ab-angle->ABCF C

Percentage Accurate: 79.7% → 79.6%

Time: 28.1s

Alternatives: 12

Speedup: 1.0×

• 35.7% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Initial Program: 79.7% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Alternative 1: 79.6% accurate, 0.6× speedup

Math

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

FPCore

(FPCore (a b angle)
 :precision binary64
 (+
  (pow
   (*
    a
    (cos (* (cbrt PI) (* (pow (cbrt PI) 2.0) (* angle 0.005555555555555556)))))
   2.0)
  (pow (* b (sin (* PI (/ angle 180.0)))) 2.0)))

C

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

Java

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

Julia

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

Wolfram

code[a_, b_, angle_] := N[(N[Power[N[(a * N[Cos[N[(N[Power[Pi, 1/3], $MachinePrecision] * N[(N[Power[N[Power[Pi, 1/3], $MachinePrecision], 2.0], $MachinePrecision] * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Sin[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 83.1%

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

   1. associate-*r/ 83.2%

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

   2. clear-num 83.2%

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

3. Applied egg-rr 83.2%

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

   1. clear-num 83.2%

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

   2. associate-*r/ 83.1%

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

   3. *-commutative 83.1%

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

   4. add-cube-cbrt 83.2%

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

   5. associate-*r* 83.3%

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

   6. div-inv 83.3%

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

   7. metadata-eval 83.3%

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

   8. pow2 83.3%

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

5. Applied egg-rr 83.3%

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

6. Final simplification 83.3%

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

Alternative 2: 79.6% accurate, 0.7× speedup

Math

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

FPCore

(FPCore (a b angle)
 :precision binary64
 (+
  (pow (* b (sin (* PI (/ angle 180.0)))) 2.0)
  (pow (* a (cos (* (/ (sqrt PI) 180.0) (/ (sqrt PI) (/ 1.0 angle))))) 2.0)))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

code[a_, b_, angle_] := N[(N[Power[N[(b * N[Sin[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(a * N[Cos[N[(N[(N[Sqrt[Pi], $MachinePrecision] / 180.0), $MachinePrecision] * N[(N[Sqrt[Pi], $MachinePrecision] / N[(1.0 / angle), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 83.1%

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

   1. associate-*r/ 83.2%

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

   2. clear-num 83.2%

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

3. Applied egg-rr 83.2%

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

   1. clear-num 83.2%

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

   2. associate-/l* 83.1%

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

   3. add-sqr-sqrt 83.3%

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

   4. div-inv 83.2%

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

   5. times-frac 83.2%

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

5. Applied egg-rr 83.2%

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

6. Final simplification 83.2%

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

Alternative 3: 79.6% accurate, 1.0× speedup

Math

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

FPCore

(FPCore (a b angle)
 :precision binary64
 (+
  (pow (* b (sin (* PI (/ angle 180.0)))) 2.0)
  (pow (* a (cos (* 0.005555555555555556 (/ PI (/ 1.0 angle))))) 2.0)))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

code[a_, b_, angle_] := N[(N[Power[N[(b * N[Sin[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(a * N[Cos[N[(0.005555555555555556 * N[(Pi / N[(1.0 / angle), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 83.1%

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

   1. associate-*r/ 83.2%

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

   2. clear-num 83.2%

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

3. Applied egg-rr 83.2%

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

   1. clear-num 83.2%

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

   2. associate-/l* 83.1%

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

   3. *-un-lft-identity 83.1%

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

   4. div-inv 83.2%

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

   5. times-frac 83.2%

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

   6. metadata-eval 83.2%

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

5. Applied egg-rr 83.2%

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

6. Final simplification 83.2%

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

Alternative 4: 79.7% accurate, 1.0× speedup

Math

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

FPCore

(FPCore (a b angle)
 :precision binary64
 (+
  (pow (* b (sin (* PI (/ angle 180.0)))) 2.0)
  (pow (* a (cos (/ 1.0 (/ 180.0 (* angle PI))))) 2.0)))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

code[a_, b_, angle_] := N[(N[Power[N[(b * N[Sin[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(a * N[Cos[N[(1.0 / N[(180.0 / N[(angle * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 83.1%

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

   1. associate-*r/ 83.2%

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

   2. clear-num 83.2%

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

3. Applied egg-rr 83.2%

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

4. Final simplification 83.2%

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

Alternative 5: 79.7% accurate, 1.0× speedup

Math

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

FPCore

(FPCore (a b angle)
 :precision binary64
 (+
  (pow (* b (sin (* PI (/ angle 180.0)))) 2.0)
  (pow (* a (cos (* PI (* angle 0.005555555555555556)))) 2.0)))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

code[a_, b_, angle_] := N[(N[Power[N[(b * N[Sin[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(a * N[Cos[N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 83.1%

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

2. Taylor expanded in angle around inf 83.1%

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

   1. associate-*r* 83.1%

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

   2. *-commutative 83.1%

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

   3. *-commutative 83.1%

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

   4. *-commutative 83.1%

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

4. Simplified 83.1%

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

5. Final simplification 83.1%

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

Alternative 6: 79.8% accurate, 1.0× speedup

Math

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

FPCore

(FPCore (a b angle)
 :precision binary64
 (+
  (pow (* a (cos (* PI (/ angle 180.0)))) 2.0)
  (pow (* b (sin (* angle (* 0.005555555555555556 PI)))) 2.0)))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

code[a_, b_, angle_] := N[(N[Power[N[(a * N[Cos[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Sin[N[(angle * N[(0.005555555555555556 * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 83.1%

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

2. Taylor expanded in angle around inf 83.1%

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

   1. associate-*r* 83.0%

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

   2. *-commutative 83.0%

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

   3. associate-*l* 83.1%

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

4. Simplified 83.1%

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

5. Final simplification 83.1%

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

Alternative 7: 79.7% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 83.1%

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

2. Final simplification 83.1%

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

Alternative 8: 79.7% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 83.1%

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

   1. associate-*r/ 83.2%

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

3. Applied egg-rr 83.2%

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

4. Final simplification 83.2%

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

Alternative 9: 79.6% accurate, 1.5× speedup

Math

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

FPCore

(FPCore (a b angle)
 :precision binary64
 (+ (pow (* b (sin (* angle (* 0.005555555555555556 PI)))) 2.0) (pow a 2.0)))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

code[a_, b_, angle_] := N[(N[Power[N[(b * N[Sin[N[(angle * N[(0.005555555555555556 * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 83.1%

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

2. Taylor expanded in angle around 0 83.1%

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

3. Taylor expanded in angle around inf 83.0%

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

   1. associate-*r* 83.0%

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

   2. *-commutative 83.0%

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

   3. associate-*l* 83.1%

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

5. Simplified 83.1%

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

6. Final simplification 83.1%

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

Alternative 10: 67.2% accurate, 2.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;b \leq 3.9 \cdot 10^{-114}:\\ \;\;\;\;{a}^{2} + {\left(b \cdot 0\right)}^{2}\\ \mathbf{else}:\\ \;\;\;\;{a}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\left(angle \cdot \left(\pi \cdot b\right)\right)}^{2}\\ \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (if (<= b 3.9e-114)
   (+ (pow a 2.0) (pow (* b 0.0) 2.0))
   (+ (pow a 2.0) (* 3.08641975308642e-5 (pow (* angle (* PI b)) 2.0)))))

C

double code(double a, double b, double angle) {
	double tmp;
	if (b <= 3.9e-114) {
		tmp = pow(a, 2.0) + pow((b * 0.0), 2.0);
	} else {
		tmp = pow(a, 2.0) + (3.08641975308642e-5 * pow((angle * (((double) M_PI) * b)), 2.0));
	}
	return tmp;
}

Java

public static double code(double a, double b, double angle) {
	double tmp;
	if (b <= 3.9e-114) {
		tmp = Math.pow(a, 2.0) + Math.pow((b * 0.0), 2.0);
	} else {
		tmp = Math.pow(a, 2.0) + (3.08641975308642e-5 * Math.pow((angle * (Math.PI * b)), 2.0));
	}
	return tmp;
}

Python

def code(a, b, angle):
	tmp = 0
	if b <= 3.9e-114:
		tmp = math.pow(a, 2.0) + math.pow((b * 0.0), 2.0)
	else:
		tmp = math.pow(a, 2.0) + (3.08641975308642e-5 * math.pow((angle * (math.pi * b)), 2.0))
	return tmp

Julia

function code(a, b, angle)
	tmp = 0.0
	if (b <= 3.9e-114)
		tmp = Float64((a ^ 2.0) + (Float64(b * 0.0) ^ 2.0));
	else
		tmp = Float64((a ^ 2.0) + Float64(3.08641975308642e-5 * (Float64(angle * Float64(pi * b)) ^ 2.0)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, b, angle)
	tmp = 0.0;
	if (b <= 3.9e-114)
		tmp = (a ^ 2.0) + ((b * 0.0) ^ 2.0);
	else
		tmp = (a ^ 2.0) + (3.08641975308642e-5 * ((angle * (pi * b)) ^ 2.0));
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, b_, angle_] := If[LessEqual[b, 3.9e-114], N[(N[Power[a, 2.0], $MachinePrecision] + N[Power[N[(b * 0.0), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(N[Power[a, 2.0], $MachinePrecision] + N[(3.08641975308642e-5 * N[Power[N[(angle * N[(Pi * b), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if b < 3.90000000000000002e-114

   1. Initial program 81.0%

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

   2. Taylor expanded in angle around 0 81.0%

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

      1. add-cube-cbrt 80.7%

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

      2. pow3 80.6%

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

      3. pow-to-exp 39.3%

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

      4. div-inv 39.3%

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

      5. metadata-eval 39.3%

         \[\leadsto {\left(a \cdot 1\right)}^{2} + {\left(b \cdot \sin \left(e^{\log \left(\sqrt[3]{\pi \cdot \left(angle \cdot \color{blue}{0.005555555555555556}\right)}\right) \cdot 3}\right)\right)}^{2} \]

   4. Applied egg-rr 39.3%

      \[\leadsto {\left(a \cdot 1\right)}^{2} + {\left(b \cdot \sin \color{blue}{\left(e^{\log \left(\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right) \cdot 3}\right)}\right)}^{2} \]

   5. Taylor expanded in angle around 0 61.7%

      \[\leadsto {\left(a \cdot 1\right)}^{2} + {\left(b \cdot \color{blue}{0}\right)}^{2} \]

   if 3.90000000000000002e-114 < b

   1. Initial program 87.3%

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

   2. Taylor expanded in angle around 0 87.4%

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

   3. Taylor expanded in angle around 0 85.2%

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

      1. *-commutative 85.2%

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

   5. Simplified 85.2%

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

      1. *-commutative 85.2%

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

      2. unpow-prod-down 85.2%

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

      3. metadata-eval 85.2%

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

   7. Applied egg-rr 85.2%

      \[\leadsto {\left(a \cdot 1\right)}^{2} + \color{blue}{{\left(angle \cdot \left(\pi \cdot b\right)\right)}^{2} \cdot 3.08641975308642 \cdot 10^{-5}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 69.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq 3.9 \cdot 10^{-114}:\\ \;\;\;\;{a}^{2} + {\left(b \cdot 0\right)}^{2}\\ \mathbf{else}:\\ \;\;\;\;{a}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\left(angle \cdot \left(\pi \cdot b\right)\right)}^{2}\\ \end{array} \]

Alternative 11: 67.2% accurate, 2.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;b \leq 3.4 \cdot 10^{-114}:\\ \;\;\;\;{a}^{2} + {\left(b \cdot 0\right)}^{2}\\ \mathbf{else}:\\ \;\;\;\;{a}^{2} + {\left(\pi \cdot \left(angle \cdot b\right)\right)}^{2} \cdot 3.08641975308642 \cdot 10^{-5}\\ \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (if (<= b 3.4e-114)
   (+ (pow a 2.0) (pow (* b 0.0) 2.0))
   (+ (pow a 2.0) (* (pow (* PI (* angle b)) 2.0) 3.08641975308642e-5))))

C

double code(double a, double b, double angle) {
	double tmp;
	if (b <= 3.4e-114) {
		tmp = pow(a, 2.0) + pow((b * 0.0), 2.0);
	} else {
		tmp = pow(a, 2.0) + (pow((((double) M_PI) * (angle * b)), 2.0) * 3.08641975308642e-5);
	}
	return tmp;
}

Java

public static double code(double a, double b, double angle) {
	double tmp;
	if (b <= 3.4e-114) {
		tmp = Math.pow(a, 2.0) + Math.pow((b * 0.0), 2.0);
	} else {
		tmp = Math.pow(a, 2.0) + (Math.pow((Math.PI * (angle * b)), 2.0) * 3.08641975308642e-5);
	}
	return tmp;
}

Python

def code(a, b, angle):
	tmp = 0
	if b <= 3.4e-114:
		tmp = math.pow(a, 2.0) + math.pow((b * 0.0), 2.0)
	else:
		tmp = math.pow(a, 2.0) + (math.pow((math.pi * (angle * b)), 2.0) * 3.08641975308642e-5)
	return tmp

Julia

function code(a, b, angle)
	tmp = 0.0
	if (b <= 3.4e-114)
		tmp = Float64((a ^ 2.0) + (Float64(b * 0.0) ^ 2.0));
	else
		tmp = Float64((a ^ 2.0) + Float64((Float64(pi * Float64(angle * b)) ^ 2.0) * 3.08641975308642e-5));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, b, angle)
	tmp = 0.0;
	if (b <= 3.4e-114)
		tmp = (a ^ 2.0) + ((b * 0.0) ^ 2.0);
	else
		tmp = (a ^ 2.0) + (((pi * (angle * b)) ^ 2.0) * 3.08641975308642e-5);
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, b_, angle_] := If[LessEqual[b, 3.4e-114], N[(N[Power[a, 2.0], $MachinePrecision] + N[Power[N[(b * 0.0), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(N[Power[a, 2.0], $MachinePrecision] + N[(N[Power[N[(Pi * N[(angle * b), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * 3.08641975308642e-5), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if b < 3.39999999999999981e-114

   1. Initial program 81.0%

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

   2. Taylor expanded in angle around 0 81.0%

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

      1. add-cube-cbrt 80.7%

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

      2. pow3 80.6%

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

      3. pow-to-exp 39.3%

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

      4. div-inv 39.3%

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

      5. metadata-eval 39.3%

         \[\leadsto {\left(a \cdot 1\right)}^{2} + {\left(b \cdot \sin \left(e^{\log \left(\sqrt[3]{\pi \cdot \left(angle \cdot \color{blue}{0.005555555555555556}\right)}\right) \cdot 3}\right)\right)}^{2} \]

   4. Applied egg-rr 39.3%

      \[\leadsto {\left(a \cdot 1\right)}^{2} + {\left(b \cdot \sin \color{blue}{\left(e^{\log \left(\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right) \cdot 3}\right)}\right)}^{2} \]

   5. Taylor expanded in angle around 0 61.7%

      \[\leadsto {\left(a \cdot 1\right)}^{2} + {\left(b \cdot \color{blue}{0}\right)}^{2} \]

   if 3.39999999999999981e-114 < b

   1. Initial program 87.3%

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

   2. Taylor expanded in angle around 0 87.4%

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

   3. Taylor expanded in angle around 0 85.2%

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

      1. *-commutative 85.2%

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

   5. Simplified 85.2%

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

      1. *-commutative 85.2%

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

      2. unpow-prod-down 85.2%

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

      3. metadata-eval 85.2%

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

   7. Applied egg-rr 85.2%

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

   8. Taylor expanded in angle around 0 85.2%

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

      1. associate-*r* 85.2%

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

   10. Simplified 85.2%

       \[\leadsto {\left(a \cdot 1\right)}^{2} + {\color{blue}{\left(\left(angle \cdot b\right) \cdot \pi\right)}}^{2} \cdot 3.08641975308642 \cdot 10^{-5} \]
3. Recombined 2 regimes into one program.

4. Final simplification 69.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq 3.4 \cdot 10^{-114}:\\ \;\;\;\;{a}^{2} + {\left(b \cdot 0\right)}^{2}\\ \mathbf{else}:\\ \;\;\;\;{a}^{2} + {\left(\pi \cdot \left(angle \cdot b\right)\right)}^{2} \cdot 3.08641975308642 \cdot 10^{-5}\\ \end{array} \]

Alternative 12: 57.0% accurate, 3.0× speedup

Math

\[\begin{array}{l} \\ {a}^{2} + {\left(b \cdot 0\right)}^{2} \end{array} \]

FPCore

(FPCore (a b angle) :precision binary64 (+ (pow a 2.0) (pow (* b 0.0) 2.0)))

C

double code(double a, double b, double angle) {
	return pow(a, 2.0) + pow((b * 0.0), 2.0);
}

Fortran

real(8) function code(a, b, angle)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: angle
    code = (a ** 2.0d0) + ((b * 0.0d0) ** 2.0d0)
end function

Java

public static double code(double a, double b, double angle) {
	return Math.pow(a, 2.0) + Math.pow((b * 0.0), 2.0);
}

Python

def code(a, b, angle):
	return math.pow(a, 2.0) + math.pow((b * 0.0), 2.0)

Julia

function code(a, b, angle)
	return Float64((a ^ 2.0) + (Float64(b * 0.0) ^ 2.0))
end

MATLAB

function tmp = code(a, b, angle)
	tmp = (a ^ 2.0) + ((b * 0.0) ^ 2.0);
end

Wolfram

code[a_, b_, angle_] := N[(N[Power[a, 2.0], $MachinePrecision] + N[Power[N[(b * 0.0), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 83.1%

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

2. Taylor expanded in angle around 0 83.1%

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

   1. add-cube-cbrt 82.9%

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

   2. pow3 82.8%

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

   3. pow-to-exp 38.6%

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

   4. div-inv 38.6%

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

   5. metadata-eval 38.6%

      \[\leadsto {\left(a \cdot 1\right)}^{2} + {\left(b \cdot \sin \left(e^{\log \left(\sqrt[3]{\pi \cdot \left(angle \cdot \color{blue}{0.005555555555555556}\right)}\right) \cdot 3}\right)\right)}^{2} \]

4. Applied egg-rr 38.6%

   \[\leadsto {\left(a \cdot 1\right)}^{2} + {\left(b \cdot \sin \color{blue}{\left(e^{\log \left(\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right) \cdot 3}\right)}\right)}^{2} \]

5. Taylor expanded in angle around 0 60.8%

   \[\leadsto {\left(a \cdot 1\right)}^{2} + {\left(b \cdot \color{blue}{0}\right)}^{2} \]

6. Final simplification 60.8%

   \[\leadsto {a}^{2} + {\left(b \cdot 0\right)}^{2} \]

Reproduce

herbie shell --seed 2023240 
(FPCore (a b angle)
  :name "ab-angle->ABCF C"
  :precision binary64
  (+ (pow (* a (cos (* PI (/ angle 180.0)))) 2.0) (pow (* b (sin (* PI (/ angle 180.0)))) 2.0)))