ab-angle->ABCF B

Percentage Accurate: 53.8% → 57.1%

Time: 32.8s

Alternatives: 10

Speedup: 3.6×

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

Specification

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \pi \cdot \frac{angle}{180}\\ \left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin t\_0\right) \cdot \cos t\_0 \end{array} \end{array} \]

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Initial Program: 53.8% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \pi \cdot \frac{angle}{180}\\ \left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin t\_0\right) \cdot \cos t\_0 \end{array} \end{array} \]

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Alternative 1: 57.1% accurate, 0.5× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := e^{\log \left(\pi \cdot \left(angle\_m \cdot 0.005555555555555556\right)\right)}\\ t_1 := \cos t\_0\\ t_2 := 2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;\frac{angle\_m}{180} \leq 5 \cdot 10^{+218}:\\ \;\;\;\;\left(t\_2 \cdot \sin \left(\frac{angle\_m}{180} \cdot \pi\right)\right) \cdot t\_1\\ \mathbf{else}:\\ \;\;\;\;t\_1 \cdot \left(t\_2 \cdot \mathsf{expm1}\left(\mathsf{log1p}\left(\sin t\_0\right)\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (exp (log (* PI (* angle_m 0.005555555555555556)))))
        (t_1 (cos t_0))
        (t_2 (* 2.0 (* (- b a) (+ b a)))))
   (*
    angle_s
    (if (<= (/ angle_m 180.0) 5e+218)
      (* (* t_2 (sin (* (/ angle_m 180.0) PI))) t_1)
      (* t_1 (* t_2 (expm1 (log1p (sin t_0)))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = exp(log((((double) M_PI) * (angle_m * 0.005555555555555556))));
	double t_1 = cos(t_0);
	double t_2 = 2.0 * ((b - a) * (b + a));
	double tmp;
	if ((angle_m / 180.0) <= 5e+218) {
		tmp = (t_2 * sin(((angle_m / 180.0) * ((double) M_PI)))) * t_1;
	} else {
		tmp = t_1 * (t_2 * expm1(log1p(sin(t_0))));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = Math.exp(Math.log((Math.PI * (angle_m * 0.005555555555555556))));
	double t_1 = Math.cos(t_0);
	double t_2 = 2.0 * ((b - a) * (b + a));
	double tmp;
	if ((angle_m / 180.0) <= 5e+218) {
		tmp = (t_2 * Math.sin(((angle_m / 180.0) * Math.PI))) * t_1;
	} else {
		tmp = t_1 * (t_2 * Math.expm1(Math.log1p(Math.sin(t_0))));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = math.exp(math.log((math.pi * (angle_m * 0.005555555555555556))))
	t_1 = math.cos(t_0)
	t_2 = 2.0 * ((b - a) * (b + a))
	tmp = 0
	if (angle_m / 180.0) <= 5e+218:
		tmp = (t_2 * math.sin(((angle_m / 180.0) * math.pi))) * t_1
	else:
		tmp = t_1 * (t_2 * math.expm1(math.log1p(math.sin(t_0))))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = exp(log(Float64(pi * Float64(angle_m * 0.005555555555555556))))
	t_1 = cos(t_0)
	t_2 = Float64(2.0 * Float64(Float64(b - a) * Float64(b + a)))
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 5e+218)
		tmp = Float64(Float64(t_2 * sin(Float64(Float64(angle_m / 180.0) * pi))) * t_1);
	else
		tmp = Float64(t_1 * Float64(t_2 * expm1(log1p(sin(t_0)))));
	end
	return Float64(angle_s * tmp)
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[Exp[N[Log[N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]}, Block[{t$95$1 = N[Cos[t$95$0], $MachinePrecision]}, Block[{t$95$2 = N[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 5e+218], N[(N[(t$95$2 * N[Sin[N[(N[(angle$95$m / 180.0), $MachinePrecision] * Pi), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * t$95$1), $MachinePrecision], N[(t$95$1 * N[(t$95$2 * N[(Exp[N[Log[1 + N[Sin[t$95$0], $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 angle 180) < 4.99999999999999983e218

   1. Initial program 51.9%

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

      1. unpow2 51.9%

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

      2. unpow2 51.9%

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

      3. difference-of-squares 56.2%

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

   4. Applied egg-rr 56.2%

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

      1. div-inv 58.7%

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

      2. metadata-eval 58.7%

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

      3. add-exp-log 27.4%

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

   6. Applied egg-rr 27.4%

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

   if 4.99999999999999983e218 < (/.f64 angle 180)

   1. Initial program 29.6%

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

      1. unpow2 29.6%

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

      2. unpow2 29.6%

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

      3. difference-of-squares 29.6%

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

   4. Applied egg-rr 29.6%

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

      1. div-inv 28.7%

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

      2. metadata-eval 28.7%

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

      3. add-exp-log 21.4%

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

   6. Applied egg-rr 21.4%

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

      1. expm1-log1p-u 21.4%

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

      2. div-inv 22.7%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \mathsf{expm1}\left(\mathsf{log1p}\left(\sin \left(\pi \cdot \color{blue}{\left(angle \cdot \frac{1}{180}\right)}\right)\right)\right)\right) \cdot \cos \left(e^{\log \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}\right) \]

      3. metadata-eval 22.7%

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

   8. Applied egg-rr 22.7%

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

      1. rem-exp-log 29.6%

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

   10. Applied egg-rr 29.6%

       \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \mathsf{expm1}\left(\mathsf{log1p}\left(\sin \color{blue}{\left(e^{\log \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}\right)}\right)\right)\right) \cdot \cos \left(e^{\log \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 27.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 5 \cdot 10^{+218}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right) \cdot \cos \left(e^{\log \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(e^{\log \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}\right) \cdot \left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \mathsf{expm1}\left(\mathsf{log1p}\left(\sin \left(e^{\log \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}\right)\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 57.6% accurate, 1.0× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\\ t_1 := \pi \cdot \left(angle\_m \cdot 0.005555555555555556\right)\\ t_2 := \cos \left(\frac{angle\_m}{180} \cdot \pi\right)\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;\frac{angle\_m}{180} \leq 5 \cdot 10^{+201}:\\ \;\;\;\;t\_2 \cdot \left(t\_0 \cdot \sin \left(\mathsf{expm1}\left(\mathsf{log1p}\left(t\_1\right)\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(t\_0 \cdot \sin t\_1\right) \cdot t\_2\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* 2.0 (* (- b a) (+ b a))))
        (t_1 (* PI (* angle_m 0.005555555555555556)))
        (t_2 (cos (* (/ angle_m 180.0) PI))))
   (*
    angle_s
    (if (<= (/ angle_m 180.0) 5e+201)
      (* t_2 (* t_0 (sin (expm1 (log1p t_1)))))
      (* (* t_0 (sin t_1)) t_2)))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b - a) * (b + a));
	double t_1 = ((double) M_PI) * (angle_m * 0.005555555555555556);
	double t_2 = cos(((angle_m / 180.0) * ((double) M_PI)));
	double tmp;
	if ((angle_m / 180.0) <= 5e+201) {
		tmp = t_2 * (t_0 * sin(expm1(log1p(t_1))));
	} else {
		tmp = (t_0 * sin(t_1)) * t_2;
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = 2.0 * ((b - a) * (b + a));
	double t_1 = Math.PI * (angle_m * 0.005555555555555556);
	double t_2 = Math.cos(((angle_m / 180.0) * Math.PI));
	double tmp;
	if ((angle_m / 180.0) <= 5e+201) {
		tmp = t_2 * (t_0 * Math.sin(Math.expm1(Math.log1p(t_1))));
	} else {
		tmp = (t_0 * Math.sin(t_1)) * t_2;
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = 2.0 * ((b - a) * (b + a))
	t_1 = math.pi * (angle_m * 0.005555555555555556)
	t_2 = math.cos(((angle_m / 180.0) * math.pi))
	tmp = 0
	if (angle_m / 180.0) <= 5e+201:
		tmp = t_2 * (t_0 * math.sin(math.expm1(math.log1p(t_1))))
	else:
		tmp = (t_0 * math.sin(t_1)) * t_2
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(2.0 * Float64(Float64(b - a) * Float64(b + a)))
	t_1 = Float64(pi * Float64(angle_m * 0.005555555555555556))
	t_2 = cos(Float64(Float64(angle_m / 180.0) * pi))
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 5e+201)
		tmp = Float64(t_2 * Float64(t_0 * sin(expm1(log1p(t_1)))));
	else
		tmp = Float64(Float64(t_0 * sin(t_1)) * t_2);
	end
	return Float64(angle_s * tmp)
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[Cos[N[(N[(angle$95$m / 180.0), $MachinePrecision] * Pi), $MachinePrecision]], $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 5e+201], N[(t$95$2 * N[(t$95$0 * N[Sin[N[(Exp[N[Log[1 + t$95$1], $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(t$95$0 * N[Sin[t$95$1], $MachinePrecision]), $MachinePrecision] * t$95$2), $MachinePrecision]]), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 angle 180) < 4.9999999999999995e201

   1. Initial program 52.6%

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

      1. unpow2 52.6%

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

      2. unpow2 52.6%

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

      3. difference-of-squares 57.1%

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

   4. Applied egg-rr 57.1%

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

      1. div-inv 58.8%

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

      2. metadata-eval 58.8%

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

      3. expm1-log1p-u 50.9%

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

   6. Applied egg-rr 50.9%

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

   if 4.9999999999999995e201 < (/.f64 angle 180)

   1. Initial program 28.5%

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

      1. unpow2 28.5%

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

      2. unpow2 28.5%

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

      3. difference-of-squares 28.5%

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

   4. Applied egg-rr 28.5%

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

   5. Taylor expanded in angle around inf 23.1%

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

      1. *-commutative 17.1%

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

      2. *-commutative 17.1%

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

      3. associate-*r* 20.8%

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

   7. Simplified 29.9%

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

4. Final simplification 49.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 5 \cdot 10^{+201}:\\ \;\;\;\;\cos \left(\frac{angle}{180} \cdot \pi\right) \cdot \left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sin \left(\mathsf{expm1}\left(\mathsf{log1p}\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \cos \left(\frac{angle}{180} \cdot \pi\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 57.3% accurate, 1.0× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \frac{angle\_m}{180} \cdot \pi\\ t_1 := \pi \cdot \left(angle\_m \cdot 0.005555555555555556\right)\\ t_2 := 2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;\frac{angle\_m}{180} \leq 4 \cdot 10^{+210}:\\ \;\;\;\;\left(t\_2 \cdot \sin t\_0\right) \cdot \cos \left(e^{\log t\_1}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(t\_2 \cdot \sin t\_1\right) \cdot \cos t\_0\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* (/ angle_m 180.0) PI))
        (t_1 (* PI (* angle_m 0.005555555555555556)))
        (t_2 (* 2.0 (* (- b a) (+ b a)))))
   (*
    angle_s
    (if (<= (/ angle_m 180.0) 4e+210)
      (* (* t_2 (sin t_0)) (cos (exp (log t_1))))
      (* (* t_2 (sin t_1)) (cos t_0))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (angle_m / 180.0) * ((double) M_PI);
	double t_1 = ((double) M_PI) * (angle_m * 0.005555555555555556);
	double t_2 = 2.0 * ((b - a) * (b + a));
	double tmp;
	if ((angle_m / 180.0) <= 4e+210) {
		tmp = (t_2 * sin(t_0)) * cos(exp(log(t_1)));
	} else {
		tmp = (t_2 * sin(t_1)) * cos(t_0);
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (angle_m / 180.0) * Math.PI;
	double t_1 = Math.PI * (angle_m * 0.005555555555555556);
	double t_2 = 2.0 * ((b - a) * (b + a));
	double tmp;
	if ((angle_m / 180.0) <= 4e+210) {
		tmp = (t_2 * Math.sin(t_0)) * Math.cos(Math.exp(Math.log(t_1)));
	} else {
		tmp = (t_2 * Math.sin(t_1)) * Math.cos(t_0);
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = (angle_m / 180.0) * math.pi
	t_1 = math.pi * (angle_m * 0.005555555555555556)
	t_2 = 2.0 * ((b - a) * (b + a))
	tmp = 0
	if (angle_m / 180.0) <= 4e+210:
		tmp = (t_2 * math.sin(t_0)) * math.cos(math.exp(math.log(t_1)))
	else:
		tmp = (t_2 * math.sin(t_1)) * math.cos(t_0)
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(Float64(angle_m / 180.0) * pi)
	t_1 = Float64(pi * Float64(angle_m * 0.005555555555555556))
	t_2 = Float64(2.0 * Float64(Float64(b - a) * Float64(b + a)))
	tmp = 0.0
	if (Float64(angle_m / 180.0) <= 4e+210)
		tmp = Float64(Float64(t_2 * sin(t_0)) * cos(exp(log(t_1))));
	else
		tmp = Float64(Float64(t_2 * sin(t_1)) * cos(t_0));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = (angle_m / 180.0) * pi;
	t_1 = pi * (angle_m * 0.005555555555555556);
	t_2 = 2.0 * ((b - a) * (b + a));
	tmp = 0.0;
	if ((angle_m / 180.0) <= 4e+210)
		tmp = (t_2 * sin(t_0)) * cos(exp(log(t_1)));
	else
		tmp = (t_2 * sin(t_1)) * cos(t_0);
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(N[(angle$95$m / 180.0), $MachinePrecision] * Pi), $MachinePrecision]}, Block[{t$95$1 = N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[(angle$95$m / 180.0), $MachinePrecision], 4e+210], N[(N[(t$95$2 * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision] * N[Cos[N[Exp[N[Log[t$95$1], $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[(t$95$2 * N[Sin[t$95$1], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 angle 180) < 3.99999999999999971e210

   1. Initial program 52.1%

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

      1. unpow2 52.1%

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

      2. unpow2 52.1%

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

      3. difference-of-squares 56.5%

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

   4. Applied egg-rr 56.5%

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

      1. div-inv 59.0%

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

      2. metadata-eval 59.0%

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

      3. add-exp-log 27.2%

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

   6. Applied egg-rr 27.2%

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

   if 3.99999999999999971e210 < (/.f64 angle 180)

   1. Initial program 31.7%

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

      1. unpow2 31.7%

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

      2. unpow2 31.7%

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

      3. difference-of-squares 31.7%

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

   4. Applied egg-rr 31.7%

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

   5. Taylor expanded in angle around inf 25.4%

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

      1. *-commutative 16.5%

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

      2. *-commutative 16.5%

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

      3. associate-*r* 21.5%

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

   7. Simplified 32.5%

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

4. Final simplification 27.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{angle}{180} \leq 4 \cdot 10^{+210}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right) \cdot \cos \left(e^{\log \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \cos \left(\frac{angle}{180} \cdot \pi\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 55.6% accurate, 1.0× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \frac{angle\_m}{180} \cdot \pi\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;b \leq 4.4 \cdot 10^{+156}:\\ \;\;\;\;\left(\sin t\_0 \cdot \left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right)\right) \cdot \cos \left(\pi \cdot \left(angle\_m \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\cos t\_0 \cdot \left(0.011111111111111112 \cdot \left(angle\_m \cdot \left(\pi \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right)\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* (/ angle_m 180.0) PI)))
   (*
    angle_s
    (if (<= b 4.4e+156)
      (*
       (* (sin t_0) (* 2.0 (- (pow b 2.0) (pow a 2.0))))
       (cos (* PI (* angle_m 0.005555555555555556))))
      (*
       (cos t_0)
       (* 0.011111111111111112 (* angle_m (* PI (* (- b a) (+ b a))))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (angle_m / 180.0) * ((double) M_PI);
	double tmp;
	if (b <= 4.4e+156) {
		tmp = (sin(t_0) * (2.0 * (pow(b, 2.0) - pow(a, 2.0)))) * cos((((double) M_PI) * (angle_m * 0.005555555555555556)));
	} else {
		tmp = cos(t_0) * (0.011111111111111112 * (angle_m * (((double) M_PI) * ((b - a) * (b + a)))));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (angle_m / 180.0) * Math.PI;
	double tmp;
	if (b <= 4.4e+156) {
		tmp = (Math.sin(t_0) * (2.0 * (Math.pow(b, 2.0) - Math.pow(a, 2.0)))) * Math.cos((Math.PI * (angle_m * 0.005555555555555556)));
	} else {
		tmp = Math.cos(t_0) * (0.011111111111111112 * (angle_m * (Math.PI * ((b - a) * (b + a)))));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = (angle_m / 180.0) * math.pi
	tmp = 0
	if b <= 4.4e+156:
		tmp = (math.sin(t_0) * (2.0 * (math.pow(b, 2.0) - math.pow(a, 2.0)))) * math.cos((math.pi * (angle_m * 0.005555555555555556)))
	else:
		tmp = math.cos(t_0) * (0.011111111111111112 * (angle_m * (math.pi * ((b - a) * (b + a)))))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(Float64(angle_m / 180.0) * pi)
	tmp = 0.0
	if (b <= 4.4e+156)
		tmp = Float64(Float64(sin(t_0) * Float64(2.0 * Float64((b ^ 2.0) - (a ^ 2.0)))) * cos(Float64(pi * Float64(angle_m * 0.005555555555555556))));
	else
		tmp = Float64(cos(t_0) * Float64(0.011111111111111112 * Float64(angle_m * Float64(pi * Float64(Float64(b - a) * Float64(b + a))))));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = (angle_m / 180.0) * pi;
	tmp = 0.0;
	if (b <= 4.4e+156)
		tmp = (sin(t_0) * (2.0 * ((b ^ 2.0) - (a ^ 2.0)))) * cos((pi * (angle_m * 0.005555555555555556)));
	else
		tmp = cos(t_0) * (0.011111111111111112 * (angle_m * (pi * ((b - a) * (b + a)))));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(N[(angle$95$m / 180.0), $MachinePrecision] * Pi), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[b, 4.4e+156], N[(N[(N[Sin[t$95$0], $MachinePrecision] * N[(2.0 * N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Cos[N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[Cos[t$95$0], $MachinePrecision] * N[(0.011111111111111112 * N[(angle$95$m * N[(Pi * N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if b < 4.40000000000000008e156

   1. Initial program 54.3%

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

   3. Taylor expanded in angle around inf 55.9%

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

      1. *-commutative 57.7%

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

      2. *-commutative 57.7%

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

      3. associate-*r* 58.3%

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

   5. Simplified 56.0%

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

   if 4.40000000000000008e156 < b

   1. Initial program 21.3%

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

      1. unpow2 21.3%

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

      2. unpow2 21.3%

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

      3. difference-of-squares 39.1%

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

   4. Applied egg-rr 39.1%

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

   5. Taylor expanded in angle around 0 55.7%

      \[\leadsto \color{blue}{\left(0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right)\right)\right)} \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 56.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq 4.4 \cdot 10^{+156}:\\ \;\;\;\;\left(\sin \left(\frac{angle}{180} \cdot \pi\right) \cdot \left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\frac{angle}{180} \cdot \pi\right) \cdot \left(0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 56.5% accurate, 1.9× speedup

Math

\[\begin{array}{l} angle_m = \left|angle\right| \\ angle_s = \mathsf{copysign}\left(1, angle\right) \\ \begin{array}{l} t_0 := \left(b - a\right) \cdot \left(b + a\right)\\ angle\_s \cdot \begin{array}{l} \mathbf{if}\;{b}^{2} \leq 2 \cdot 10^{+293}:\\ \;\;\;\;\left(2 \cdot t\_0\right) \cdot \sin \left(\pi \cdot \left(angle\_m \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\frac{angle\_m}{180} \cdot \pi\right) \cdot \left(0.011111111111111112 \cdot \left(angle\_m \cdot \left(\pi \cdot t\_0\right)\right)\right)\\ \end{array} \end{array} \end{array} \]

FPCore

angle_m = (fabs.f64 angle)
angle_s = (copysign.f64 1 angle)
(FPCore (angle_s a b angle_m)
 :precision binary64
 (let* ((t_0 (* (- b a) (+ b a))))
   (*
    angle_s
    (if (<= (pow b 2.0) 2e+293)
      (* (* 2.0 t_0) (sin (* PI (* angle_m 0.005555555555555556))))
      (*
       (cos (* (/ angle_m 180.0) PI))
       (* 0.011111111111111112 (* angle_m (* PI t_0))))))))

C

angle_m = fabs(angle);
angle_s = copysign(1.0, angle);
double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (b - a) * (b + a);
	double tmp;
	if (pow(b, 2.0) <= 2e+293) {
		tmp = (2.0 * t_0) * sin((((double) M_PI) * (angle_m * 0.005555555555555556)));
	} else {
		tmp = cos(((angle_m / 180.0) * ((double) M_PI))) * (0.011111111111111112 * (angle_m * (((double) M_PI) * t_0)));
	}
	return angle_s * tmp;
}

Java

angle_m = Math.abs(angle);
angle_s = Math.copySign(1.0, angle);
public static double code(double angle_s, double a, double b, double angle_m) {
	double t_0 = (b - a) * (b + a);
	double tmp;
	if (Math.pow(b, 2.0) <= 2e+293) {
		tmp = (2.0 * t_0) * Math.sin((Math.PI * (angle_m * 0.005555555555555556)));
	} else {
		tmp = Math.cos(((angle_m / 180.0) * Math.PI)) * (0.011111111111111112 * (angle_m * (Math.PI * t_0)));
	}
	return angle_s * tmp;
}

Python

angle_m = math.fabs(angle)
angle_s = math.copysign(1.0, angle)
def code(angle_s, a, b, angle_m):
	t_0 = (b - a) * (b + a)
	tmp = 0
	if math.pow(b, 2.0) <= 2e+293:
		tmp = (2.0 * t_0) * math.sin((math.pi * (angle_m * 0.005555555555555556)))
	else:
		tmp = math.cos(((angle_m / 180.0) * math.pi)) * (0.011111111111111112 * (angle_m * (math.pi * t_0)))
	return angle_s * tmp

Julia

angle_m = abs(angle)
angle_s = copysign(1.0, angle)
function code(angle_s, a, b, angle_m)
	t_0 = Float64(Float64(b - a) * Float64(b + a))
	tmp = 0.0
	if ((b ^ 2.0) <= 2e+293)
		tmp = Float64(Float64(2.0 * t_0) * sin(Float64(pi * Float64(angle_m * 0.005555555555555556))));
	else
		tmp = Float64(cos(Float64(Float64(angle_m / 180.0) * pi)) * Float64(0.011111111111111112 * Float64(angle_m * Float64(pi * t_0))));
	end
	return Float64(angle_s * tmp)
end

MATLAB

angle_m = abs(angle);
angle_s = sign(angle) * abs(1.0);
function tmp_2 = code(angle_s, a, b, angle_m)
	t_0 = (b - a) * (b + a);
	tmp = 0.0;
	if ((b ^ 2.0) <= 2e+293)
		tmp = (2.0 * t_0) * sin((pi * (angle_m * 0.005555555555555556)));
	else
		tmp = cos(((angle_m / 180.0) * pi)) * (0.011111111111111112 * (angle_m * (pi * t_0)));
	end
	tmp_2 = angle_s * tmp;
end

Wolfram

angle_m = N[Abs[angle], $MachinePrecision]
angle_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[angle]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[angle$95$s_, a_, b_, angle$95$m_] := Block[{t$95$0 = N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]}, N[(angle$95$s * If[LessEqual[N[Power[b, 2.0], $MachinePrecision], 2e+293], N[(N[(2.0 * t$95$0), $MachinePrecision] * N[Sin[N[(Pi * N[(angle$95$m * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[Cos[N[(N[(angle$95$m / 180.0), $MachinePrecision] * Pi), $MachinePrecision]], $MachinePrecision] * N[(0.011111111111111112 * N[(angle$95$m * N[(Pi * t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 b 2) < 1.9999999999999998e293

   1. Initial program 57.6%

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

      1. unpow2 57.6%

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

      2. unpow2 57.6%

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

      3. difference-of-squares 57.6%

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

   4. Applied egg-rr 57.6%

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

   5. Taylor expanded in angle around 0 55.7%

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

   6. Taylor expanded in angle around inf 55.7%

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

      1. *-commutative 55.7%

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

      2. *-commutative 55.7%

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

      3. associate-*r* 56.1%

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

   8. Simplified 56.1%

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

   if 1.9999999999999998e293 < (pow.f64 b 2)

   1. Initial program 28.5%

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

      1. unpow2 28.5%

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

      2. unpow2 28.5%

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

      3. difference-of-squares 45.1%

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

   4. Applied egg-rr 45.1%

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

   5. Taylor expanded in angle around 0 54.7%

      \[\leadsto \color{blue}{\left(0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right)\right)\right)} \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 55.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{b}^{2} \leq 2 \cdot 10^{+293}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\frac{angle}{180} \cdot \pi\right) \cdot \left(0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 57.5% accurate, 1.9× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 50.5%

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

   1. unpow2 50.5%

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

   2. unpow2 50.5%

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

   3. difference-of-squares 54.6%

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

4. Applied egg-rr 54.6%

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

5. Taylor expanded in angle around inf 55.5%

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

   1. *-commutative 53.1%

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

   2. *-commutative 53.1%

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

   3. associate-*r* 52.6%

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

7. Simplified 56.2%

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

8. Final simplification 56.2%

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

Alternative 7: 57.8% accurate, 1.9× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 50.5%

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

   1. unpow2 50.5%

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

   2. unpow2 50.5%

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

   3. difference-of-squares 54.6%

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

4. Applied egg-rr 54.6%

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

5. Taylor expanded in angle around inf 55.5%

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

   1. *-commutative 55.5%

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

   2. *-commutative 55.5%

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

   3. associate-*r* 56.8%

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

7. Simplified 56.8%

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

8. Final simplification 56.8%

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

Alternative 8: 56.4% accurate, 3.6× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 50.5%

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

   1. unpow2 50.5%

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

   2. unpow2 50.5%

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

   3. difference-of-squares 54.6%

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

4. Applied egg-rr 54.6%

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

5. Taylor expanded in angle around 0 53.5%

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

6. Taylor expanded in angle around inf 53.1%

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

7. Final simplification 53.1%

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

Alternative 9: 56.2% accurate, 3.6× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 50.5%

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

   1. unpow2 50.5%

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

   2. unpow2 50.5%

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

   3. difference-of-squares 54.6%

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

4. Applied egg-rr 54.6%

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

5. Taylor expanded in angle around 0 53.5%

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

6. Final simplification 53.5%

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

Alternative 10: 53.7% accurate, 32.2× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 50.5%

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

   1. unpow2 50.5%

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

   2. unpow2 50.5%

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

   3. difference-of-squares 54.6%

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

4. Applied egg-rr 54.6%

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

5. Taylor expanded in angle around 0 53.5%

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

6. Taylor expanded in angle around 0 52.1%

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

7. Final simplification 52.1%

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

Reproduce

herbie shell --seed 2024026 
(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)))))