ab-angle->ABCF A

Percentage Accurate: 79.9% → 79.9%

Time: 24.8s

Alternatives: 14

Speedup: 1.0×

• 36.4% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Initial Program: 79.9% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Alternative 1: 79.9% accurate, 0.5× speedup

Math

\[\begin{array}{l} angle = |angle|\\ \\ \begin{array}{l} t_0 := 1 + \pi \cdot \left(angle \cdot 0.005555555555555556\right)\\ {\left(a \cdot \sin \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} + {\left(b \cdot \mathsf{fma}\left(\cos t_0, \cos 1, \sin t_0 \cdot \sin 1\right)\right)}^{2} \end{array} \end{array} \]

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (+ 1.0 (* PI (* angle 0.005555555555555556)))))
   (+
    (pow (* a (sin (/ PI (/ 180.0 angle)))) 2.0)
    (pow (* b (fma (cos t_0) (cos 1.0) (* (sin t_0) (sin 1.0)))) 2.0))))

C

angle = abs(angle);
double code(double a, double b, double angle) {
	double t_0 = 1.0 + (((double) M_PI) * (angle * 0.005555555555555556));
	return pow((a * sin((((double) M_PI) / (180.0 / angle)))), 2.0) + pow((b * fma(cos(t_0), cos(1.0), (sin(t_0) * sin(1.0)))), 2.0);
}

Julia

angle = abs(angle)
function code(a, b, angle)
	t_0 = Float64(1.0 + Float64(pi * Float64(angle * 0.005555555555555556)))
	return Float64((Float64(a * sin(Float64(pi / Float64(180.0 / angle)))) ^ 2.0) + (Float64(b * fma(cos(t_0), cos(1.0), Float64(sin(t_0) * sin(1.0)))) ^ 2.0))
end

Wolfram

NOTE: angle should be positive before calling this function
code[a_, b_, angle_] := Block[{t$95$0 = N[(1.0 + N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(N[Power[N[(a * N[Sin[N[(Pi / N[(180.0 / angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[(N[Cos[t$95$0], $MachinePrecision] * N[Cos[1.0], $MachinePrecision] + N[(N[Sin[t$95$0], $MachinePrecision] * N[Sin[1.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Initial program 78.5%

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

   1. *-commutative 78.5%

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

   2. clear-num 78.5%

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

   3. un-div-inv 78.5%

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

3. Applied egg-rr 78.5%

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

   1. associate-*l/ 78.5%

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

   2. expm1-log1p-u 65.5%

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

   3. associate-*r/ 65.5%

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

   4. div-inv 65.5%

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

   5. metadata-eval 65.5%

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

5. Applied egg-rr 65.5%

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

   1. expm1-udef 65.6%

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

   2. cos-diff 65.5%

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

   3. log1p-udef 65.5%

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

   4. add-exp-log 65.5%

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

   5. log1p-udef 65.5%

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

   6. add-exp-log 78.5%

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

7. Applied egg-rr 78.5%

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

   1. fma-def 78.5%

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

   2. *-commutative 78.5%

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

   3. associate-*l* 78.4%

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

   4. *-commutative 78.4%

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

   5. associate-*l* 78.5%

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

9. Simplified 78.5%

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

10. Final simplification 78.5%

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

Alternative 2: 79.9% accurate, 0.6× speedup

Math

\[\begin{array}{l} angle = |angle|\\ \\ \begin{array}{l} t_0 := 1 + angle \cdot \left(\pi \cdot 0.005555555555555556\right)\\ {\left(a \cdot \sin \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} + {\left(b \cdot \left(\cos 1 \cdot \cos t_0 + \sin 1 \cdot \sin t_0\right)\right)}^{2} \end{array} \end{array} \]

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (+ 1.0 (* angle (* PI 0.005555555555555556)))))
   (+
    (pow (* a (sin (/ PI (/ 180.0 angle)))) 2.0)
    (pow (* b (+ (* (cos 1.0) (cos t_0)) (* (sin 1.0) (sin t_0)))) 2.0))))

C

angle = abs(angle);
double code(double a, double b, double angle) {
	double t_0 = 1.0 + (angle * (((double) M_PI) * 0.005555555555555556));
	return pow((a * sin((((double) M_PI) / (180.0 / angle)))), 2.0) + pow((b * ((cos(1.0) * cos(t_0)) + (sin(1.0) * sin(t_0)))), 2.0);
}

Java

angle = Math.abs(angle);
public static double code(double a, double b, double angle) {
	double t_0 = 1.0 + (angle * (Math.PI * 0.005555555555555556));
	return Math.pow((a * Math.sin((Math.PI / (180.0 / angle)))), 2.0) + Math.pow((b * ((Math.cos(1.0) * Math.cos(t_0)) + (Math.sin(1.0) * Math.sin(t_0)))), 2.0);
}

Python

angle = abs(angle)
def code(a, b, angle):
	t_0 = 1.0 + (angle * (math.pi * 0.005555555555555556))
	return math.pow((a * math.sin((math.pi / (180.0 / angle)))), 2.0) + math.pow((b * ((math.cos(1.0) * math.cos(t_0)) + (math.sin(1.0) * math.sin(t_0)))), 2.0)

Julia

angle = abs(angle)
function code(a, b, angle)
	t_0 = Float64(1.0 + Float64(angle * Float64(pi * 0.005555555555555556)))
	return Float64((Float64(a * sin(Float64(pi / Float64(180.0 / angle)))) ^ 2.0) + (Float64(b * Float64(Float64(cos(1.0) * cos(t_0)) + Float64(sin(1.0) * sin(t_0)))) ^ 2.0))
end

MATLAB

angle = abs(angle)
function tmp = code(a, b, angle)
	t_0 = 1.0 + (angle * (pi * 0.005555555555555556));
	tmp = ((a * sin((pi / (180.0 / angle)))) ^ 2.0) + ((b * ((cos(1.0) * cos(t_0)) + (sin(1.0) * sin(t_0)))) ^ 2.0);
end

Wolfram

NOTE: angle should be positive before calling this function
code[a_, b_, angle_] := Block[{t$95$0 = N[(1.0 + N[(angle * N[(Pi * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(N[Power[N[(a * N[Sin[N[(Pi / N[(180.0 / angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[(N[(N[Cos[1.0], $MachinePrecision] * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision] + N[(N[Sin[1.0], $MachinePrecision] * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Initial program 78.5%

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

   1. *-commutative 78.5%

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

   2. clear-num 78.5%

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

   3. un-div-inv 78.5%

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

3. Applied egg-rr 78.5%

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

   1. associate-*l/ 78.5%

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

   2. expm1-log1p-u 65.5%

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

   3. associate-*r/ 65.5%

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

   4. div-inv 65.5%

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

   5. metadata-eval 65.5%

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

5. Applied egg-rr 65.5%

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

   1. expm1-udef 65.6%

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

   2. cos-diff 65.5%

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

   3. log1p-udef 65.5%

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

   4. add-exp-log 65.5%

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

   5. log1p-udef 65.5%

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

   6. add-exp-log 78.5%

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

7. Applied egg-rr 78.5%

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

8. Final simplification 78.5%

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

Alternative 3: 79.9% accurate, 0.8× speedup

Math

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

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (+
  (pow (* a (sin (* PI (/ angle 180.0)))) 2.0)
  (pow (* b (cos (exp (log (* angle (* PI 0.005555555555555556)))))) 2.0)))

C

angle = abs(angle);
double code(double a, double b, double angle) {
	return pow((a * sin((((double) M_PI) * (angle / 180.0)))), 2.0) + pow((b * cos(exp(log((angle * (((double) M_PI) * 0.005555555555555556)))))), 2.0);
}

Java

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

Python

angle = abs(angle)
def code(a, b, angle):
	return math.pow((a * math.sin((math.pi * (angle / 180.0)))), 2.0) + math.pow((b * math.cos(math.exp(math.log((angle * (math.pi * 0.005555555555555556)))))), 2.0)

Julia

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

MATLAB

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

Wolfram

NOTE: angle should be positive before calling this function
code[a_, b_, angle_] := N[(N[Power[N[(a * N[Sin[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Cos[N[Exp[N[Log[N[(angle * N[(Pi * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 78.5%

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

   1. associate-*l/ 78.5%

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

   2. add-exp-log 37.1%

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

   3. associate-*r/ 37.1%

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

   4. div-inv 37.1%

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

   5. metadata-eval 37.1%

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

3. Applied egg-rr 37.1%

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

4. Final simplification 37.1%

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

Alternative 4: 79.9% accurate, 0.8× speedup

Math

\[\begin{array}{l} angle = |angle|\\ \\ {\left(a \cdot \sin \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} + {\left(b \cdot \cos \left(\mathsf{expm1}\left(\mathsf{log1p}\left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\right)\right)\right)}^{2} \end{array} \]

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (+
  (pow (* a (sin (/ PI (/ 180.0 angle)))) 2.0)
  (pow (* b (cos (expm1 (log1p (* angle (* PI 0.005555555555555556)))))) 2.0)))

C

angle = abs(angle);
double code(double a, double b, double angle) {
	return pow((a * sin((((double) M_PI) / (180.0 / angle)))), 2.0) + pow((b * cos(expm1(log1p((angle * (((double) M_PI) * 0.005555555555555556)))))), 2.0);
}

Java

angle = Math.abs(angle);
public static double code(double a, double b, double angle) {
	return Math.pow((a * Math.sin((Math.PI / (180.0 / angle)))), 2.0) + Math.pow((b * Math.cos(Math.expm1(Math.log1p((angle * (Math.PI * 0.005555555555555556)))))), 2.0);
}

Python

angle = abs(angle)
def code(a, b, angle):
	return math.pow((a * math.sin((math.pi / (180.0 / angle)))), 2.0) + math.pow((b * math.cos(math.expm1(math.log1p((angle * (math.pi * 0.005555555555555556)))))), 2.0)

Julia

angle = abs(angle)
function code(a, b, angle)
	return Float64((Float64(a * sin(Float64(pi / Float64(180.0 / angle)))) ^ 2.0) + (Float64(b * cos(expm1(log1p(Float64(angle * Float64(pi * 0.005555555555555556)))))) ^ 2.0))
end

Wolfram

NOTE: angle should be positive before calling this function
code[a_, b_, angle_] := N[(N[Power[N[(a * N[Sin[N[(Pi / N[(180.0 / angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Cos[N[(Exp[N[Log[1 + N[(angle * N[(Pi * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 78.5%

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

   1. *-commutative 78.5%

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

   2. clear-num 78.5%

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

   3. un-div-inv 78.5%

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

3. Applied egg-rr 78.5%

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

   1. associate-*l/ 78.5%

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

   2. expm1-log1p-u 65.5%

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

   3. associate-*r/ 65.5%

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

   4. div-inv 65.5%

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

   5. metadata-eval 65.5%

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

5. Applied egg-rr 65.5%

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

6. Final simplification 65.5%

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

Alternative 5: 79.9% accurate, 1.0× speedup

Math

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

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* angle (/ PI 180.0))))
   (+ (pow (* a (sin t_0)) 2.0) (pow (* b (cos t_0)) 2.0))))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 78.5%

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

   1. associate-*l/ 78.5%

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

   2. associate-*r/ 78.5%

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

   3. associate-*l/ 78.5%

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

   4. associate-*r/ 78.4%

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

3. Simplified 78.4%

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

4. Final simplification 78.4%

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

Alternative 6: 79.8% accurate, 1.0× speedup

Math

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

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (+
  (pow (* b (cos (/ (* PI angle) 180.0))) 2.0)
  (pow (* a (sin (* PI (* angle 0.005555555555555556)))) 2.0)))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

NOTE: angle should be positive before calling this function
code[a_, b_, angle_] := N[(N[Power[N[(b * N[Cos[N[(N[(Pi * angle), $MachinePrecision] / 180.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(a * N[Sin[N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 78.5%

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

   1. associate-*l/ 78.5%

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

   2. associate-*l/ 78.5%

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

3. Simplified 78.5%

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

   1. associate-*l/ 78.5%

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

   2. div-inv 78.5%

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

   3. metadata-eval 78.5%

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

5. Applied egg-rr 78.5%

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

6. Final simplification 78.5%

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

Alternative 7: 79.8% accurate, 1.0× speedup

Math

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

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (+
  (pow (* a (sin (* 0.005555555555555556 (* PI angle)))) 2.0)
  (pow (* b (cos (/ (* PI angle) 180.0))) 2.0)))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

NOTE: angle should be positive before calling this function
code[a_, b_, angle_] := N[(N[Power[N[(a * N[Sin[N[(0.005555555555555556 * N[(Pi * angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Cos[N[(N[(Pi * angle), $MachinePrecision] / 180.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 78.5%

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

   1. associate-*l/ 78.5%

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

   2. associate-*l/ 78.5%

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

3. Simplified 78.5%

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

   1. div-inv 78.5%

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

   2. metadata-eval 78.5%

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

5. Applied egg-rr 78.5%

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

6. Final simplification 78.5%

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

Alternative 8: 79.9% accurate, 1.0× speedup

Math

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

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (+
  (pow (* a (sin (/ PI (/ 180.0 angle)))) 2.0)
  (pow (* b (cos (* PI (/ angle 180.0)))) 2.0)))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

NOTE: angle should be positive before calling this function
code[a_, b_, angle_] := N[(N[Power[N[(a * N[Sin[N[(Pi / N[(180.0 / angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Cos[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 78.5%

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

   1. *-commutative 78.5%

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

   2. clear-num 78.5%

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

   3. un-div-inv 78.5%

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

3. Applied egg-rr 78.5%

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

4. Final simplification 78.5%

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

Alternative 9: 79.7% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

NOTE: angle should be positive before calling this function
code[a_, b_, angle_] := N[(N[Power[N[(a * N[Sin[N[(0.005555555555555556 * N[(Pi * angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[b, 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 78.5%

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

   1. associate-*l/ 78.5%

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

   2. associate-*r/ 78.5%

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

   3. associate-*l/ 78.5%

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

   4. associate-*r/ 78.4%

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

3. Simplified 78.4%

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

4. Taylor expanded in angle around 0 78.0%

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

5. Taylor expanded in angle around inf 78.0%

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

6. Final simplification 78.0%

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

Alternative 10: 79.8% accurate, 1.5× speedup

Math

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

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (+ (pow (* a (sin (* angle (/ PI 180.0)))) 2.0) (pow b 2.0)))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

NOTE: angle should be positive before calling this function
code[a_, b_, angle_] := N[(N[Power[N[(a * N[Sin[N[(angle * N[(Pi / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[b, 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 78.5%

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

   1. associate-*l/ 78.5%

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

   2. associate-*r/ 78.5%

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

   3. associate-*l/ 78.5%

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

   4. associate-*r/ 78.4%

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

3. Simplified 78.4%

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

4. Taylor expanded in angle around 0 78.0%

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

5. Final simplification 78.0%

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

Alternative 11: 79.8% accurate, 1.5× speedup

Math

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

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (+ (pow (* a (sin (/ PI (/ 180.0 angle)))) 2.0) (pow b 2.0)))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

NOTE: angle should be positive before calling this function
code[a_, b_, angle_] := N[(N[Power[N[(a * N[Sin[N[(Pi / N[(180.0 / angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[b, 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 78.5%

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

   1. *-commutative 78.5%

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

   2. clear-num 78.5%

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

   3. un-div-inv 78.5%

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

3. Applied egg-rr 78.5%

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

4. Taylor expanded in angle around 0 78.0%

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

5. Final simplification 78.0%

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

Alternative 12: 74.8% accurate, 1.9× speedup

Math

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

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (+
  (pow b 2.0)
  (*
   (* PI (* a angle))
   (* 0.005555555555555556 (* (* angle 0.005555555555555556) (* a PI))))))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

NOTE: angle should be positive before calling this function
code[a_, b_, angle_] := N[(N[Power[b, 2.0], $MachinePrecision] + N[(N[(Pi * N[(a * angle), $MachinePrecision]), $MachinePrecision] * N[(0.005555555555555556 * N[(N[(angle * 0.005555555555555556), $MachinePrecision] * N[(a * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 78.5%

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

   1. associate-*l/ 78.5%

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

   2. associate-*r/ 78.5%

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

   3. associate-*l/ 78.5%

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

   4. associate-*r/ 78.4%

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

3. Simplified 78.4%

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

4. Taylor expanded in angle around 0 78.0%

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

5. Taylor expanded in angle around 0 72.1%

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

   1. *-commutative 72.1%

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

7. Simplified 72.1%

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

   1. unpow2 72.1%

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

   2. associate-*r* 72.1%

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

   3. *-commutative 72.1%

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

   4. *-commutative 72.1%

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

   5. associate-*l* 72.1%

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

   6. associate-*r* 72.2%

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

   7. *-commutative 72.2%

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

   8. associate-*l* 72.2%

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

9. Applied egg-rr 72.2%

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

10. Final simplification 72.2%

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

Alternative 13: 74.8% accurate, 1.9× speedup

Math

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

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (+
  (pow b 2.0)
  (*
   0.005555555555555556
   (* (* (* angle 0.005555555555555556) (* a PI)) (* PI (* a angle))))))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

NOTE: angle should be positive before calling this function
code[a_, b_, angle_] := N[(N[Power[b, 2.0], $MachinePrecision] + N[(0.005555555555555556 * N[(N[(N[(angle * 0.005555555555555556), $MachinePrecision] * N[(a * Pi), $MachinePrecision]), $MachinePrecision] * N[(Pi * N[(a * angle), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 78.5%

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

   1. associate-*l/ 78.5%

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

   2. associate-*r/ 78.5%

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

   3. associate-*l/ 78.5%

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

   4. associate-*r/ 78.4%

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

3. Simplified 78.4%

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

4. Taylor expanded in angle around 0 78.0%

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

5. Taylor expanded in angle around 0 72.1%

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

   1. *-commutative 72.1%

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

7. Simplified 72.1%

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

   1. unpow2 72.1%

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

   2. *-commutative 72.1%

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

   3. associate-*r* 72.2%

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

   4. *-commutative 72.2%

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

   5. *-commutative 72.2%

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

   6. associate-*l* 72.2%

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

   7. associate-*r* 72.2%

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

   8. *-commutative 72.2%

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

   9. associate-*l* 72.2%

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

9. Applied egg-rr 72.2%

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

10. Final simplification 72.2%

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

Alternative 14: 74.7% accurate, 2.0× speedup

Math

\[\begin{array}{l} angle = |angle|\\ \\ {b}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\left(angle \cdot \left(a \cdot \pi\right)\right)}^{2} \end{array} \]

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (+ (pow b 2.0) (* 3.08641975308642e-5 (pow (* angle (* a PI)) 2.0))))

C

angle = abs(angle);
double code(double a, double b, double angle) {
	return pow(b, 2.0) + (3.08641975308642e-5 * pow((angle * (a * ((double) M_PI))), 2.0));
}

Java

angle = Math.abs(angle);
public static double code(double a, double b, double angle) {
	return Math.pow(b, 2.0) + (3.08641975308642e-5 * Math.pow((angle * (a * Math.PI)), 2.0));
}

Python

angle = abs(angle)
def code(a, b, angle):
	return math.pow(b, 2.0) + (3.08641975308642e-5 * math.pow((angle * (a * math.pi)), 2.0))

Julia

angle = abs(angle)
function code(a, b, angle)
	return Float64((b ^ 2.0) + Float64(3.08641975308642e-5 * (Float64(angle * Float64(a * pi)) ^ 2.0)))
end

MATLAB

angle = abs(angle)
function tmp = code(a, b, angle)
	tmp = (b ^ 2.0) + (3.08641975308642e-5 * ((angle * (a * pi)) ^ 2.0));
end

Wolfram

NOTE: angle should be positive before calling this function
code[a_, b_, angle_] := N[(N[Power[b, 2.0], $MachinePrecision] + N[(3.08641975308642e-5 * N[Power[N[(angle * N[(a * Pi), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 78.5%

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

   1. associate-*l/ 78.5%

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

   2. associate-*r/ 78.5%

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

   3. associate-*l/ 78.5%

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

   4. associate-*r/ 78.4%

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

3. Simplified 78.4%

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

4. Taylor expanded in angle around 0 78.0%

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

5. Taylor expanded in angle around 0 72.1%

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

   1. *-commutative 72.1%

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

7. Simplified 72.1%

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

8. Taylor expanded in angle around 0 59.2%

   \[\leadsto \color{blue}{3.08641975308642 \cdot 10^{-5} \cdot \left({angle}^{2} \cdot \left({a}^{2} \cdot {\pi}^{2}\right)\right)} + {\left(b \cdot 1\right)}^{2} \]
9. Step-by-step derivation

   1. associate-*r* 59.2%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot \color{blue}{\left(\left({angle}^{2} \cdot {a}^{2}\right) \cdot {\pi}^{2}\right)} + {\left(b \cdot 1\right)}^{2} \]

   2. unpow2 59.2%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot \left(\left(\color{blue}{\left(angle \cdot angle\right)} \cdot {a}^{2}\right) \cdot {\pi}^{2}\right) + {\left(b \cdot 1\right)}^{2} \]

   3. unpow2 59.2%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot \left(\left(\left(angle \cdot angle\right) \cdot \color{blue}{\left(a \cdot a\right)}\right) \cdot {\pi}^{2}\right) + {\left(b \cdot 1\right)}^{2} \]

   4. unswap-sqr 72.2%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot \left(\color{blue}{\left(\left(angle \cdot a\right) \cdot \left(angle \cdot a\right)\right)} \cdot {\pi}^{2}\right) + {\left(b \cdot 1\right)}^{2} \]

   5. *-commutative 72.2%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot \color{blue}{\left({\pi}^{2} \cdot \left(\left(angle \cdot a\right) \cdot \left(angle \cdot a\right)\right)\right)} + {\left(b \cdot 1\right)}^{2} \]

   6. unpow2 72.2%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot \left(\color{blue}{\left(\pi \cdot \pi\right)} \cdot \left(\left(angle \cdot a\right) \cdot \left(angle \cdot a\right)\right)\right) + {\left(b \cdot 1\right)}^{2} \]

   7. swap-sqr 72.2%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot \color{blue}{\left(\left(\pi \cdot \left(angle \cdot a\right)\right) \cdot \left(\pi \cdot \left(angle \cdot a\right)\right)\right)} + {\left(b \cdot 1\right)}^{2} \]

   8. unpow2 72.2%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot \color{blue}{{\left(\pi \cdot \left(angle \cdot a\right)\right)}^{2}} + {\left(b \cdot 1\right)}^{2} \]

   9. *-commutative 72.2%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot {\color{blue}{\left(\left(angle \cdot a\right) \cdot \pi\right)}}^{2} + {\left(b \cdot 1\right)}^{2} \]

   10. associate-*r* 72.2%

       \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot {\color{blue}{\left(angle \cdot \left(a \cdot \pi\right)\right)}}^{2} + {\left(b \cdot 1\right)}^{2} \]

   11. *-commutative 72.2%

       \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot {\left(angle \cdot \color{blue}{\left(\pi \cdot a\right)}\right)}^{2} + {\left(b \cdot 1\right)}^{2} \]

10. Simplified 72.2%

    \[\leadsto \color{blue}{3.08641975308642 \cdot 10^{-5} \cdot {\left(angle \cdot \left(\pi \cdot a\right)\right)}^{2}} + {\left(b \cdot 1\right)}^{2} \]

11. Final simplification 72.2%

    \[\leadsto {b}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\left(angle \cdot \left(a \cdot \pi\right)\right)}^{2} \]

Reproduce

herbie shell --seed 2023257 
(FPCore (a b angle)
  :name "ab-angle->ABCF A"
  :precision binary64
  (+ (pow (* a (sin (* (/ angle 180.0) PI))) 2.0) (pow (* b (cos (* (/ angle 180.0) PI))) 2.0)))