ab-angle->ABCF B

Percentage Accurate: 53.3% → 56.4%

Time: 47.3s

Alternatives: 10

Speedup: 32.2×

• 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: 53.3% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Alternative 1: 56.4% accurate, 0.8× speedup

Math

\[\begin{array}{l} a_m = \left|a\right| \\ \begin{array}{l} t_0 := \pi \cdot \frac{angle}{180}\\ t_1 := 2 \cdot \left(\left(a_m + b\right) \cdot \left(b - a_m\right)\right)\\ \mathbf{if}\;a_m \leq 2.7 \cdot 10^{+25}:\\ \;\;\;\;\cos t_0 \cdot \left(t_1 \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\\ \mathbf{elif}\;a_m \leq 1.2 \cdot 10^{+193}:\\ \;\;\;\;\left(t_1 \cdot \sin t_0\right) \cdot \cos \left(\frac{angle}{180} \cdot {\left(\sqrt[3]{{\pi}^{1.5}}\right)}^{2}\right)\\ \mathbf{else}:\\ \;\;\;\;t_1 \cdot \left(-2.8577960676726107 \cdot 10^{-8} \cdot \left({angle}^{3} \cdot {\pi}^{3}\right) + 0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\\ \end{array} \end{array} \]

FPCore

a_m = (fabs.f64 a)
(FPCore (a_m b angle)
 :precision binary64
 (let* ((t_0 (* PI (/ angle 180.0))) (t_1 (* 2.0 (* (+ a_m b) (- b a_m)))))
   (if (<= a_m 2.7e+25)
     (* (cos t_0) (* t_1 (sin (* PI (* angle 0.005555555555555556)))))
     (if (<= a_m 1.2e+193)
       (*
        (* t_1 (sin t_0))
        (cos (* (/ angle 180.0) (pow (cbrt (pow PI 1.5)) 2.0))))
       (*
        t_1
        (+
         (* -2.8577960676726107e-8 (* (pow angle 3.0) (pow PI 3.0)))
         (* 0.005555555555555556 (* PI angle))))))))

C

a_m = fabs(a);
double code(double a_m, double b, double angle) {
	double t_0 = ((double) M_PI) * (angle / 180.0);
	double t_1 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (a_m <= 2.7e+25) {
		tmp = cos(t_0) * (t_1 * sin((((double) M_PI) * (angle * 0.005555555555555556))));
	} else if (a_m <= 1.2e+193) {
		tmp = (t_1 * sin(t_0)) * cos(((angle / 180.0) * pow(cbrt(pow(((double) M_PI), 1.5)), 2.0)));
	} else {
		tmp = t_1 * ((-2.8577960676726107e-8 * (pow(angle, 3.0) * pow(((double) M_PI), 3.0))) + (0.005555555555555556 * (((double) M_PI) * angle)));
	}
	return tmp;
}

Java

a_m = Math.abs(a);
public static double code(double a_m, double b, double angle) {
	double t_0 = Math.PI * (angle / 180.0);
	double t_1 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (a_m <= 2.7e+25) {
		tmp = Math.cos(t_0) * (t_1 * Math.sin((Math.PI * (angle * 0.005555555555555556))));
	} else if (a_m <= 1.2e+193) {
		tmp = (t_1 * Math.sin(t_0)) * Math.cos(((angle / 180.0) * Math.pow(Math.cbrt(Math.pow(Math.PI, 1.5)), 2.0)));
	} else {
		tmp = t_1 * ((-2.8577960676726107e-8 * (Math.pow(angle, 3.0) * Math.pow(Math.PI, 3.0))) + (0.005555555555555556 * (Math.PI * angle)));
	}
	return tmp;
}

Julia

a_m = abs(a)
function code(a_m, b, angle)
	t_0 = Float64(pi * Float64(angle / 180.0))
	t_1 = Float64(2.0 * Float64(Float64(a_m + b) * Float64(b - a_m)))
	tmp = 0.0
	if (a_m <= 2.7e+25)
		tmp = Float64(cos(t_0) * Float64(t_1 * sin(Float64(pi * Float64(angle * 0.005555555555555556)))));
	elseif (a_m <= 1.2e+193)
		tmp = Float64(Float64(t_1 * sin(t_0)) * cos(Float64(Float64(angle / 180.0) * (cbrt((pi ^ 1.5)) ^ 2.0))));
	else
		tmp = Float64(t_1 * Float64(Float64(-2.8577960676726107e-8 * Float64((angle ^ 3.0) * (pi ^ 3.0))) + Float64(0.005555555555555556 * Float64(pi * angle))));
	end
	return tmp
end

Wolfram

a_m = N[Abs[a], $MachinePrecision]
code[a$95$m_, b_, angle_] := Block[{t$95$0 = N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(2.0 * N[(N[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[a$95$m, 2.7e+25], N[(N[Cos[t$95$0], $MachinePrecision] * N[(t$95$1 * N[Sin[N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[a$95$m, 1.2e+193], N[(N[(t$95$1 * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(N[(angle / 180.0), $MachinePrecision] * N[Power[N[Power[N[Power[Pi, 1.5], $MachinePrecision], 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(t$95$1 * N[(N[(-2.8577960676726107e-8 * N[(N[Power[angle, 3.0], $MachinePrecision] * N[Power[Pi, 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(0.005555555555555556 * N[(Pi * angle), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if a < 2.7e25

   1. Initial program 56.1%

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

      1. unpow2 56.1%

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

      2. unpow2 56.1%

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

      3. difference-of-squares 59.6%

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

   4. Applied egg-rr 59.6%

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

      1. add-sqr-sqrt 28.6%

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

      2. sqrt-unprod 37.7%

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

      3. pow2 37.7%

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

      4. div-inv 37.7%

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

      5. metadata-eval 37.7%

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

   6. Applied egg-rr 37.7%

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

      1. sqrt-pow1 60.8%

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

      2. metadata-eval 60.8%

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

      3. pow1 60.8%

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

   8. Applied egg-rr 60.8%

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

   if 2.7e25 < a < 1.2e193

   1. Initial program 56.6%

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

      1. unpow2 56.6%

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

      2. unpow2 56.6%

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

      3. difference-of-squares 59.0%

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

   4. Applied egg-rr 59.0%

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

      1. add-sqr-sqrt 60.8%

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

      2. pow2 60.8%

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

   6. Applied egg-rr 60.8%

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

      1. add-cbrt-cube 60.8%

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

      2. sqrt-unprod 60.8%

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

      3. sqrt-prod 67.3%

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

      4. unpow3 67.3%

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

      5. sqrt-pow1 67.3%

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

      6. pow1/3 60.8%

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

      7. metadata-eval 60.8%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left({\left({\left({\pi}^{\color{blue}{1.5}}\right)}^{0.3333333333333333}\right)}^{2} \cdot \frac{angle}{180}\right) \]

   8. Applied egg-rr 60.8%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left({\color{blue}{\left({\left({\pi}^{1.5}\right)}^{0.3333333333333333}\right)}}^{2} \cdot \frac{angle}{180}\right) \]
   9. Step-by-step derivation

      1. unpow1/3 67.3%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left({\color{blue}{\left(\sqrt[3]{{\pi}^{1.5}}\right)}}^{2} \cdot \frac{angle}{180}\right) \]

   10. Simplified 67.3%

       \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left({\color{blue}{\left(\sqrt[3]{{\pi}^{1.5}}\right)}}^{2} \cdot \frac{angle}{180}\right) \]

   if 1.2e193 < a

   1. Initial program 41.7%

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

      1. unpow2 41.7%

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

      2. unpow2 41.7%

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

      3. difference-of-squares 42.5%

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

   4. Applied egg-rr 42.5%

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

   5. Taylor expanded in angle around 0 42.5%

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

   6. Taylor expanded in angle around 0 62.5%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \color{blue}{\left(-2.8577960676726107 \cdot 10^{-8} \cdot \left({angle}^{3} \cdot {\pi}^{3}\right) + 0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}\right) \cdot 1 \]
3. Recombined 3 regimes into one program.

4. Final simplification 62.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 2.7 \cdot 10^{+25}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\\ \mathbf{elif}\;a \leq 1.2 \cdot 10^{+193}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\frac{angle}{180} \cdot {\left(\sqrt[3]{{\pi}^{1.5}}\right)}^{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \left(-2.8577960676726107 \cdot 10^{-8} \cdot \left({angle}^{3} \cdot {\pi}^{3}\right) + 0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 57.3% accurate, 0.8× speedup

Math

\[\begin{array}{l} a_m = \left|a\right| \\ \begin{array}{l} t_0 := \cos \left(\pi \cdot \frac{angle}{180}\right)\\ t_1 := 2 \cdot \left(\left(a_m + b\right) \cdot \left(b - a_m\right)\right)\\ \mathbf{if}\;a_m \leq 2 \cdot 10^{+173}:\\ \;\;\;\;t_0 \cdot \left(t_1 \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;t_0 \cdot \left(t_1 \cdot \sin \left(\frac{angle}{180} \cdot \left(\sqrt[3]{\pi} \cdot {\left(\sqrt[3]{\pi}\right)}^{2}\right)\right)\right)\\ \end{array} \end{array} \]

FPCore

a_m = (fabs.f64 a)
(FPCore (a_m b angle)
 :precision binary64
 (let* ((t_0 (cos (* PI (/ angle 180.0))))
        (t_1 (* 2.0 (* (+ a_m b) (- b a_m)))))
   (if (<= a_m 2e+173)
     (* t_0 (* t_1 (sin (* PI (* angle 0.005555555555555556)))))
     (*
      t_0
      (* t_1 (sin (* (/ angle 180.0) (* (cbrt PI) (pow (cbrt PI) 2.0)))))))))

C

a_m = fabs(a);
double code(double a_m, double b, double angle) {
	double t_0 = cos((((double) M_PI) * (angle / 180.0)));
	double t_1 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (a_m <= 2e+173) {
		tmp = t_0 * (t_1 * sin((((double) M_PI) * (angle * 0.005555555555555556))));
	} else {
		tmp = t_0 * (t_1 * sin(((angle / 180.0) * (cbrt(((double) M_PI)) * pow(cbrt(((double) M_PI)), 2.0)))));
	}
	return tmp;
}

Java

a_m = Math.abs(a);
public static double code(double a_m, double b, double angle) {
	double t_0 = Math.cos((Math.PI * (angle / 180.0)));
	double t_1 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (a_m <= 2e+173) {
		tmp = t_0 * (t_1 * Math.sin((Math.PI * (angle * 0.005555555555555556))));
	} else {
		tmp = t_0 * (t_1 * Math.sin(((angle / 180.0) * (Math.cbrt(Math.PI) * Math.pow(Math.cbrt(Math.PI), 2.0)))));
	}
	return tmp;
}

Julia

a_m = abs(a)
function code(a_m, b, angle)
	t_0 = cos(Float64(pi * Float64(angle / 180.0)))
	t_1 = Float64(2.0 * Float64(Float64(a_m + b) * Float64(b - a_m)))
	tmp = 0.0
	if (a_m <= 2e+173)
		tmp = Float64(t_0 * Float64(t_1 * sin(Float64(pi * Float64(angle * 0.005555555555555556)))));
	else
		tmp = Float64(t_0 * Float64(t_1 * sin(Float64(Float64(angle / 180.0) * Float64(cbrt(pi) * (cbrt(pi) ^ 2.0))))));
	end
	return tmp
end

Wolfram

a_m = N[Abs[a], $MachinePrecision]
code[a$95$m_, b_, angle_] := Block[{t$95$0 = N[Cos[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]}, Block[{t$95$1 = N[(2.0 * N[(N[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[a$95$m, 2e+173], N[(t$95$0 * N[(t$95$1 * N[Sin[N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(t$95$0 * N[(t$95$1 * N[Sin[N[(N[(angle / 180.0), $MachinePrecision] * N[(N[Power[Pi, 1/3], $MachinePrecision] * N[Power[N[Power[Pi, 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if a < 2e173

   1. Initial program 57.0%

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

      1. unpow2 57.0%

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

      2. unpow2 57.0%

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

      3. difference-of-squares 60.5%

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

   4. Applied egg-rr 60.5%

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

      1. add-sqr-sqrt 28.8%

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

      2. sqrt-unprod 34.8%

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

      3. pow2 34.8%

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

      4. div-inv 34.8%

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

      5. metadata-eval 34.8%

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

   6. Applied egg-rr 34.8%

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

      1. sqrt-pow1 61.4%

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

      2. metadata-eval 61.4%

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

      3. pow1 61.4%

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

   8. Applied egg-rr 61.4%

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

   if 2e173 < a

   1. Initial program 33.8%

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

      1. unpow2 33.8%

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

      2. unpow2 33.8%

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

      3. difference-of-squares 34.4%

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

   4. Applied egg-rr 34.4%

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

      1. add-cube-cbrt 66.0%

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

      2. pow2 66.0%

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

   6. Applied egg-rr 66.0%

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

4. Final simplification 61.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 2 \cdot 10^{+173}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{angle}{180} \cdot \left(\sqrt[3]{\pi} \cdot {\left(\sqrt[3]{\pi}\right)}^{2}\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 57.1% accurate, 0.8× speedup

Math

\[\begin{array}{l} a_m = \left|a\right| \\ \begin{array}{l} t_0 := \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ t_1 := 2 \cdot \left(\left(a_m + b\right) \cdot \left(b - a_m\right)\right)\\ \mathbf{if}\;a_m \leq 2.6 \cdot 10^{+14}:\\ \;\;\;\;\left(t_1 \cdot \sin \left(\frac{\pi}{\frac{180}{angle}}\right)\right) \cdot t_0\\ \mathbf{else}:\\ \;\;\;\;\left(t_1 \cdot \sin \left(\frac{angle}{180} \cdot \left(\sqrt[3]{\pi} \cdot {\left(\sqrt[3]{\pi}\right)}^{2}\right)\right)\right) \cdot t_0\\ \end{array} \end{array} \]

FPCore

a_m = (fabs.f64 a)
(FPCore (a_m b angle)
 :precision binary64
 (let* ((t_0 (cos (* PI (* angle 0.005555555555555556))))
        (t_1 (* 2.0 (* (+ a_m b) (- b a_m)))))
   (if (<= a_m 2.6e+14)
     (* (* t_1 (sin (/ PI (/ 180.0 angle)))) t_0)
     (*
      (* t_1 (sin (* (/ angle 180.0) (* (cbrt PI) (pow (cbrt PI) 2.0)))))
      t_0))))

C

a_m = fabs(a);
double code(double a_m, double b, double angle) {
	double t_0 = cos((((double) M_PI) * (angle * 0.005555555555555556)));
	double t_1 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (a_m <= 2.6e+14) {
		tmp = (t_1 * sin((((double) M_PI) / (180.0 / angle)))) * t_0;
	} else {
		tmp = (t_1 * sin(((angle / 180.0) * (cbrt(((double) M_PI)) * pow(cbrt(((double) M_PI)), 2.0))))) * t_0;
	}
	return tmp;
}

Java

a_m = Math.abs(a);
public static double code(double a_m, double b, double angle) {
	double t_0 = Math.cos((Math.PI * (angle * 0.005555555555555556)));
	double t_1 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (a_m <= 2.6e+14) {
		tmp = (t_1 * Math.sin((Math.PI / (180.0 / angle)))) * t_0;
	} else {
		tmp = (t_1 * Math.sin(((angle / 180.0) * (Math.cbrt(Math.PI) * Math.pow(Math.cbrt(Math.PI), 2.0))))) * t_0;
	}
	return tmp;
}

Julia

a_m = abs(a)
function code(a_m, b, angle)
	t_0 = cos(Float64(pi * Float64(angle * 0.005555555555555556)))
	t_1 = Float64(2.0 * Float64(Float64(a_m + b) * Float64(b - a_m)))
	tmp = 0.0
	if (a_m <= 2.6e+14)
		tmp = Float64(Float64(t_1 * sin(Float64(pi / Float64(180.0 / angle)))) * t_0);
	else
		tmp = Float64(Float64(t_1 * sin(Float64(Float64(angle / 180.0) * Float64(cbrt(pi) * (cbrt(pi) ^ 2.0))))) * t_0);
	end
	return tmp
end

Wolfram

a_m = N[Abs[a], $MachinePrecision]
code[a$95$m_, b_, angle_] := Block[{t$95$0 = N[Cos[N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]}, Block[{t$95$1 = N[(2.0 * N[(N[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[a$95$m, 2.6e+14], N[(N[(t$95$1 * N[Sin[N[(Pi / N[(180.0 / angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * t$95$0), $MachinePrecision], N[(N[(t$95$1 * N[Sin[N[(N[(angle / 180.0), $MachinePrecision] * N[(N[Power[Pi, 1/3], $MachinePrecision] * N[Power[N[Power[Pi, 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * t$95$0), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if a < 2.6e14

   1. Initial program 56.8%

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

      1. unpow2 56.8%

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

      2. unpow2 56.8%

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

      3. difference-of-squares 60.5%

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

   4. Applied egg-rr 60.5%

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

   5. Taylor expanded in angle around inf 60.3%

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

      1. associate-*r* 60.6%

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

      2. *-commutative 60.6%

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

      3. *-commutative 60.6%

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

      4. *-commutative 60.6%

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

   7. Simplified 60.6%

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

      1. clear-num 60.9%

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

      2. un-div-inv 60.8%

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

   9. Applied egg-rr 60.8%

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

   if 2.6e14 < a

   1. Initial program 50.7%

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

      1. unpow2 50.7%

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

      2. unpow2 50.7%

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

      3. difference-of-squares 52.5%

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

   4. Applied egg-rr 52.5%

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

   5. Taylor expanded in angle around inf 53.5%

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

      1. associate-*r* 53.7%

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

      2. *-commutative 53.7%

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

      3. *-commutative 53.7%

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

      4. *-commutative 53.7%

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

   7. Simplified 53.7%

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

      1. add-cube-cbrt 62.7%

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

      2. pow2 62.7%

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

   9. Applied egg-rr 61.3%

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

4. Final simplification 60.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 2.6 \cdot 10^{+14}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi}{\frac{180}{angle}}\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{angle}{180} \cdot \left(\sqrt[3]{\pi} \cdot {\left(\sqrt[3]{\pi}\right)}^{2}\right)\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 57.1% accurate, 1.0× speedup

Math

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

FPCore

a_m = (fabs.f64 a)
(FPCore (a_m b angle)
 :precision binary64
 (let* ((t_0 (* PI (* angle 0.005555555555555556)))
        (t_1 (* 2.0 (* (+ a_m b) (- b a_m)))))
   (if (<= (- (pow b 2.0) (pow a_m 2.0)) (- INFINITY))
     (* t_1 (sin (* 0.005555555555555556 (* PI angle))))
     (* (cos t_0) (* t_1 (sin t_0))))))

C

a_m = fabs(a);
double code(double a_m, double b, double angle) {
	double t_0 = ((double) M_PI) * (angle * 0.005555555555555556);
	double t_1 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if ((pow(b, 2.0) - pow(a_m, 2.0)) <= -((double) INFINITY)) {
		tmp = t_1 * sin((0.005555555555555556 * (((double) M_PI) * angle)));
	} else {
		tmp = cos(t_0) * (t_1 * sin(t_0));
	}
	return tmp;
}

Java

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

Python

a_m = math.fabs(a)
def code(a_m, b, angle):
	t_0 = math.pi * (angle * 0.005555555555555556)
	t_1 = 2.0 * ((a_m + b) * (b - a_m))
	tmp = 0
	if (math.pow(b, 2.0) - math.pow(a_m, 2.0)) <= -math.inf:
		tmp = t_1 * math.sin((0.005555555555555556 * (math.pi * angle)))
	else:
		tmp = math.cos(t_0) * (t_1 * math.sin(t_0))
	return tmp

Julia

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

MATLAB

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

Wolfram

a_m = N[Abs[a], $MachinePrecision]
code[a$95$m_, b_, angle_] := Block[{t$95$0 = N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(2.0 * N[(N[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a$95$m, 2.0], $MachinePrecision]), $MachinePrecision], (-Infinity)], N[(t$95$1 * N[Sin[N[(0.005555555555555556 * N[(Pi * angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[Cos[t$95$0], $MachinePrecision] * N[(t$95$1 * 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)) < -inf.0

   1. Initial program 38.7%

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

      1. unpow2 38.7%

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

      2. unpow2 38.7%

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

      3. difference-of-squares 38.7%

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

   4. Applied egg-rr 38.7%

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

   5. Taylor expanded in angle around 0 38.7%

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

   6. Taylor expanded in angle around inf 45.6%

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

   if -inf.0 < (-.f64 (pow.f64 b 2) (pow.f64 a 2))

   1. Initial program 58.8%

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

      1. unpow2 58.8%

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

      2. unpow2 58.8%

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

      3. difference-of-squares 62.6%

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

   4. Applied egg-rr 62.6%

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

   5. Taylor expanded in angle around inf 61.8%

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

      1. associate-*r* 63.1%

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

      2. *-commutative 63.1%

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

      3. *-commutative 63.1%

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

      4. *-commutative 63.1%

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

   7. Simplified 63.1%

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

   8. Taylor expanded in angle around inf 62.5%

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

      1. associate-*r* 63.5%

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

      2. *-commutative 63.5%

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

      3. *-commutative 63.5%

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

      4. *-commutative 63.5%

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

   10. Simplified 63.5%

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

4. Final simplification 60.4%

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

Alternative 5: 57.2% accurate, 1.0× speedup

Math

\[\begin{array}{l} a_m = \left|a\right| \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(a_m + b\right) \cdot \left(b - a_m\right)\right)\\ \mathbf{if}\;a_m \leq 10^{+164}:\\ \;\;\;\;\left(t_0 \cdot \sin \left(\frac{\pi}{\frac{180}{angle}}\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(t_0 \cdot \sin \left(\frac{angle}{180} \cdot \sqrt[3]{{\pi}^{3}}\right)\right)\\ \end{array} \end{array} \]

FPCore

a_m = (fabs.f64 a)
(FPCore (a_m b angle)
 :precision binary64
 (let* ((t_0 (* 2.0 (* (+ a_m b) (- b a_m)))))
   (if (<= a_m 1e+164)
     (*
      (* t_0 (sin (/ PI (/ 180.0 angle))))
      (cos (* PI (* angle 0.005555555555555556))))
     (*
      (cos (* PI (/ angle 180.0)))
      (* t_0 (sin (* (/ angle 180.0) (cbrt (pow PI 3.0)))))))))

C

a_m = fabs(a);
double code(double a_m, double b, double angle) {
	double t_0 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (a_m <= 1e+164) {
		tmp = (t_0 * sin((((double) M_PI) / (180.0 / angle)))) * cos((((double) M_PI) * (angle * 0.005555555555555556)));
	} else {
		tmp = cos((((double) M_PI) * (angle / 180.0))) * (t_0 * sin(((angle / 180.0) * cbrt(pow(((double) M_PI), 3.0)))));
	}
	return tmp;
}

Java

a_m = Math.abs(a);
public static double code(double a_m, double b, double angle) {
	double t_0 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (a_m <= 1e+164) {
		tmp = (t_0 * Math.sin((Math.PI / (180.0 / angle)))) * Math.cos((Math.PI * (angle * 0.005555555555555556)));
	} else {
		tmp = Math.cos((Math.PI * (angle / 180.0))) * (t_0 * Math.sin(((angle / 180.0) * Math.cbrt(Math.pow(Math.PI, 3.0)))));
	}
	return tmp;
}

Julia

a_m = abs(a)
function code(a_m, b, angle)
	t_0 = Float64(2.0 * Float64(Float64(a_m + b) * Float64(b - a_m)))
	tmp = 0.0
	if (a_m <= 1e+164)
		tmp = Float64(Float64(t_0 * sin(Float64(pi / Float64(180.0 / angle)))) * cos(Float64(pi * Float64(angle * 0.005555555555555556))));
	else
		tmp = Float64(cos(Float64(pi * Float64(angle / 180.0))) * Float64(t_0 * sin(Float64(Float64(angle / 180.0) * cbrt((pi ^ 3.0))))));
	end
	return tmp
end

Wolfram

a_m = N[Abs[a], $MachinePrecision]
code[a$95$m_, b_, angle_] := Block[{t$95$0 = N[(2.0 * N[(N[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[a$95$m, 1e+164], N[(N[(t$95$0 * N[Sin[N[(Pi / N[(180.0 / angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Cos[N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[Cos[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(t$95$0 * N[Sin[N[(N[(angle / 180.0), $MachinePrecision] * N[Power[N[Power[Pi, 3.0], $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if a < 1e164

   1. Initial program 58.0%

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

      1. unpow2 58.0%

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

      2. unpow2 58.0%

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

      3. difference-of-squares 61.0%

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

   4. Applied egg-rr 61.0%

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

   5. Taylor expanded in angle around inf 60.7%

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

      1. associate-*r* 61.0%

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

      2. *-commutative 61.0%

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

      3. *-commutative 61.0%

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

      4. *-commutative 61.0%

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

   7. Simplified 61.0%

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

      1. clear-num 61.3%

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

      2. un-div-inv 61.2%

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

   9. Applied egg-rr 61.2%

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

   if 1e164 < a

   1. Initial program 28.3%

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

      1. unpow2 28.3%

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

      2. unpow2 28.3%

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

      3. difference-of-squares 33.2%

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

   4. Applied egg-rr 33.2%

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

      1. add-cbrt-cube 50.6%

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

      2. pow3 50.6%

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

   6. Applied egg-rr 50.6%

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

4. Final simplification 60.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 10^{+164}:\\ \;\;\;\;\left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{\pi}{\frac{180}{angle}}\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\frac{angle}{180} \cdot \sqrt[3]{{\pi}^{3}}\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 56.5% accurate, 1.8× speedup

Math

\[\begin{array}{l} a_m = \left|a\right| \\ \begin{array}{l} t_0 := 2 \cdot \left(\left(a_m + b\right) \cdot \left(b - a_m\right)\right)\\ \mathbf{if}\;a_m \leq 2 \cdot 10^{+175}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(t_0 \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;t_0 \cdot \left(-2.8577960676726107 \cdot 10^{-8} \cdot \left({angle}^{3} \cdot {\pi}^{3}\right) + 0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\\ \end{array} \end{array} \]

FPCore

a_m = (fabs.f64 a)
(FPCore (a_m b angle)
 :precision binary64
 (let* ((t_0 (* 2.0 (* (+ a_m b) (- b a_m)))))
   (if (<= a_m 2e+175)
     (*
      (cos (* PI (/ angle 180.0)))
      (* t_0 (sin (* PI (* angle 0.005555555555555556)))))
     (*
      t_0
      (+
       (* -2.8577960676726107e-8 (* (pow angle 3.0) (pow PI 3.0)))
       (* 0.005555555555555556 (* PI angle)))))))

C

a_m = fabs(a);
double code(double a_m, double b, double angle) {
	double t_0 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (a_m <= 2e+175) {
		tmp = cos((((double) M_PI) * (angle / 180.0))) * (t_0 * sin((((double) M_PI) * (angle * 0.005555555555555556))));
	} else {
		tmp = t_0 * ((-2.8577960676726107e-8 * (pow(angle, 3.0) * pow(((double) M_PI), 3.0))) + (0.005555555555555556 * (((double) M_PI) * angle)));
	}
	return tmp;
}

Java

a_m = Math.abs(a);
public static double code(double a_m, double b, double angle) {
	double t_0 = 2.0 * ((a_m + b) * (b - a_m));
	double tmp;
	if (a_m <= 2e+175) {
		tmp = Math.cos((Math.PI * (angle / 180.0))) * (t_0 * Math.sin((Math.PI * (angle * 0.005555555555555556))));
	} else {
		tmp = t_0 * ((-2.8577960676726107e-8 * (Math.pow(angle, 3.0) * Math.pow(Math.PI, 3.0))) + (0.005555555555555556 * (Math.PI * angle)));
	}
	return tmp;
}

Python

a_m = math.fabs(a)
def code(a_m, b, angle):
	t_0 = 2.0 * ((a_m + b) * (b - a_m))
	tmp = 0
	if a_m <= 2e+175:
		tmp = math.cos((math.pi * (angle / 180.0))) * (t_0 * math.sin((math.pi * (angle * 0.005555555555555556))))
	else:
		tmp = t_0 * ((-2.8577960676726107e-8 * (math.pow(angle, 3.0) * math.pow(math.pi, 3.0))) + (0.005555555555555556 * (math.pi * angle)))
	return tmp

Julia

a_m = abs(a)
function code(a_m, b, angle)
	t_0 = Float64(2.0 * Float64(Float64(a_m + b) * Float64(b - a_m)))
	tmp = 0.0
	if (a_m <= 2e+175)
		tmp = Float64(cos(Float64(pi * Float64(angle / 180.0))) * Float64(t_0 * sin(Float64(pi * Float64(angle * 0.005555555555555556)))));
	else
		tmp = Float64(t_0 * Float64(Float64(-2.8577960676726107e-8 * Float64((angle ^ 3.0) * (pi ^ 3.0))) + Float64(0.005555555555555556 * Float64(pi * angle))));
	end
	return tmp
end

MATLAB

a_m = abs(a);
function tmp_2 = code(a_m, b, angle)
	t_0 = 2.0 * ((a_m + b) * (b - a_m));
	tmp = 0.0;
	if (a_m <= 2e+175)
		tmp = cos((pi * (angle / 180.0))) * (t_0 * sin((pi * (angle * 0.005555555555555556))));
	else
		tmp = t_0 * ((-2.8577960676726107e-8 * ((angle ^ 3.0) * (pi ^ 3.0))) + (0.005555555555555556 * (pi * angle)));
	end
	tmp_2 = tmp;
end

Wolfram

a_m = N[Abs[a], $MachinePrecision]
code[a$95$m_, b_, angle_] := Block[{t$95$0 = N[(2.0 * N[(N[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[a$95$m, 2e+175], N[(N[Cos[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(t$95$0 * N[Sin[N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(t$95$0 * N[(N[(-2.8577960676726107e-8 * N[(N[Power[angle, 3.0], $MachinePrecision] * N[Power[Pi, 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(0.005555555555555556 * N[(Pi * angle), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if a < 1.9999999999999999e175

   1. Initial program 56.8%

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

      1. unpow2 56.8%

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

      2. unpow2 56.8%

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

      3. difference-of-squares 60.2%

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

   4. Applied egg-rr 60.2%

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

      1. add-sqr-sqrt 28.7%

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

      2. sqrt-unprod 34.7%

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

      3. pow2 34.7%

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

      4. div-inv 34.7%

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

      5. metadata-eval 34.7%

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

   6. Applied egg-rr 34.7%

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

      1. sqrt-pow1 61.2%

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

      2. metadata-eval 61.2%

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

      3. pow1 61.2%

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

   8. Applied egg-rr 61.2%

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

   if 1.9999999999999999e175 < a

   1. Initial program 35.4%

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

      1. unpow2 35.4%

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

      2. unpow2 35.4%

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

      3. difference-of-squares 36.1%

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

   4. Applied egg-rr 36.1%

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

   5. Taylor expanded in angle around 0 41.7%

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

   6. Taylor expanded in angle around 0 52.8%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \color{blue}{\left(-2.8577960676726107 \cdot 10^{-8} \cdot \left({angle}^{3} \cdot {\pi}^{3}\right) + 0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}\right) \cdot 1 \]
3. Recombined 2 regimes into one program.

4. Final simplification 60.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 2 \cdot 10^{+175}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right) \cdot \left(-2.8577960676726107 \cdot 10^{-8} \cdot \left({angle}^{3} \cdot {\pi}^{3}\right) + 0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 57.5% accurate, 1.9× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

a_m = N[Abs[a], $MachinePrecision]
code[a$95$m_, b_, angle_] := N[(N[Cos[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[(2.0 * N[(N[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(0.005555555555555556 * N[(Pi * angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 55.3%

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

   1. unpow2 55.3%

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

   2. unpow2 55.3%

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

   3. difference-of-squares 58.5%

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

4. Applied egg-rr 58.5%

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

5. Taylor expanded in angle around inf 58.5%

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

6. Final simplification 58.5%

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

Alternative 8: 57.2% accurate, 1.9× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

a_m = N[Abs[a], $MachinePrecision]
code[a$95$m_, b_, angle_] := N[(N[Cos[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[(2.0 * N[(N[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 55.3%

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

   1. unpow2 55.3%

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

   2. unpow2 55.3%

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

   3. difference-of-squares 58.5%

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

4. Applied egg-rr 58.5%

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

   1. add-sqr-sqrt 28.0%

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

   2. sqrt-unprod 34.2%

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

   3. pow2 34.2%

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

   4. div-inv 34.2%

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

   5. metadata-eval 34.2%

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

6. Applied egg-rr 34.2%

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

   1. sqrt-pow1 59.8%

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

   2. metadata-eval 59.8%

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

   3. pow1 59.8%

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

8. Applied egg-rr 59.8%

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

9. Final simplification 59.8%

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

Alternative 9: 55.7% accurate, 3.6× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

a_m = N[Abs[a], $MachinePrecision]
code[a$95$m_, b_, angle_] := N[(N[(2.0 * N[(N[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(0.005555555555555556 * N[(Pi * angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 55.3%

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

   1. unpow2 55.3%

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

   2. unpow2 55.3%

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

   3. difference-of-squares 58.5%

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

4. Applied egg-rr 58.5%

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

5. Taylor expanded in angle around 0 55.4%

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

6. Taylor expanded in angle around inf 57.8%

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

7. Final simplification 57.8%

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

Alternative 10: 54.2% accurate, 32.2× speedup

Math

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

FPCore

a_m = (fabs.f64 a)
(FPCore (a_m b angle)
 :precision binary64
 (* 0.011111111111111112 (* angle (* (* (+ a_m b) (- b a_m)) PI))))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

a_m = N[Abs[a], $MachinePrecision]
code[a$95$m_, b_, angle_] := N[(0.011111111111111112 * N[(angle * N[(N[(N[(a$95$m + b), $MachinePrecision] * N[(b - a$95$m), $MachinePrecision]), $MachinePrecision] * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 55.3%

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

   1. unpow2 55.3%

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

   2. unpow2 55.3%

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

   3. difference-of-squares 58.5%

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

4. Applied egg-rr 58.5%

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

5. Taylor expanded in angle around 0 55.4%

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

6. Taylor expanded in angle around 0 54.9%

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

7. Final simplification 54.9%

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

Reproduce

herbie shell --seed 2024019 
(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)))))