55.6%

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

       1. associate-*r* 56.6%

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

       2. *-commutative 56.6%

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

       3. *-commutative 56.6%

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

   11. Simplified 56.6%

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

       1. add-cbrt-cube 57.4%

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

       2. pow3 57.4%

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

   13. Applied egg-rr 57.4%

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

   if 3.7999999999999998e148 < a

   1. Initial program 30.3%

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

   3. Simplified 35.5%

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

      1. unpow2 35.5%

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

      2. unpow2 35.5%

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

      3. difference-of-squares 46.3%

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

   5. Applied egg-rr 46.3%

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

      1. add-cbrt-cube 49.0%

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

      2. pow3 49.0%

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

   7. Applied egg-rr 51.6%

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

      1. add-cbrt-cube 49.0%

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

      2. pow3 49.0%

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

   9. Applied egg-rr 56.9%

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

4. Final simplification 57.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 3.8 \cdot 10^{+148}:\\ \;\;\;\;\left(2 \cdot \sin \left(\left(angle \cdot \sqrt[3]{{\pi}^{3}}\right) \cdot -0.005555555555555556\right)\right) \cdot \left(\cos \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \sin \left(angle \cdot \frac{\sqrt[3]{{\pi}^{3}}}{-180}\right)\right) \cdot \left(\left(\left(a + b\right) \cdot \left(a - b\right)\right) \cdot \cos \left(angle \cdot \frac{\sqrt[3]{{\pi}^{3}}}{-180}\right)\right)\\ \end{array} \]

Alternative 2: 58.0% accurate, 0.7× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} t_0 := \pi \cdot \left(angle \cdot -0.005555555555555556\right)\\ t_1 := \left(a + b\right) \cdot \left(a - b\right)\\ \mathbf{if}\;{b}^{2} - {a}^{2} \leq -5 \cdot 10^{+303}:\\ \;\;\;\;\left(2 \cdot \sin \left(angle \cdot \frac{\pi}{-180}\right)\right) \cdot \left(t_1 \cdot \cos \left(angle \cdot \frac{{\left(\sqrt{\pi}\right)}^{2}}{-180}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\cos t_0 \cdot t_1\right) \cdot \left(2 \cdot \sin t_0\right)\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* PI (* angle -0.005555555555555556))) (t_1 (* (+ a b) (- a b))))
   (if (<= (- (pow b 2.0) (pow a 2.0)) -5e+303)
     (*
      (* 2.0 (sin (* angle (/ PI -180.0))))
      (* t_1 (cos (* angle (/ (pow (sqrt PI) 2.0) -180.0)))))
     (* (* (cos t_0) t_1) (* 2.0 (sin t_0))))))

C

a = abs(a);
double code(double a, double b, double angle) {
	double t_0 = ((double) M_PI) * (angle * -0.005555555555555556);
	double t_1 = (a + b) * (a - b);
	double tmp;
	if ((pow(b, 2.0) - pow(a, 2.0)) <= -5e+303) {
		tmp = (2.0 * sin((angle * (((double) M_PI) / -180.0)))) * (t_1 * cos((angle * (pow(sqrt(((double) M_PI)), 2.0) / -180.0))));
	} else {
		tmp = (cos(t_0) * t_1) * (2.0 * sin(t_0));
	}
	return tmp;
}

Java

a = Math.abs(a);
public static double code(double a, double b, double angle) {
	double t_0 = Math.PI * (angle * -0.005555555555555556);
	double t_1 = (a + b) * (a - b);
	double tmp;
	if ((Math.pow(b, 2.0) - Math.pow(a, 2.0)) <= -5e+303) {
		tmp = (2.0 * Math.sin((angle * (Math.PI / -180.0)))) * (t_1 * Math.cos((angle * (Math.pow(Math.sqrt(Math.PI), 2.0) / -180.0))));
	} else {
		tmp = (Math.cos(t_0) * t_1) * (2.0 * Math.sin(t_0));
	}
	return tmp;
}

Python

a = abs(a)
def code(a, b, angle):
	t_0 = math.pi * (angle * -0.005555555555555556)
	t_1 = (a + b) * (a - b)
	tmp = 0
	if (math.pow(b, 2.0) - math.pow(a, 2.0)) <= -5e+303:
		tmp = (2.0 * math.sin((angle * (math.pi / -180.0)))) * (t_1 * math.cos((angle * (math.pow(math.sqrt(math.pi), 2.0) / -180.0))))
	else:
		tmp = (math.cos(t_0) * t_1) * (2.0 * math.sin(t_0))
	return tmp

Julia

a = abs(a)
function code(a, b, angle)
	t_0 = Float64(pi * Float64(angle * -0.005555555555555556))
	t_1 = Float64(Float64(a + b) * Float64(a - b))
	tmp = 0.0
	if (Float64((b ^ 2.0) - (a ^ 2.0)) <= -5e+303)
		tmp = Float64(Float64(2.0 * sin(Float64(angle * Float64(pi / -180.0)))) * Float64(t_1 * cos(Float64(angle * Float64((sqrt(pi) ^ 2.0) / -180.0)))));
	else
		tmp = Float64(Float64(cos(t_0) * t_1) * Float64(2.0 * sin(t_0)));
	end
	return tmp
end

MATLAB

a = abs(a)
function tmp_2 = code(a, b, angle)
	t_0 = pi * (angle * -0.005555555555555556);
	t_1 = (a + b) * (a - b);
	tmp = 0.0;
	if (((b ^ 2.0) - (a ^ 2.0)) <= -5e+303)
		tmp = (2.0 * sin((angle * (pi / -180.0)))) * (t_1 * cos((angle * ((sqrt(pi) ^ 2.0) / -180.0))));
	else
		tmp = (cos(t_0) * t_1) * (2.0 * sin(t_0));
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := Block[{t$95$0 = N[(Pi * N[(angle * -0.005555555555555556), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision], -5e+303], N[(N[(2.0 * N[Sin[N[(angle * N[(Pi / -180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[(t$95$1 * N[Cos[N[(angle * N[(N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision] / -180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Cos[t$95$0], $MachinePrecision] * t$95$1), $MachinePrecision] * N[(2.0 * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if (-.f64 (pow.f64 b 2) (pow.f64 a 2)) < -4.9999999999999997e303

   1. Initial program 47.3%

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

   3. Simplified 49.5%

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

      1. unpow2 49.5%

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

      2. unpow2 49.5%

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

      3. difference-of-squares 49.5%

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

   5. Applied egg-rr 49.5%

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

      1. add-sqr-sqrt 57.1%

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

      2. pow2 57.1%

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

   7. Applied egg-rr 57.1%

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

   if -4.9999999999999997e303 < (-.f64 (pow.f64 b 2) (pow.f64 a 2))

   1. Initial program 51.7%

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

   3. Simplified 51.2%

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

      1. unpow2 51.2%

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

      2. unpow2 51.2%

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

      3. difference-of-squares 55.6%

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

   5. Applied egg-rr 55.6%

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

   6. Taylor expanded in angle around inf 55.5%

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

      1. *-commutative 55.5%

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

   8. Simplified 55.5%

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

   9. Taylor expanded in angle around inf 55.9%

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

       1. associate-*r* 56.5%

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

       2. *-commutative 56.5%

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

       3. *-commutative 56.5%

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

   11. Simplified 56.5%

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

   12. Taylor expanded in angle around inf 56.5%

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

       1. associate-*r* 57.1%

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

       2. *-commutative 57.1%

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

       3. *-commutative 57.1%

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

   14. Simplified 57.1%

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

4. Final simplification 57.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq -5 \cdot 10^{+303}:\\ \;\;\;\;\left(2 \cdot \sin \left(angle \cdot \frac{\pi}{-180}\right)\right) \cdot \left(\left(\left(a + b\right) \cdot \left(a - b\right)\right) \cdot \cos \left(angle \cdot \frac{{\left(\sqrt{\pi}\right)}^{2}}{-180}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\cos \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \cdot \left(2 \cdot \sin \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right)\right)\\ \end{array} \]

Alternative 3: 57.7% accurate, 0.7× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} t_0 := \left(a + b\right) \cdot \left(a - b\right)\\ t_1 := angle \cdot \frac{\pi}{-180}\\ \mathbf{if}\;{b}^{2} - {a}^{2} \leq 5 \cdot 10^{+281}:\\ \;\;\;\;\left(2 \cdot \sin t_1\right) \cdot \left(t_0 \cdot \cos \left(angle \cdot \frac{{\left(\sqrt{\pi}\right)}^{2}}{-180}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \sin \left(angle \cdot \frac{\sqrt[3]{{\pi}^{3}}}{-180}\right)\right) \cdot \left(t_0 \cdot \cos t_1\right)\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* (+ a b) (- a b))) (t_1 (* angle (/ PI -180.0))))
   (if (<= (- (pow b 2.0) (pow a 2.0)) 5e+281)
     (*
      (* 2.0 (sin t_1))
      (* t_0 (cos (* angle (/ (pow (sqrt PI) 2.0) -180.0)))))
     (*
      (* 2.0 (sin (* angle (/ (cbrt (pow PI 3.0)) -180.0))))
      (* t_0 (cos t_1))))))

C

a = abs(a);
double code(double a, double b, double angle) {
	double t_0 = (a + b) * (a - b);
	double t_1 = angle * (((double) M_PI) / -180.0);
	double tmp;
	if ((pow(b, 2.0) - pow(a, 2.0)) <= 5e+281) {
		tmp = (2.0 * sin(t_1)) * (t_0 * cos((angle * (pow(sqrt(((double) M_PI)), 2.0) / -180.0))));
	} else {
		tmp = (2.0 * sin((angle * (cbrt(pow(((double) M_PI), 3.0)) / -180.0)))) * (t_0 * cos(t_1));
	}
	return tmp;
}

Java

a = Math.abs(a);
public static double code(double a, double b, double angle) {
	double t_0 = (a + b) * (a - b);
	double t_1 = angle * (Math.PI / -180.0);
	double tmp;
	if ((Math.pow(b, 2.0) - Math.pow(a, 2.0)) <= 5e+281) {
		tmp = (2.0 * Math.sin(t_1)) * (t_0 * Math.cos((angle * (Math.pow(Math.sqrt(Math.PI), 2.0) / -180.0))));
	} else {
		tmp = (2.0 * Math.sin((angle * (Math.cbrt(Math.pow(Math.PI, 3.0)) / -180.0)))) * (t_0 * Math.cos(t_1));
	}
	return tmp;
}

Julia

a = abs(a)
function code(a, b, angle)
	t_0 = Float64(Float64(a + b) * Float64(a - b))
	t_1 = Float64(angle * Float64(pi / -180.0))
	tmp = 0.0
	if (Float64((b ^ 2.0) - (a ^ 2.0)) <= 5e+281)
		tmp = Float64(Float64(2.0 * sin(t_1)) * Float64(t_0 * cos(Float64(angle * Float64((sqrt(pi) ^ 2.0) / -180.0)))));
	else
		tmp = Float64(Float64(2.0 * sin(Float64(angle * Float64(cbrt((pi ^ 3.0)) / -180.0)))) * Float64(t_0 * cos(t_1)));
	end
	return tmp
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := Block[{t$95$0 = N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(angle * N[(Pi / -180.0), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision], 5e+281], N[(N[(2.0 * N[Sin[t$95$1], $MachinePrecision]), $MachinePrecision] * N[(t$95$0 * N[Cos[N[(angle * N[(N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision] / -180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(2.0 * N[Sin[N[(angle * N[(N[Power[N[Power[Pi, 3.0], $MachinePrecision], 1/3], $MachinePrecision] / -180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[(t$95$0 * N[Cos[t$95$1], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

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

   1. Initial program 56.4%

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

   3. Simplified 56.9%

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

      1. unpow2 56.9%

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

      2. unpow2 56.9%

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

      3. difference-of-squares 56.9%

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

   5. Applied egg-rr 56.9%

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

      1. add-sqr-sqrt 58.3%

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

      2. pow2 58.3%

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

   7. Applied egg-rr 58.3%

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

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

   1. Initial program 35.8%

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

   3. Simplified 34.6%

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

      1. unpow2 34.6%

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

      2. unpow2 34.6%

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

      3. difference-of-squares 47.6%

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

   5. Applied egg-rr 47.6%

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

      1. add-cbrt-cube 58.0%

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

      2. pow3 58.0%

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

   7. Applied egg-rr 55.3%

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

4. Final simplification 57.5%

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

Alternative 4: 57.8% accurate, 0.7× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} t_0 := \left(a + b\right) \cdot \left(a - b\right)\\ \mathbf{if}\;{b}^{2} - {a}^{2} \leq 5 \cdot 10^{+281}:\\ \;\;\;\;\left(2 \cdot \sin \left(angle \cdot \frac{\pi}{-180}\right)\right) \cdot \left(t_0 \cdot \cos \left(angle \cdot \frac{{\left(\sqrt{\pi}\right)}^{2}}{-180}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\cos \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right) \cdot t_0\right) \cdot \left(2 \cdot \sin \left(angle \cdot \frac{\sqrt[3]{{\pi}^{3}}}{-180}\right)\right)\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* (+ a b) (- a b))))
   (if (<= (- (pow b 2.0) (pow a 2.0)) 5e+281)
     (*
      (* 2.0 (sin (* angle (/ PI -180.0))))
      (* t_0 (cos (* angle (/ (pow (sqrt PI) 2.0) -180.0)))))
     (*
      (* (cos (* PI (* angle -0.005555555555555556))) t_0)
      (* 2.0 (sin (* angle (/ (cbrt (pow PI 3.0)) -180.0))))))))

C

a = abs(a);
double code(double a, double b, double angle) {
	double t_0 = (a + b) * (a - b);
	double tmp;
	if ((pow(b, 2.0) - pow(a, 2.0)) <= 5e+281) {
		tmp = (2.0 * sin((angle * (((double) M_PI) / -180.0)))) * (t_0 * cos((angle * (pow(sqrt(((double) M_PI)), 2.0) / -180.0))));
	} else {
		tmp = (cos((((double) M_PI) * (angle * -0.005555555555555556))) * t_0) * (2.0 * sin((angle * (cbrt(pow(((double) M_PI), 3.0)) / -180.0))));
	}
	return tmp;
}

Java

a = Math.abs(a);
public static double code(double a, double b, double angle) {
	double t_0 = (a + b) * (a - b);
	double tmp;
	if ((Math.pow(b, 2.0) - Math.pow(a, 2.0)) <= 5e+281) {
		tmp = (2.0 * Math.sin((angle * (Math.PI / -180.0)))) * (t_0 * Math.cos((angle * (Math.pow(Math.sqrt(Math.PI), 2.0) / -180.0))));
	} else {
		tmp = (Math.cos((Math.PI * (angle * -0.005555555555555556))) * t_0) * (2.0 * Math.sin((angle * (Math.cbrt(Math.pow(Math.PI, 3.0)) / -180.0))));
	}
	return tmp;
}

Julia

a = abs(a)
function code(a, b, angle)
	t_0 = Float64(Float64(a + b) * Float64(a - b))
	tmp = 0.0
	if (Float64((b ^ 2.0) - (a ^ 2.0)) <= 5e+281)
		tmp = Float64(Float64(2.0 * sin(Float64(angle * Float64(pi / -180.0)))) * Float64(t_0 * cos(Float64(angle * Float64((sqrt(pi) ^ 2.0) / -180.0)))));
	else
		tmp = Float64(Float64(cos(Float64(pi * Float64(angle * -0.005555555555555556))) * t_0) * Float64(2.0 * sin(Float64(angle * Float64(cbrt((pi ^ 3.0)) / -180.0)))));
	end
	return tmp
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := Block[{t$95$0 = N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision], 5e+281], N[(N[(2.0 * N[Sin[N[(angle * N[(Pi / -180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[(t$95$0 * N[Cos[N[(angle * N[(N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision] / -180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Cos[N[(Pi * N[(angle * -0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * t$95$0), $MachinePrecision] * N[(2.0 * N[Sin[N[(angle * N[(N[Power[N[Power[Pi, 3.0], $MachinePrecision], 1/3], $MachinePrecision] / -180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

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

   1. Initial program 56.4%

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

   3. Simplified 56.9%

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

      1. unpow2 56.9%

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

      2. unpow2 56.9%

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

      3. difference-of-squares 56.9%

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

   5. Applied egg-rr 56.9%

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

      1. add-sqr-sqrt 58.3%

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

      2. pow2 58.3%

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

   7. Applied egg-rr 58.3%

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

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

   1. Initial program 35.8%

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

   3. Simplified 34.6%

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

      1. unpow2 34.6%

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

      2. unpow2 34.6%

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

      3. difference-of-squares 47.6%

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

   5. Applied egg-rr 47.6%

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

      1. add-cbrt-cube 58.0%

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

      2. pow3 58.0%

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

   7. Applied egg-rr 55.3%

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

   8. Taylor expanded in angle around inf 56.8%

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

      1. associate-*r* 50.4%

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

      2. *-commutative 50.4%

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

      3. *-commutative 50.4%

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

   10. Simplified 58.2%

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

4. Final simplification 58.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq 5 \cdot 10^{+281}:\\ \;\;\;\;\left(2 \cdot \sin \left(angle \cdot \frac{\pi}{-180}\right)\right) \cdot \left(\left(\left(a + b\right) \cdot \left(a - b\right)\right) \cdot \cos \left(angle \cdot \frac{{\left(\sqrt{\pi}\right)}^{2}}{-180}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\cos \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \cdot \left(2 \cdot \sin \left(angle \cdot \frac{\sqrt[3]{{\pi}^{3}}}{-180}\right)\right)\\ \end{array} \]

Alternative 5: 57.5% accurate, 0.7× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} t_0 := \left(a + b\right) \cdot \left(a - b\right)\\ \mathbf{if}\;{b}^{2} - {a}^{2} \leq -5 \cdot 10^{+303}:\\ \;\;\;\;\left(2 \cdot \sin \left(angle \cdot \frac{\pi}{-180}\right)\right) \cdot \left(t_0 \cdot \cos \left(angle \cdot \frac{{\left(\sqrt{\pi}\right)}^{2}}{-180}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \sin \left(\left(angle \cdot \sqrt[3]{{\pi}^{3}}\right) \cdot -0.005555555555555556\right)\right) \cdot \left(\cos \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right) \cdot t_0\right)\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* (+ a b) (- a b))))
   (if (<= (- (pow b 2.0) (pow a 2.0)) -5e+303)
     (*
      (* 2.0 (sin (* angle (/ PI -180.0))))
      (* t_0 (cos (* angle (/ (pow (sqrt PI) 2.0) -180.0)))))
     (*
      (* 2.0 (sin (* (* angle (cbrt (pow PI 3.0))) -0.005555555555555556)))
      (* (cos (* PI (* angle -0.005555555555555556))) t_0)))))

C

a = abs(a);
double code(double a, double b, double angle) {
	double t_0 = (a + b) * (a - b);
	double tmp;
	if ((pow(b, 2.0) - pow(a, 2.0)) <= -5e+303) {
		tmp = (2.0 * sin((angle * (((double) M_PI) / -180.0)))) * (t_0 * cos((angle * (pow(sqrt(((double) M_PI)), 2.0) / -180.0))));
	} else {
		tmp = (2.0 * sin(((angle * cbrt(pow(((double) M_PI), 3.0))) * -0.005555555555555556))) * (cos((((double) M_PI) * (angle * -0.005555555555555556))) * t_0);
	}
	return tmp;
}

Java

a = Math.abs(a);
public static double code(double a, double b, double angle) {
	double t_0 = (a + b) * (a - b);
	double tmp;
	if ((Math.pow(b, 2.0) - Math.pow(a, 2.0)) <= -5e+303) {
		tmp = (2.0 * Math.sin((angle * (Math.PI / -180.0)))) * (t_0 * Math.cos((angle * (Math.pow(Math.sqrt(Math.PI), 2.0) / -180.0))));
	} else {
		tmp = (2.0 * Math.sin(((angle * Math.cbrt(Math.pow(Math.PI, 3.0))) * -0.005555555555555556))) * (Math.cos((Math.PI * (angle * -0.005555555555555556))) * t_0);
	}
	return tmp;
}

Julia

a = abs(a)
function code(a, b, angle)
	t_0 = Float64(Float64(a + b) * Float64(a - b))
	tmp = 0.0
	if (Float64((b ^ 2.0) - (a ^ 2.0)) <= -5e+303)
		tmp = Float64(Float64(2.0 * sin(Float64(angle * Float64(pi / -180.0)))) * Float64(t_0 * cos(Float64(angle * Float64((sqrt(pi) ^ 2.0) / -180.0)))));
	else
		tmp = Float64(Float64(2.0 * sin(Float64(Float64(angle * cbrt((pi ^ 3.0))) * -0.005555555555555556))) * Float64(cos(Float64(pi * Float64(angle * -0.005555555555555556))) * t_0));
	end
	return tmp
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := Block[{t$95$0 = N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision], -5e+303], N[(N[(2.0 * N[Sin[N[(angle * N[(Pi / -180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[(t$95$0 * N[Cos[N[(angle * N[(N[Power[N[Sqrt[Pi], $MachinePrecision], 2.0], $MachinePrecision] / -180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(2.0 * N[Sin[N[(N[(angle * N[Power[N[Power[Pi, 3.0], $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision] * -0.005555555555555556), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[(N[Cos[N[(Pi * N[(angle * -0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * t$95$0), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (-.f64 (pow.f64 b 2) (pow.f64 a 2)) < -4.9999999999999997e303

   1. Initial program 47.3%

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

   3. Simplified 49.5%

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

      1. unpow2 49.5%

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

      2. unpow2 49.5%

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

      3. difference-of-squares 49.5%

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

   5. Applied egg-rr 49.5%

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

      1. add-sqr-sqrt 57.1%

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

      2. pow2 57.1%

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

   7. Applied egg-rr 57.1%

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

   if -4.9999999999999997e303 < (-.f64 (pow.f64 b 2) (pow.f64 a 2))

   1. Initial program 51.7%

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

   3. Simplified 51.2%

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

      1. unpow2 51.2%

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

      2. unpow2 51.2%

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

      3. difference-of-squares 55.6%

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

   5. Applied egg-rr 55.6%

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

   6. Taylor expanded in angle around inf 55.5%

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

      1. *-commutative 55.5%

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

   8. Simplified 55.5%

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

   9. Taylor expanded in angle around inf 55.9%

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

       1. associate-*r* 56.5%

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

       2. *-commutative 56.5%

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

       3. *-commutative 56.5%

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

   11. Simplified 56.5%

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

       1. add-cbrt-cube 58.9%

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

       2. pow3 58.9%

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

   13. Applied egg-rr 58.9%

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

4. Final simplification 58.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} - {a}^{2} \leq -5 \cdot 10^{+303}:\\ \;\;\;\;\left(2 \cdot \sin \left(angle \cdot \frac{\pi}{-180}\right)\right) \cdot \left(\left(\left(a + b\right) \cdot \left(a - b\right)\right) \cdot \cos \left(angle \cdot \frac{{\left(\sqrt{\pi}\right)}^{2}}{-180}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \sin \left(\left(angle \cdot \sqrt[3]{{\pi}^{3}}\right) \cdot -0.005555555555555556\right)\right) \cdot \left(\cos \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right)\\ \end{array} \]

Alternative 6: 57.4% accurate, 1.2× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} t_0 := \pi \cdot \left(angle \cdot -0.005555555555555556\right)\\ \mathbf{if}\;{a}^{2} \leq 4 \cdot 10^{+240}:\\ \;\;\;\;2 \cdot \left(\left({b}^{2} - {a}^{2}\right) \cdot \frac{\sin \left(\left(angle \cdot \pi\right) \cdot 0.011111111111111112\right)}{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\cos t_0 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \cdot \left(2 \cdot t_0\right)\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* PI (* angle -0.005555555555555556))))
   (if (<= (pow a 2.0) 4e+240)
     (*
      2.0
      (*
       (- (pow b 2.0) (pow a 2.0))
       (/ (sin (* (* angle PI) 0.011111111111111112)) 2.0)))
     (* (* (cos t_0) (* (+ a b) (- a b))) (* 2.0 t_0)))))

C

a = abs(a);
double code(double a, double b, double angle) {
	double t_0 = ((double) M_PI) * (angle * -0.005555555555555556);
	double tmp;
	if (pow(a, 2.0) <= 4e+240) {
		tmp = 2.0 * ((pow(b, 2.0) - pow(a, 2.0)) * (sin(((angle * ((double) M_PI)) * 0.011111111111111112)) / 2.0));
	} else {
		tmp = (cos(t_0) * ((a + b) * (a - b))) * (2.0 * t_0);
	}
	return tmp;
}

Java

a = Math.abs(a);
public static double code(double a, double b, double angle) {
	double t_0 = Math.PI * (angle * -0.005555555555555556);
	double tmp;
	if (Math.pow(a, 2.0) <= 4e+240) {
		tmp = 2.0 * ((Math.pow(b, 2.0) - Math.pow(a, 2.0)) * (Math.sin(((angle * Math.PI) * 0.011111111111111112)) / 2.0));
	} else {
		tmp = (Math.cos(t_0) * ((a + b) * (a - b))) * (2.0 * t_0);
	}
	return tmp;
}

Python

a = abs(a)
def code(a, b, angle):
	t_0 = math.pi * (angle * -0.005555555555555556)
	tmp = 0
	if math.pow(a, 2.0) <= 4e+240:
		tmp = 2.0 * ((math.pow(b, 2.0) - math.pow(a, 2.0)) * (math.sin(((angle * math.pi) * 0.011111111111111112)) / 2.0))
	else:
		tmp = (math.cos(t_0) * ((a + b) * (a - b))) * (2.0 * t_0)
	return tmp

Julia

a = abs(a)
function code(a, b, angle)
	t_0 = Float64(pi * Float64(angle * -0.005555555555555556))
	tmp = 0.0
	if ((a ^ 2.0) <= 4e+240)
		tmp = Float64(2.0 * Float64(Float64((b ^ 2.0) - (a ^ 2.0)) * Float64(sin(Float64(Float64(angle * pi) * 0.011111111111111112)) / 2.0)));
	else
		tmp = Float64(Float64(cos(t_0) * Float64(Float64(a + b) * Float64(a - b))) * Float64(2.0 * t_0));
	end
	return tmp
end

MATLAB

a = abs(a)
function tmp_2 = code(a, b, angle)
	t_0 = pi * (angle * -0.005555555555555556);
	tmp = 0.0;
	if ((a ^ 2.0) <= 4e+240)
		tmp = 2.0 * (((b ^ 2.0) - (a ^ 2.0)) * (sin(((angle * pi) * 0.011111111111111112)) / 2.0));
	else
		tmp = (cos(t_0) * ((a + b) * (a - b))) * (2.0 * t_0);
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := Block[{t$95$0 = N[(Pi * N[(angle * -0.005555555555555556), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[a, 2.0], $MachinePrecision], 4e+240], N[(2.0 * N[(N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision] * N[(N[Sin[N[(N[(angle * Pi), $MachinePrecision] * 0.011111111111111112), $MachinePrecision]], $MachinePrecision] / 2.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Cos[t$95$0], $MachinePrecision] * N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(2.0 * t$95$0), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 a 2) < 4.00000000000000006e240

   1. Initial program 57.1%

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

      1. associate-*l* 57.1%

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

      2. associate-*l* 57.1%

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

      3. cos-neg 57.1%

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

      4. distribute-rgt-neg-out 57.1%

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

      5. distribute-frac-neg 57.1%

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

      6. neg-mul-1 57.1%

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

      7. associate-/l* 56.7%

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

      8. associate-*r/ 56.9%

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

      9. associate-/r/ 57.1%

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

      10. associate-/l* 57.1%

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

      11. metadata-eval 57.1%

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

   3. Simplified 57.1%

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

      1. sin-cos-mult 57.0%

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

      2. associate-*r/ 57.0%

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

   5. Applied egg-rr 58.8%

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

      1. *-commutative 58.8%

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

      2. associate-/l* 58.8%

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

      3. associate-/r/ 58.8%

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

   7. Simplified 57.5%

      \[\leadsto 2 \cdot \color{blue}{\left(\frac{\sin \left(\left(angle \cdot \pi\right) \cdot 0.011111111111111112\right) + 0}{2} \cdot \left({b}^{2} - {a}^{2}\right)\right)} \]

   if 4.00000000000000006e240 < (pow.f64 a 2)

   1. Initial program 38.8%

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

   3. Simplified 39.0%

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

      1. unpow2 39.0%

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

      2. unpow2 39.0%

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

      3. difference-of-squares 49.4%

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

   5. Applied egg-rr 49.4%

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

   6. Taylor expanded in angle around inf 51.4%

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

      1. *-commutative 51.4%

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

   8. Simplified 51.4%

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

   9. Taylor expanded in angle around inf 49.1%

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

       1. associate-*r* 51.6%

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

       2. *-commutative 51.6%

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

       3. *-commutative 51.6%

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

   11. Simplified 51.6%

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

   12. Taylor expanded in angle around 0 52.8%

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

       1. associate-*r* 52.8%

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

       2. *-commutative 52.8%

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

       3. *-commutative 52.8%

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

   14. Simplified 52.8%

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

4. Final simplification 55.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{a}^{2} \leq 4 \cdot 10^{+240}:\\ \;\;\;\;2 \cdot \left(\left({b}^{2} - {a}^{2}\right) \cdot \frac{\sin \left(\left(angle \cdot \pi\right) \cdot 0.011111111111111112\right)}{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\cos \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \cdot \left(2 \cdot \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right)\right)\\ \end{array} \]

Alternative 7: 57.8% accurate, 1.5× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} t_0 := angle \cdot \left(\pi \cdot -0.005555555555555556\right)\\ t_1 := \left(a + b\right) \cdot \left(a - b\right)\\ \mathbf{if}\;b \leq 9 \cdot 10^{+184}:\\ \;\;\;\;2 \cdot \left(t_1 \cdot \left(\cos t_0 \cdot \sin t_0\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\cos \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right) \cdot t_1\right) \cdot \left(2 \cdot \left(-0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* angle (* PI -0.005555555555555556))) (t_1 (* (+ a b) (- a b))))
   (if (<= b 9e+184)
     (* 2.0 (* t_1 (* (cos t_0) (sin t_0))))
     (*
      (* (cos (* PI (* angle -0.005555555555555556))) t_1)
      (* 2.0 (* -0.005555555555555556 (* angle PI)))))))

C

a = abs(a);
double code(double a, double b, double angle) {
	double t_0 = angle * (((double) M_PI) * -0.005555555555555556);
	double t_1 = (a + b) * (a - b);
	double tmp;
	if (b <= 9e+184) {
		tmp = 2.0 * (t_1 * (cos(t_0) * sin(t_0)));
	} else {
		tmp = (cos((((double) M_PI) * (angle * -0.005555555555555556))) * t_1) * (2.0 * (-0.005555555555555556 * (angle * ((double) M_PI))));
	}
	return tmp;
}

Java

a = Math.abs(a);
public static double code(double a, double b, double angle) {
	double t_0 = angle * (Math.PI * -0.005555555555555556);
	double t_1 = (a + b) * (a - b);
	double tmp;
	if (b <= 9e+184) {
		tmp = 2.0 * (t_1 * (Math.cos(t_0) * Math.sin(t_0)));
	} else {
		tmp = (Math.cos((Math.PI * (angle * -0.005555555555555556))) * t_1) * (2.0 * (-0.005555555555555556 * (angle * Math.PI)));
	}
	return tmp;
}

Python

a = abs(a)
def code(a, b, angle):
	t_0 = angle * (math.pi * -0.005555555555555556)
	t_1 = (a + b) * (a - b)
	tmp = 0
	if b <= 9e+184:
		tmp = 2.0 * (t_1 * (math.cos(t_0) * math.sin(t_0)))
	else:
		tmp = (math.cos((math.pi * (angle * -0.005555555555555556))) * t_1) * (2.0 * (-0.005555555555555556 * (angle * math.pi)))
	return tmp

Julia

a = abs(a)
function code(a, b, angle)
	t_0 = Float64(angle * Float64(pi * -0.005555555555555556))
	t_1 = Float64(Float64(a + b) * Float64(a - b))
	tmp = 0.0
	if (b <= 9e+184)
		tmp = Float64(2.0 * Float64(t_1 * Float64(cos(t_0) * sin(t_0))));
	else
		tmp = Float64(Float64(cos(Float64(pi * Float64(angle * -0.005555555555555556))) * t_1) * Float64(2.0 * Float64(-0.005555555555555556 * Float64(angle * pi))));
	end
	return tmp
end

MATLAB

a = abs(a)
function tmp_2 = code(a, b, angle)
	t_0 = angle * (pi * -0.005555555555555556);
	t_1 = (a + b) * (a - b);
	tmp = 0.0;
	if (b <= 9e+184)
		tmp = 2.0 * (t_1 * (cos(t_0) * sin(t_0)));
	else
		tmp = (cos((pi * (angle * -0.005555555555555556))) * t_1) * (2.0 * (-0.005555555555555556 * (angle * pi)));
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := Block[{t$95$0 = N[(angle * N[(Pi * -0.005555555555555556), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[b, 9e+184], N[(2.0 * N[(t$95$1 * N[(N[Cos[t$95$0], $MachinePrecision] * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Cos[N[(Pi * N[(angle * -0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * t$95$1), $MachinePrecision] * N[(2.0 * N[(-0.005555555555555556 * N[(angle * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if b < 9.00000000000000072e184

   1. Initial program 53.9%

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

   3. Simplified 53.9%

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

      1. unpow2 53.9%

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

      2. unpow2 53.9%

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

      3. difference-of-squares 55.7%

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

   5. Applied egg-rr 55.7%

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

      1. add-cbrt-cube 55.9%

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

      2. pow3 55.9%

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

   7. Applied egg-rr 56.0%

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

   8. Taylor expanded in angle around inf 55.5%

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

      1. associate-*r* 55.5%

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

      2. *-commutative 55.5%

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

      3. associate-*l* 56.4%

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

      4. *-commutative 56.4%

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

      5. associate-*l* 55.7%

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

   10. Simplified 55.7%

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

   if 9.00000000000000072e184 < b

   1. Initial program 19.2%

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

   3. Simplified 19.2%

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

      1. unpow2 19.2%

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

      2. unpow2 19.2%

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

      3. difference-of-squares 40.9%

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

   5. Applied egg-rr 40.9%

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

   6. Taylor expanded in angle around inf 40.9%

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

      1. *-commutative 40.9%

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

   8. Simplified 40.9%

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

   9. Taylor expanded in angle around inf 45.3%

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

       1. associate-*r* 45.3%

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

       2. *-commutative 45.3%

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

       3. *-commutative 45.3%

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

   11. Simplified 45.3%

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

   12. Taylor expanded in angle around 0 54.0%

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

4. Final simplification 55.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq 9 \cdot 10^{+184}:\\ \;\;\;\;2 \cdot \left(\left(\left(a + b\right) \cdot \left(a - b\right)\right) \cdot \left(\cos \left(angle \cdot \left(\pi \cdot -0.005555555555555556\right)\right) \cdot \sin \left(angle \cdot \left(\pi \cdot -0.005555555555555556\right)\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\cos \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \cdot \left(2 \cdot \left(-0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\\ \end{array} \]

Alternative 8: 57.8% accurate, 1.5× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} t_0 := \left(a + b\right) \cdot \left(a - b\right)\\ t_1 := angle \cdot \frac{\pi}{-180}\\ \mathbf{if}\;b \leq 10^{+186}:\\ \;\;\;\;\left(2 \cdot \sin t_1\right) \cdot \left(t_0 \cdot \cos t_1\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\cos \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right) \cdot t_0\right) \cdot \left(2 \cdot \left(-0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* (+ a b) (- a b))) (t_1 (* angle (/ PI -180.0))))
   (if (<= b 1e+186)
     (* (* 2.0 (sin t_1)) (* t_0 (cos t_1)))
     (*
      (* (cos (* PI (* angle -0.005555555555555556))) t_0)
      (* 2.0 (* -0.005555555555555556 (* angle PI)))))))

C

a = abs(a);
double code(double a, double b, double angle) {
	double t_0 = (a + b) * (a - b);
	double t_1 = angle * (((double) M_PI) / -180.0);
	double tmp;
	if (b <= 1e+186) {
		tmp = (2.0 * sin(t_1)) * (t_0 * cos(t_1));
	} else {
		tmp = (cos((((double) M_PI) * (angle * -0.005555555555555556))) * t_0) * (2.0 * (-0.005555555555555556 * (angle * ((double) M_PI))));
	}
	return tmp;
}

Java

a = Math.abs(a);
public static double code(double a, double b, double angle) {
	double t_0 = (a + b) * (a - b);
	double t_1 = angle * (Math.PI / -180.0);
	double tmp;
	if (b <= 1e+186) {
		tmp = (2.0 * Math.sin(t_1)) * (t_0 * Math.cos(t_1));
	} else {
		tmp = (Math.cos((Math.PI * (angle * -0.005555555555555556))) * t_0) * (2.0 * (-0.005555555555555556 * (angle * Math.PI)));
	}
	return tmp;
}

Python

a = abs(a)
def code(a, b, angle):
	t_0 = (a + b) * (a - b)
	t_1 = angle * (math.pi / -180.0)
	tmp = 0
	if b <= 1e+186:
		tmp = (2.0 * math.sin(t_1)) * (t_0 * math.cos(t_1))
	else:
		tmp = (math.cos((math.pi * (angle * -0.005555555555555556))) * t_0) * (2.0 * (-0.005555555555555556 * (angle * math.pi)))
	return tmp

Julia

a = abs(a)
function code(a, b, angle)
	t_0 = Float64(Float64(a + b) * Float64(a - b))
	t_1 = Float64(angle * Float64(pi / -180.0))
	tmp = 0.0
	if (b <= 1e+186)
		tmp = Float64(Float64(2.0 * sin(t_1)) * Float64(t_0 * cos(t_1)));
	else
		tmp = Float64(Float64(cos(Float64(pi * Float64(angle * -0.005555555555555556))) * t_0) * Float64(2.0 * Float64(-0.005555555555555556 * Float64(angle * pi))));
	end
	return tmp
end

MATLAB

a = abs(a)
function tmp_2 = code(a, b, angle)
	t_0 = (a + b) * (a - b);
	t_1 = angle * (pi / -180.0);
	tmp = 0.0;
	if (b <= 1e+186)
		tmp = (2.0 * sin(t_1)) * (t_0 * cos(t_1));
	else
		tmp = (cos((pi * (angle * -0.005555555555555556))) * t_0) * (2.0 * (-0.005555555555555556 * (angle * pi)));
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := Block[{t$95$0 = N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(angle * N[(Pi / -180.0), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[b, 1e+186], N[(N[(2.0 * N[Sin[t$95$1], $MachinePrecision]), $MachinePrecision] * N[(t$95$0 * N[Cos[t$95$1], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Cos[N[(Pi * N[(angle * -0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * t$95$0), $MachinePrecision] * N[(2.0 * N[(-0.005555555555555556 * N[(angle * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if b < 9.9999999999999998e185

   1. Initial program 53.9%

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

   3. Simplified 53.9%

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

      1. unpow2 53.9%

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

      2. unpow2 53.9%

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

      3. difference-of-squares 55.7%

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

   5. Applied egg-rr 55.7%

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

   if 9.9999999999999998e185 < b

   1. Initial program 19.2%

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

   3. Simplified 19.2%

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

      1. unpow2 19.2%

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

      2. unpow2 19.2%

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

      3. difference-of-squares 40.9%

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

   5. Applied egg-rr 40.9%

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

   6. Taylor expanded in angle around inf 40.9%

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

      1. *-commutative 40.9%

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

   8. Simplified 40.9%

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

   9. Taylor expanded in angle around inf 45.3%

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

       1. associate-*r* 45.3%

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

       2. *-commutative 45.3%

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

       3. *-commutative 45.3%

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

   11. Simplified 45.3%

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

   12. Taylor expanded in angle around 0 54.0%

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

4. Final simplification 55.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq 10^{+186}:\\ \;\;\;\;\left(2 \cdot \sin \left(angle \cdot \frac{\pi}{-180}\right)\right) \cdot \left(\left(\left(a + b\right) \cdot \left(a - b\right)\right) \cdot \cos \left(angle \cdot \frac{\pi}{-180}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\cos \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \cdot \left(2 \cdot \left(-0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\\ \end{array} \]

Alternative 9: 58.0% accurate, 1.5× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} t_0 := \cos \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\\ \mathbf{if}\;b \leq 4 \cdot 10^{+184}:\\ \;\;\;\;t_0 \cdot \left(2 \cdot \sin \left(angle \cdot \frac{\pi}{-180}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;t_0 \cdot \left(2 \cdot \left(-0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0
         (* (cos (* PI (* angle -0.005555555555555556))) (* (+ a b) (- a b)))))
   (if (<= b 4e+184)
     (* t_0 (* 2.0 (sin (* angle (/ PI -180.0)))))
     (* t_0 (* 2.0 (* -0.005555555555555556 (* angle PI)))))))

C

a = abs(a);
double code(double a, double b, double angle) {
	double t_0 = cos((((double) M_PI) * (angle * -0.005555555555555556))) * ((a + b) * (a - b));
	double tmp;
	if (b <= 4e+184) {
		tmp = t_0 * (2.0 * sin((angle * (((double) M_PI) / -180.0))));
	} else {
		tmp = t_0 * (2.0 * (-0.005555555555555556 * (angle * ((double) M_PI))));
	}
	return tmp;
}

Java

a = Math.abs(a);
public static double code(double a, double b, double angle) {
	double t_0 = Math.cos((Math.PI * (angle * -0.005555555555555556))) * ((a + b) * (a - b));
	double tmp;
	if (b <= 4e+184) {
		tmp = t_0 * (2.0 * Math.sin((angle * (Math.PI / -180.0))));
	} else {
		tmp = t_0 * (2.0 * (-0.005555555555555556 * (angle * Math.PI)));
	}
	return tmp;
}

Python

a = abs(a)
def code(a, b, angle):
	t_0 = math.cos((math.pi * (angle * -0.005555555555555556))) * ((a + b) * (a - b))
	tmp = 0
	if b <= 4e+184:
		tmp = t_0 * (2.0 * math.sin((angle * (math.pi / -180.0))))
	else:
		tmp = t_0 * (2.0 * (-0.005555555555555556 * (angle * math.pi)))
	return tmp

Julia

a = abs(a)
function code(a, b, angle)
	t_0 = Float64(cos(Float64(pi * Float64(angle * -0.005555555555555556))) * Float64(Float64(a + b) * Float64(a - b)))
	tmp = 0.0
	if (b <= 4e+184)
		tmp = Float64(t_0 * Float64(2.0 * sin(Float64(angle * Float64(pi / -180.0)))));
	else
		tmp = Float64(t_0 * Float64(2.0 * Float64(-0.005555555555555556 * Float64(angle * pi))));
	end
	return tmp
end

MATLAB

a = abs(a)
function tmp_2 = code(a, b, angle)
	t_0 = cos((pi * (angle * -0.005555555555555556))) * ((a + b) * (a - b));
	tmp = 0.0;
	if (b <= 4e+184)
		tmp = t_0 * (2.0 * sin((angle * (pi / -180.0))));
	else
		tmp = t_0 * (2.0 * (-0.005555555555555556 * (angle * pi)));
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := Block[{t$95$0 = N[(N[Cos[N[(Pi * N[(angle * -0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[b, 4e+184], N[(t$95$0 * N[(2.0 * N[Sin[N[(angle * N[(Pi / -180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(t$95$0 * N[(2.0 * N[(-0.005555555555555556 * N[(angle * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if b < 4.00000000000000007e184

   1. Initial program 53.9%

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

   3. Simplified 53.9%

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

      1. unpow2 53.9%

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

      2. unpow2 53.9%

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

      3. difference-of-squares 55.7%

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

   5. Applied egg-rr 55.7%

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

   6. Taylor expanded in angle around inf 55.4%

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

      1. associate-*r* 56.4%

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

      2. *-commutative 56.4%

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

      3. *-commutative 56.4%

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

   8. Simplified 56.4%

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

   if 4.00000000000000007e184 < b

   1. Initial program 19.2%

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

   3. Simplified 19.2%

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

      1. unpow2 19.2%

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

      2. unpow2 19.2%

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

      3. difference-of-squares 40.9%

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

   5. Applied egg-rr 40.9%

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

   6. Taylor expanded in angle around inf 40.9%

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

      1. *-commutative 40.9%

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

   8. Simplified 40.9%

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

   9. Taylor expanded in angle around inf 45.3%

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

       1. associate-*r* 45.3%

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

       2. *-commutative 45.3%

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

       3. *-commutative 45.3%

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

   11. Simplified 45.3%

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

   12. Taylor expanded in angle around 0 54.0%

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

4. Final simplification 56.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq 4 \cdot 10^{+184}:\\ \;\;\;\;\left(\cos \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \cdot \left(2 \cdot \sin \left(angle \cdot \frac{\pi}{-180}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\cos \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \cdot \left(2 \cdot \left(-0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\\ \end{array} \]

Alternative 10: 57.8% accurate, 1.5× speedup

Math

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

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* PI (* angle -0.005555555555555556))))
   (* (* (cos t_0) (* (+ a b) (- a b))) (* 2.0 (sin t_0)))))

C

a = abs(a);
double code(double a, double b, double angle) {
	double t_0 = ((double) M_PI) * (angle * -0.005555555555555556);
	return (cos(t_0) * ((a + b) * (a - b))) * (2.0 * sin(t_0));
}

Java

a = Math.abs(a);
public static double code(double a, double b, double angle) {
	double t_0 = Math.PI * (angle * -0.005555555555555556);
	return (Math.cos(t_0) * ((a + b) * (a - b))) * (2.0 * Math.sin(t_0));
}

Python

a = abs(a)
def code(a, b, angle):
	t_0 = math.pi * (angle * -0.005555555555555556)
	return (math.cos(t_0) * ((a + b) * (a - b))) * (2.0 * math.sin(t_0))

Julia

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

MATLAB

a = abs(a)
function tmp = code(a, b, angle)
	t_0 = pi * (angle * -0.005555555555555556);
	tmp = (cos(t_0) * ((a + b) * (a - b))) * (2.0 * sin(t_0));
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := Block[{t$95$0 = N[(Pi * N[(angle * -0.005555555555555556), $MachinePrecision]), $MachinePrecision]}, N[(N[(N[Cos[t$95$0], $MachinePrecision] * N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(2.0 * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Initial program 50.8%

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

3. Simplified 50.8%

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

   1. unpow2 50.8%

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

   2. unpow2 50.8%

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

   3. difference-of-squares 54.4%

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

5. Applied egg-rr 54.4%

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

6. Taylor expanded in angle around inf 55.0%

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

   1. *-commutative 55.0%

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

8. Simplified 55.0%

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

9. Taylor expanded in angle around inf 54.6%

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

    1. associate-*r* 55.4%

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

    2. *-commutative 55.4%

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

    3. *-commutative 55.4%

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

11. Simplified 55.4%

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

12. Taylor expanded in angle around inf 55.4%

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

    1. associate-*r* 55.5%

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

    2. *-commutative 55.5%

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

    3. *-commutative 55.5%

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

14. Simplified 55.5%

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

15. Final simplification 55.5%

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

Alternative 11: 56.4% accurate, 1.9× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} t_0 := \left(a + b\right) \cdot \left(a - b\right)\\ t_1 := -0.005555555555555556 \cdot \left(angle \cdot \pi\right)\\ \mathbf{if}\;b \leq 5 \cdot 10^{+162}:\\ \;\;\;\;t_0 \cdot \left(2 \cdot \sin t_1\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\cos \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right) \cdot t_0\right) \cdot \left(2 \cdot t_1\right)\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* (+ a b) (- a b))) (t_1 (* -0.005555555555555556 (* angle PI))))
   (if (<= b 5e+162)
     (* t_0 (* 2.0 (sin t_1)))
     (* (* (cos (* PI (* angle -0.005555555555555556))) t_0) (* 2.0 t_1)))))

C

a = abs(a);
double code(double a, double b, double angle) {
	double t_0 = (a + b) * (a - b);
	double t_1 = -0.005555555555555556 * (angle * ((double) M_PI));
	double tmp;
	if (b <= 5e+162) {
		tmp = t_0 * (2.0 * sin(t_1));
	} else {
		tmp = (cos((((double) M_PI) * (angle * -0.005555555555555556))) * t_0) * (2.0 * t_1);
	}
	return tmp;
}

Java

a = Math.abs(a);
public static double code(double a, double b, double angle) {
	double t_0 = (a + b) * (a - b);
	double t_1 = -0.005555555555555556 * (angle * Math.PI);
	double tmp;
	if (b <= 5e+162) {
		tmp = t_0 * (2.0 * Math.sin(t_1));
	} else {
		tmp = (Math.cos((Math.PI * (angle * -0.005555555555555556))) * t_0) * (2.0 * t_1);
	}
	return tmp;
}

Python

a = abs(a)
def code(a, b, angle):
	t_0 = (a + b) * (a - b)
	t_1 = -0.005555555555555556 * (angle * math.pi)
	tmp = 0
	if b <= 5e+162:
		tmp = t_0 * (2.0 * math.sin(t_1))
	else:
		tmp = (math.cos((math.pi * (angle * -0.005555555555555556))) * t_0) * (2.0 * t_1)
	return tmp

Julia

a = abs(a)
function code(a, b, angle)
	t_0 = Float64(Float64(a + b) * Float64(a - b))
	t_1 = Float64(-0.005555555555555556 * Float64(angle * pi))
	tmp = 0.0
	if (b <= 5e+162)
		tmp = Float64(t_0 * Float64(2.0 * sin(t_1)));
	else
		tmp = Float64(Float64(cos(Float64(pi * Float64(angle * -0.005555555555555556))) * t_0) * Float64(2.0 * t_1));
	end
	return tmp
end

MATLAB

a = abs(a)
function tmp_2 = code(a, b, angle)
	t_0 = (a + b) * (a - b);
	t_1 = -0.005555555555555556 * (angle * pi);
	tmp = 0.0;
	if (b <= 5e+162)
		tmp = t_0 * (2.0 * sin(t_1));
	else
		tmp = (cos((pi * (angle * -0.005555555555555556))) * t_0) * (2.0 * t_1);
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := Block[{t$95$0 = N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(-0.005555555555555556 * N[(angle * Pi), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[b, 5e+162], N[(t$95$0 * N[(2.0 * N[Sin[t$95$1], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Cos[N[(Pi * N[(angle * -0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * t$95$0), $MachinePrecision] * N[(2.0 * t$95$1), $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if b < 4.9999999999999997e162

   1. Initial program 54.6%

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

   3. Simplified 54.6%

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

      1. unpow2 54.6%

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

      2. unpow2 54.6%

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

      3. difference-of-squares 56.0%

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

   5. Applied egg-rr 56.0%

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

   6. Taylor expanded in angle around inf 56.7%

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

      1. *-commutative 56.7%

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

   8. Simplified 56.7%

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

   9. Taylor expanded in angle around 0 55.6%

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

   if 4.9999999999999997e162 < b

   1. Initial program 19.7%

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

   3. Simplified 19.7%

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

      1. unpow2 19.7%

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

      2. unpow2 19.7%

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

      3. difference-of-squares 41.1%

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

   5. Applied egg-rr 41.1%

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

   6. Taylor expanded in angle around inf 41.1%

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

      1. *-commutative 41.1%

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

   8. Simplified 41.1%

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

   9. Taylor expanded in angle around inf 44.7%

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

       1. associate-*r* 44.7%

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

       2. *-commutative 44.7%

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

       3. *-commutative 44.7%

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

   11. Simplified 44.7%

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

   12. Taylor expanded in angle around 0 44.7%

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

4. Final simplification 54.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq 5 \cdot 10^{+162}:\\ \;\;\;\;\left(\left(a + b\right) \cdot \left(a - b\right)\right) \cdot \left(2 \cdot \sin \left(-0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\cos \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \cdot \left(2 \cdot \left(-0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\\ \end{array} \]

Alternative 12: 56.3% accurate, 1.9× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} t_0 := \left(a + b\right) \cdot \left(a - b\right)\\ \mathbf{if}\;{b}^{2} \leq 10^{+290}:\\ \;\;\;\;t_0 \cdot \left(2 \cdot \sin \left(-0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;-0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot t_0\right)\right)\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* (+ a b) (- a b))))
   (if (<= (pow b 2.0) 1e+290)
     (* t_0 (* 2.0 (sin (* -0.005555555555555556 (* angle PI)))))
     (* -0.011111111111111112 (* angle (* PI t_0))))))

C

a = abs(a);
double code(double a, double b, double angle) {
	double t_0 = (a + b) * (a - b);
	double tmp;
	if (pow(b, 2.0) <= 1e+290) {
		tmp = t_0 * (2.0 * sin((-0.005555555555555556 * (angle * ((double) M_PI)))));
	} else {
		tmp = -0.011111111111111112 * (angle * (((double) M_PI) * t_0));
	}
	return tmp;
}

Java

a = Math.abs(a);
public static double code(double a, double b, double angle) {
	double t_0 = (a + b) * (a - b);
	double tmp;
	if (Math.pow(b, 2.0) <= 1e+290) {
		tmp = t_0 * (2.0 * Math.sin((-0.005555555555555556 * (angle * Math.PI))));
	} else {
		tmp = -0.011111111111111112 * (angle * (Math.PI * t_0));
	}
	return tmp;
}

Python

a = abs(a)
def code(a, b, angle):
	t_0 = (a + b) * (a - b)
	tmp = 0
	if math.pow(b, 2.0) <= 1e+290:
		tmp = t_0 * (2.0 * math.sin((-0.005555555555555556 * (angle * math.pi))))
	else:
		tmp = -0.011111111111111112 * (angle * (math.pi * t_0))
	return tmp

Julia

a = abs(a)
function code(a, b, angle)
	t_0 = Float64(Float64(a + b) * Float64(a - b))
	tmp = 0.0
	if ((b ^ 2.0) <= 1e+290)
		tmp = Float64(t_0 * Float64(2.0 * sin(Float64(-0.005555555555555556 * Float64(angle * pi)))));
	else
		tmp = Float64(-0.011111111111111112 * Float64(angle * Float64(pi * t_0)));
	end
	return tmp
end

MATLAB

a = abs(a)
function tmp_2 = code(a, b, angle)
	t_0 = (a + b) * (a - b);
	tmp = 0.0;
	if ((b ^ 2.0) <= 1e+290)
		tmp = t_0 * (2.0 * sin((-0.005555555555555556 * (angle * pi))));
	else
		tmp = -0.011111111111111112 * (angle * (pi * t_0));
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := Block[{t$95$0 = N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[b, 2.0], $MachinePrecision], 1e+290], N[(t$95$0 * N[(2.0 * N[Sin[N[(-0.005555555555555556 * N[(angle * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(-0.011111111111111112 * N[(angle * N[(Pi * t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 b 2) < 1.00000000000000006e290

   1. Initial program 56.4%

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

   3. Simplified 57.0%

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

      1. unpow2 57.0%

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

      2. unpow2 57.0%

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

      3. difference-of-squares 57.0%

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

   5. Applied egg-rr 57.0%

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

   6. Taylor expanded in angle around inf 57.4%

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

      1. *-commutative 57.4%

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

   8. Simplified 57.4%

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

   9. Taylor expanded in angle around 0 56.6%

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

   if 1.00000000000000006e290 < (pow.f64 b 2)

   1. Initial program 36.4%

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

   3. Simplified 35.0%

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

      1. unpow2 35.0%

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

      2. unpow2 35.0%

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

      3. difference-of-squares 47.6%

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

   5. Applied egg-rr 47.6%

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

   6. Taylor expanded in angle around 0 51.9%

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

4. Final simplification 55.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} \leq 10^{+290}:\\ \;\;\;\;\left(\left(a + b\right) \cdot \left(a - b\right)\right) \cdot \left(2 \cdot \sin \left(-0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;-0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right)\right)\\ \end{array} \]

Alternative 13: 56.3% accurate, 2.0× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} t_0 := \left(a + b\right) \cdot \left(a - b\right)\\ \mathbf{if}\;b \leq 9.5 \cdot 10^{+144}:\\ \;\;\;\;t_0 \cdot \left(2 \cdot \sin \left(-0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;-0.011111111111111112 \cdot \left(angle \cdot \left(\sqrt[3]{{\pi}^{3}} \cdot t_0\right)\right)\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* (+ a b) (- a b))))
   (if (<= b 9.5e+144)
     (* t_0 (* 2.0 (sin (* -0.005555555555555556 (* angle PI)))))
     (* -0.011111111111111112 (* angle (* (cbrt (pow PI 3.0)) t_0))))))

C

a = abs(a);
double code(double a, double b, double angle) {
	double t_0 = (a + b) * (a - b);
	double tmp;
	if (b <= 9.5e+144) {
		tmp = t_0 * (2.0 * sin((-0.005555555555555556 * (angle * ((double) M_PI)))));
	} else {
		tmp = -0.011111111111111112 * (angle * (cbrt(pow(((double) M_PI), 3.0)) * t_0));
	}
	return tmp;
}

Java

a = Math.abs(a);
public static double code(double a, double b, double angle) {
	double t_0 = (a + b) * (a - b);
	double tmp;
	if (b <= 9.5e+144) {
		tmp = t_0 * (2.0 * Math.sin((-0.005555555555555556 * (angle * Math.PI))));
	} else {
		tmp = -0.011111111111111112 * (angle * (Math.cbrt(Math.pow(Math.PI, 3.0)) * t_0));
	}
	return tmp;
}

Julia

a = abs(a)
function code(a, b, angle)
	t_0 = Float64(Float64(a + b) * Float64(a - b))
	tmp = 0.0
	if (b <= 9.5e+144)
		tmp = Float64(t_0 * Float64(2.0 * sin(Float64(-0.005555555555555556 * Float64(angle * pi)))));
	else
		tmp = Float64(-0.011111111111111112 * Float64(angle * Float64(cbrt((pi ^ 3.0)) * t_0)));
	end
	return tmp
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := Block[{t$95$0 = N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[b, 9.5e+144], N[(t$95$0 * N[(2.0 * N[Sin[N[(-0.005555555555555556 * N[(angle * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(-0.011111111111111112 * N[(angle * N[(N[Power[N[Power[Pi, 3.0], $MachinePrecision], 1/3], $MachinePrecision] * t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if b < 9.50000000000000031e144

   1. Initial program 55.1%

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

   3. Simplified 55.1%

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

      1. unpow2 55.1%

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

      2. unpow2 55.1%

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

      3. difference-of-squares 56.5%

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

   5. Applied egg-rr 56.5%

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

   6. Taylor expanded in angle around inf 57.2%

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

      1. *-commutative 57.2%

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

   8. Simplified 57.2%

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

   9. Taylor expanded in angle around 0 56.1%

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

   if 9.50000000000000031e144 < b

   1. Initial program 20.8%

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

   3. Simplified 20.8%

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

      1. unpow2 20.8%

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

      2. unpow2 20.8%

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

      3. difference-of-squares 39.5%

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

   5. Applied egg-rr 39.5%

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

   6. Taylor expanded in angle around 0 39.8%

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

      1. add-cbrt-cube 52.6%

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

      2. pow3 52.6%

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

   8. Applied egg-rr 39.8%

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

4. Final simplification 54.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq 9.5 \cdot 10^{+144}:\\ \;\;\;\;\left(\left(a + b\right) \cdot \left(a - b\right)\right) \cdot \left(2 \cdot \sin \left(-0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;-0.011111111111111112 \cdot \left(angle \cdot \left(\sqrt[3]{{\pi}^{3}} \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right)\right)\\ \end{array} \]

Alternative 14: 54.4% accurate, 5.5× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ -0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right)\right) \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (* -0.011111111111111112 (* angle (* PI (* (+ a b) (- a b))))))

C

a = abs(a);
double code(double a, double b, double angle) {
	return -0.011111111111111112 * (angle * (((double) M_PI) * ((a + b) * (a - b))));
}

Java

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

Python

a = abs(a)
def code(a, b, angle):
	return -0.011111111111111112 * (angle * (math.pi * ((a + b) * (a - b))))

Julia

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

MATLAB

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

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := N[(-0.011111111111111112 * N[(angle * N[(Pi * N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 50.8%

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

3. Simplified 50.8%

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

   1. unpow2 50.8%

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

   2. unpow2 50.8%

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

   3. difference-of-squares 54.4%

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

5. Applied egg-rr 54.4%

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

6. Taylor expanded in angle around 0 53.3%

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

7. Final simplification 53.3%

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

Alternative 15: 54.4% accurate, 5.5× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ -0.011111111111111112 \cdot \left(\left(\left(a + b\right) \cdot \left(a - b\right)\right) \cdot \left(angle \cdot \pi\right)\right) \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (* -0.011111111111111112 (* (* (+ a b) (- a b)) (* angle PI))))

C

a = abs(a);
double code(double a, double b, double angle) {
	return -0.011111111111111112 * (((a + b) * (a - b)) * (angle * ((double) M_PI)));
}

Java

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

Python

a = abs(a)
def code(a, b, angle):
	return -0.011111111111111112 * (((a + b) * (a - b)) * (angle * math.pi))

Julia

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

MATLAB

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

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := N[(-0.011111111111111112 * N[(N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision] * N[(angle * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 50.8%

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

3. Simplified 50.8%

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

   1. unpow2 50.8%

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

   2. unpow2 50.8%

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

   3. difference-of-squares 54.4%

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

5. Applied egg-rr 54.4%

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

   1. add-cbrt-cube 56.1%

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

   2. pow3 56.1%

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

7. Applied egg-rr 55.8%

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

8. Taylor expanded in angle around 0 53.3%

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

   1. associate-*r* 53.3%

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

   2. *-commutative 53.3%

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

10. Simplified 53.3%

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

11. Final simplification 53.3%

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

Reproduce

herbie shell --seed 2023318 
(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)))))