ab-angle->ABCF A

Percentage Accurate: 79.8% → 79.8%

Time: 41.5s

Alternatives: 15

Speedup: 1.5×

• 36.3% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

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

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: 79.8% 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: 79.8% accurate, 0.5× speedup

Math

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

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (fma (* angle 0.005555555555555556) PI 1.0)))
   (+
    (pow (* a (sin (* (/ angle 180.0) PI))) 2.0)
    (pow (* b (- (* (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 = fma((angle * 0.005555555555555556), ((double) M_PI), 1.0);
	return pow((a * sin(((angle / 180.0) * ((double) M_PI)))), 2.0) + pow((b * ((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 = fma(Float64(angle * 0.005555555555555556), pi, 1.0)
	return Float64((Float64(a * sin(Float64(Float64(angle / 180.0) * pi))) ^ 2.0) + (Float64(b * Float64(Float64(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[(N[(angle * 0.005555555555555556), $MachinePrecision] * Pi + 1.0), $MachinePrecision]}, N[(N[Power[N[(a * N[Sin[N[(N[(angle / 180.0), $MachinePrecision] * Pi), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[(N[(N[Cos[t$95$0], $MachinePrecision] * N[Cos[-1.0], $MachinePrecision]), $MachinePrecision] - N[(N[Sin[t$95$0], $MachinePrecision] * N[Sin[-1.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Initial program 81.8%

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

   1. associate-*l/ 81.3%

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

   2. associate-*r/ 81.7%

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

   3. expm1-log1p-u 63.6%

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

   4. div-inv 63.6%

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

   5. metadata-eval 63.6%

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

3. Applied egg-rr 63.6%

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

   1. expm1-udef 63.5%

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

   2. sub-neg 63.5%

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

   3. metadata-eval 63.5%

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

   4. cos-sum 63.5%

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

5. Applied egg-rr 81.8%

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

6. Final simplification 81.8%

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

Alternative 2: 79.7% accurate, 0.5× speedup

Math

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

FPCore

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

C

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

Julia

angle = abs(angle)
function code(a, b, angle)
	return Float64((Float64(a * sin(Float64(Float64(angle / 180.0) * pi))) ^ 2.0) + (Float64(b * Float64(Float64(cos(fma(Float64(angle * 0.005555555555555556), pi, 1.0)) * cos(-1.0)) - Float64(sin(-1.0) * sin(fma(angle, Float64(pi * 0.005555555555555556), 1.0))))) ^ 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[(N[(angle / 180.0), $MachinePrecision] * Pi), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[(N[(N[Cos[N[(N[(angle * 0.005555555555555556), $MachinePrecision] * Pi + 1.0), $MachinePrecision]], $MachinePrecision] * N[Cos[-1.0], $MachinePrecision]), $MachinePrecision] - N[(N[Sin[-1.0], $MachinePrecision] * N[Sin[N[(angle * N[(Pi * 0.005555555555555556), $MachinePrecision] + 1.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 81.8%

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

   1. associate-*l/ 81.3%

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

   2. associate-*r/ 81.7%

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

   3. expm1-log1p-u 63.6%

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

   4. div-inv 63.6%

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

   5. metadata-eval 63.6%

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

3. Applied egg-rr 63.6%

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

   1. expm1-udef 63.5%

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

   2. sub-neg 63.5%

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

   3. metadata-eval 63.5%

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

   4. cos-sum 63.5%

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

5. Applied egg-rr 81.8%

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

6. Taylor expanded in angle around inf 81.8%

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

   1. *-commutative 81.8%

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

   2. fma-def 81.8%

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

   3. associate-*r* 81.9%

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

   4. fma-udef 81.9%

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

   5. *-commutative 81.9%

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

8. Simplified 81.9%

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

9. Final simplification 81.9%

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

Alternative 3: 79.7% accurate, 0.6× speedup

Math

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

FPCore

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

C

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

Python

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

Julia

angle = abs(angle)
function code(a, b, angle)
	return Float64((Float64(a * sin(Float64(Float64(angle / 180.0) * pi))) ^ 2.0) + (Float64(b * log1p(expm1(cos(Float64(pi * exp(log(Float64(angle * 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[(N[(angle / 180.0), $MachinePrecision] * Pi), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Log[1 + N[(Exp[N[Cos[N[(Pi * N[Exp[N[Log[N[(angle * 0.005555555555555556), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 81.8%

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

   1. associate-*l/ 81.3%

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

   2. associate-*r/ 81.7%

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

   3. expm1-log1p-u 63.6%

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

   4. div-inv 63.6%

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

   5. metadata-eval 63.6%

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

3. Applied egg-rr 63.6%

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

   1. expm1-log1p-u 81.7%

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

   2. metadata-eval 81.7%

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

   3. div-inv 81.7%

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

   4. log1p-expm1-u 81.7%

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

   5. associate-*r/ 81.3%

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

   6. associate-*l/ 81.8%

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

   7. div-inv 81.7%

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

   8. metadata-eval 81.7%

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

5. Applied egg-rr 81.7%

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

   1. add-exp-log 39.8%

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

7. Applied egg-rr 39.8%

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

8. Final simplification 39.8%

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

Alternative 4: 79.7% accurate, 0.7× speedup

Math

\[\begin{array}{l} angle = |angle|\\ \\ {\left(a \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} + {\left(b \cdot \cos \left(\sqrt[3]{angle} \cdot \left(\left(\pi \cdot 0.005555555555555556\right) \cdot {\left(\sqrt[3]{angle}\right)}^{2}\right)\right)\right)}^{2} \end{array} \]

FPCore

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

C

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

Java

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

Julia

angle = abs(angle)
function code(a, b, angle)
	return Float64((Float64(a * sin(Float64(Float64(angle / 180.0) * pi))) ^ 2.0) + (Float64(b * cos(Float64(cbrt(angle) * Float64(Float64(pi * 0.005555555555555556) * (cbrt(angle) ^ 2.0))))) ^ 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[(N[(angle / 180.0), $MachinePrecision] * Pi), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Cos[N[(N[Power[angle, 1/3], $MachinePrecision] * N[(N[(Pi * 0.005555555555555556), $MachinePrecision] * N[Power[N[Power[angle, 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 81.8%

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

   1. associate-*l/ 81.3%

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

   2. associate-*r/ 81.7%

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

   3. expm1-log1p-u 63.6%

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

   4. div-inv 63.6%

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

   5. metadata-eval 63.6%

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

3. Applied egg-rr 63.6%

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

   1. expm1-log1p-u 81.7%

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

   2. metadata-eval 81.7%

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

   3. div-inv 81.7%

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

   4. *-commutative 81.7%

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

   5. add-cube-cbrt 81.8%

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

   6. associate-*r* 81.9%

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

   7. div-inv 81.9%

      \[\leadsto {\left(a \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} + {\left(b \cdot \cos \left(\left(\color{blue}{\left(\pi \cdot \frac{1}{180}\right)} \cdot \left(\sqrt[3]{angle} \cdot \sqrt[3]{angle}\right)\right) \cdot \sqrt[3]{angle}\right)\right)}^{2} \]

   8. metadata-eval 81.9%

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

   9. pow2 81.9%

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

5. Applied egg-rr 81.9%

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

6. Final simplification 81.9%

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

Alternative 5: 79.8% accurate, 0.8× speedup

Math

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

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (+
  (pow (* a (sin (* (/ angle 180.0) PI))) 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(((angle / 180.0) * ((double) M_PI)))), 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(((angle / 180.0) * Math.PI))), 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(((angle / 180.0) * math.pi))), 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(Float64(angle / 180.0) * pi))) ^ 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(((angle / 180.0) * pi))) ^ 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[(N[(angle / 180.0), $MachinePrecision] * Pi), $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 81.8%

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

   1. associate-*l/ 81.3%

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

   2. associate-*r/ 81.7%

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

   3. add-exp-log 39.8%

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

   4. div-inv 39.8%

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

   5. metadata-eval 39.8%

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

3. Applied egg-rr 39.8%

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

4. Final simplification 39.8%

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

Alternative 6: 79.8% accurate, 0.8× speedup

Math

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

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (+
  (pow (* a (sin (* (/ angle 180.0) PI))) 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(((angle / 180.0) * ((double) M_PI)))), 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(((angle / 180.0) * Math.PI))), 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(((angle / 180.0) * math.pi))), 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(Float64(angle / 180.0) * pi))) ^ 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[(N[(angle / 180.0), $MachinePrecision] * Pi), $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 81.8%

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

   1. associate-*l/ 81.3%

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

   2. associate-*r/ 81.7%

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

   3. expm1-log1p-u 63.6%

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

   4. div-inv 63.6%

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

   5. metadata-eval 63.6%

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

3. Applied egg-rr 63.6%

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

4. Final simplification 63.6%

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

Alternative 7: 79.8% 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 81.8%

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

   1. associate-*l/ 81.4%

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

   2. associate-*r/ 81.8%

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

   3. associate-*l/ 81.3%

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

   4. associate-*r/ 81.7%

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

3. Simplified 81.7%

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

4. Final simplification 81.7%

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

Alternative 8: 79.8% accurate, 1.0× speedup

Math

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

FPCore

NOTE: angle should be positive before calling this function
(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

angle = abs(angle);
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

angle = Math.abs(angle);
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

angle = abs(angle)
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

angle = abs(angle)
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

angle = abs(angle)
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

NOTE: angle should be positive before calling this function
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]]

Derivation

1. Initial program 81.8%

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

2. Final simplification 81.8%

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

Alternative 9: 79.8% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Python

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

Julia

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

MATLAB

angle = abs(angle)
function tmp = code(a, b, angle)
	tmp = (b ^ 2.0) + ((a * sin((0.005555555555555556 * (angle * 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[Power[N[(a * N[Sin[N[(0.005555555555555556 * N[(angle * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 81.8%

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

   1. associate-*l/ 81.4%

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

   2. associate-*r/ 81.8%

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

   3. associate-*l/ 81.3%

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

   4. associate-*r/ 81.7%

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

3. Simplified 81.7%

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

4. Taylor expanded in angle around 0 81.5%

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

5. Taylor expanded in angle around inf 81.1%

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

6. Final simplification 81.1%

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

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 81.8%

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

   1. associate-*l/ 81.4%

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

   2. associate-*r/ 81.8%

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

   3. associate-*l/ 81.3%

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

   4. associate-*r/ 81.7%

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

3. Simplified 81.7%

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

4. Taylor expanded in angle around 0 81.5%

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

5. Final simplification 81.5%

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

Alternative 11: 79.8% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Python

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

Julia

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

MATLAB

angle = abs(angle)
function tmp = code(a, b, angle)
	tmp = ((a * sin(((angle / 180.0) * pi))) ^ 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[(N[(angle / 180.0), $MachinePrecision] * Pi), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[b, 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 81.8%

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

   1. associate-*l/ 81.3%

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

   2. associate-*r/ 81.7%

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

   3. expm1-log1p-u 63.6%

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

   4. div-inv 63.6%

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

   5. metadata-eval 63.6%

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

3. Applied egg-rr 63.6%

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

4. Taylor expanded in angle around 0 81.5%

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

5. Final simplification 81.5%

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

Alternative 12: 74.8% 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 81.8%

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

   1. associate-*l/ 81.4%

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

   2. associate-*r/ 81.8%

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

   3. associate-*l/ 81.3%

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

   4. associate-*r/ 81.7%

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

3. Simplified 81.7%

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

4. Taylor expanded in angle around 0 81.5%

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

5. Taylor expanded in angle around 0 76.8%

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

   1. *-commutative 76.8%

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

7. Simplified 76.8%

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

8. Taylor expanded in angle around 0 67.9%

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

   1. associate-*r* 67.9%

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

   2. unpow2 67.9%

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

   3. unpow2 67.9%

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

   4. unswap-sqr 76.5%

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

   5. *-commutative 76.5%

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

   6. unpow2 76.5%

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

   7. swap-sqr 76.4%

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

   8. unpow2 76.4%

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

   9. *-commutative 76.4%

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

   10. associate-*r* 76.5%

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

   11. *-commutative 76.5%

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

10. Simplified 76.5%

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

11. Final simplification 76.5%

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

Alternative 13: 74.8% accurate, 2.0× speedup

Math

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

FPCore

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

C

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

Python

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

Julia

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

MATLAB

angle = abs(angle)
function tmp = code(a, b, angle)
	tmp = (b ^ 2.0) + ((0.005555555555555556 * (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[Power[N[(0.005555555555555556 * N[(angle * N[(a * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 81.8%

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

   1. associate-*l/ 81.4%

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

   2. associate-*r/ 81.8%

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

   3. associate-*l/ 81.3%

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

   4. associate-*r/ 81.7%

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

3. Simplified 81.7%

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

4. Taylor expanded in angle around 0 81.5%

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

5. Taylor expanded in angle around 0 76.8%

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

   1. *-commutative 76.8%

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

7. Simplified 76.8%

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

8. Final simplification 76.8%

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

Alternative 14: 74.8% accurate, 2.0× speedup

Math

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

FPCore

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

C

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

Python

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

Julia

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

MATLAB

angle = abs(angle)
function tmp = code(a, b, angle)
	tmp = (b ^ 2.0) + ((a * (0.005555555555555556 * (angle * 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[Power[N[(a * N[(0.005555555555555556 * N[(angle * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 81.8%

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

   1. associate-*l/ 81.4%

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

   2. associate-*r/ 81.8%

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

   3. associate-*l/ 81.3%

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

   4. associate-*r/ 81.7%

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

3. Simplified 81.7%

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

4. Taylor expanded in angle around 0 81.5%

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

5. Taylor expanded in angle around 0 76.8%

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

6. Final simplification 76.8%

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

Alternative 15: 74.8% accurate, 2.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

angle = abs(angle)
function tmp = code(a, b, angle)
	tmp = (b ^ 2.0) + ((a * (angle * (pi / 180.0))) ^ 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[Power[N[(a * N[(angle * N[(Pi / 180.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 81.8%

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

   1. associate-*l/ 81.4%

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

   2. associate-*r/ 81.8%

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

   3. associate-*l/ 81.3%

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

   4. associate-*r/ 81.7%

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

3. Simplified 81.7%

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

4. Taylor expanded in angle around 0 81.5%

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

5. Taylor expanded in angle around 0 76.8%

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

   1. associate-*r* 76.8%

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

   2. metadata-eval 76.8%

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

   3. associate-/r/ 76.8%

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

   4. associate-*l/ 76.8%

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

   5. *-lft-identity 76.8%

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

   6. associate-/r/ 76.8%

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

   7. *-commutative 76.8%

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

7. Simplified 76.8%

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

8. Final simplification 76.8%

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

Reproduce

herbie shell --seed 2023200 
(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)))