ab-angle->ABCF C

Percentage Accurate: 79.9% → 79.8%

Time: 31.4s

Alternatives: 13

Speedup: 1.0×

• 35.9% 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: 79.9% 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: 79.8% accurate, 0.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \pi \cdot \frac{angle}{180}\\ {\left(a \cdot \cos \left({\left(\sqrt[3]{{\left(\sqrt[3]{t_0}\right)}^{3}}\right)}^{3}\right)\right)}^{2} + {\left(b \cdot \sin t_0\right)}^{2} \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* PI (/ angle 180.0))))
   (+
    (pow (* a (cos (pow (cbrt (pow (cbrt t_0) 3.0)) 3.0))) 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(pow(cbrt(pow(cbrt(t_0), 3.0)), 3.0))), 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(Math.pow(Math.cbrt(Math.pow(Math.cbrt(t_0), 3.0)), 3.0))), 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((cbrt((cbrt(t_0) ^ 3.0)) ^ 3.0))) ^ 2.0) + (Float64(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[N[Power[N[Power[N[Power[N[Power[t$95$0, 1/3], $MachinePrecision], 3.0], $MachinePrecision], 1/3], $MachinePrecision], 3.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Initial program 78.3%

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

   1. add-cube-cbrt 78.4%

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

   2. pow3 78.4%

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

   3. div-inv 78.3%

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

   4. metadata-eval 78.3%

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

3. Applied egg-rr 78.3%

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

   1. add-cube-cbrt 78.4%

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

   2. pow3 78.4%

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

   3. metadata-eval 78.4%

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

   4. div-inv 78.5%

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

5. Applied egg-rr 78.5%

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

6. Final simplification 78.5%

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

Alternative 2: 65.1% accurate, 0.6× speedup

Math

\[\begin{array}{l} \\ {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(a \cdot \cos \left({\left(\sqrt[3]{\pi \cdot \mathsf{expm1}\left(\mathsf{log1p}\left(angle \cdot 0.005555555555555556\right)\right)}\right)}^{3}\right)\right)}^{2} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (+
  (pow (* b (sin (* PI (/ angle 180.0)))) 2.0)
  (pow
   (*
    a
    (cos
     (pow (cbrt (* PI (expm1 (log1p (* angle 0.005555555555555556))))) 3.0)))
   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(pow(cbrt((((double) M_PI) * expm1(log1p((angle * 0.005555555555555556))))), 3.0))), 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(Math.pow(Math.cbrt((Math.PI * Math.expm1(Math.log1p((angle * 0.005555555555555556))))), 3.0))), 2.0);
}

Julia

function code(a, b, angle)
	return Float64((Float64(b * sin(Float64(pi * Float64(angle / 180.0)))) ^ 2.0) + (Float64(a * cos((cbrt(Float64(pi * expm1(log1p(Float64(angle * 0.005555555555555556))))) ^ 3.0))) ^ 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[Power[N[Power[N[(Pi * N[(Exp[N[Log[1 + N[(angle * 0.005555555555555556), $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision], 3.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 78.3%

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

   1. add-cube-cbrt 78.4%

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

   2. pow3 78.4%

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

   3. div-inv 78.3%

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

   4. metadata-eval 78.3%

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

3. Applied egg-rr 78.3%

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

   1. expm1-log1p-u 65.6%

      \[\leadsto {\left(a \cdot \cos \left({\left(\sqrt[3]{\pi \cdot \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(angle \cdot 0.005555555555555556\right)\right)}}\right)}^{3}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

5. Applied egg-rr 65.6%

   \[\leadsto {\left(a \cdot \cos \left({\left(\sqrt[3]{\pi \cdot \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(angle \cdot 0.005555555555555556\right)\right)}}\right)}^{3}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

6. Final simplification 65.6%

   \[\leadsto {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(a \cdot \cos \left({\left(\sqrt[3]{\pi \cdot \mathsf{expm1}\left(\mathsf{log1p}\left(angle \cdot 0.005555555555555556\right)\right)}\right)}^{3}\right)\right)}^{2} \]

Alternative 3: 79.9% accurate, 0.9× speedup

Math

\[\begin{array}{l} \\ {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot \sqrt[3]{1.7146776406035666 \cdot 10^{-7}}\right)\right)\right)}^{2} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (+
  (pow (* b (sin (* PI (/ angle 180.0)))) 2.0)
  (pow (* a (cos (* PI (* angle (cbrt 1.7146776406035666e-7))))) 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 * cbrt(1.7146776406035666e-7))))), 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 * Math.cbrt(1.7146776406035666e-7))))), 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 * cbrt(1.7146776406035666e-7))))) ^ 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 * N[Power[1.7146776406035666e-7, 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 78.3%

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

   1. add-cbrt-cube 62.9%

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

   2. pow1/3 50.5%

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

   3. pow3 50.5%

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

   4. div-inv 50.5%

      \[\leadsto {\left(a \cdot \cos \left({\left({\left(\pi \cdot \color{blue}{\left(angle \cdot \frac{1}{180}\right)}\right)}^{3}\right)}^{0.3333333333333333}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

   5. metadata-eval 50.5%

      \[\leadsto {\left(a \cdot \cos \left({\left({\left(\pi \cdot \left(angle \cdot \color{blue}{0.005555555555555556}\right)\right)}^{3}\right)}^{0.3333333333333333}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

3. Applied egg-rr 50.5%

   \[\leadsto {\left(a \cdot \cos \color{blue}{\left({\left({\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{3}\right)}^{0.3333333333333333}\right)}\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

4. Taylor expanded in angle around 0 78.4%

   \[\leadsto {\left(a \cdot \cos \color{blue}{\left(\sqrt[3]{1.7146776406035666 \cdot 10^{-7}} \cdot \left(angle \cdot \pi\right)\right)}\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
5. Step-by-step derivation

   1. *-commutative 78.4%

      \[\leadsto {\left(a \cdot \cos \color{blue}{\left(\left(angle \cdot \pi\right) \cdot \sqrt[3]{1.7146776406035666 \cdot 10^{-7}}\right)}\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

   2. *-commutative 78.4%

      \[\leadsto {\left(a \cdot \cos \left(\color{blue}{\left(\pi \cdot angle\right)} \cdot \sqrt[3]{1.7146776406035666 \cdot 10^{-7}}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

   3. associate-*l* 78.4%

      \[\leadsto {\left(a \cdot \cos \color{blue}{\left(\pi \cdot \left(angle \cdot \sqrt[3]{1.7146776406035666 \cdot 10^{-7}}\right)\right)}\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

6. Simplified 78.4%

   \[\leadsto {\left(a \cdot \cos \color{blue}{\left(\pi \cdot \left(angle \cdot \sqrt[3]{1.7146776406035666 \cdot 10^{-7}}\right)\right)}\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

7. Final simplification 78.4%

   \[\leadsto {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot \sqrt[3]{1.7146776406035666 \cdot 10^{-7}}\right)\right)\right)}^{2} \]

Alternative 4: 79.8% accurate, 1.0× speedup

Math

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

FPCore

(FPCore (a b angle)
 :precision binary64
 (+
  (pow (* a (cos (* PI (/ angle 180.0)))) 2.0)
  (pow (* b (sin (* 0.005555555555555556 (* PI angle)))) 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((0.005555555555555556 * (((double) M_PI) * angle)))), 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((0.005555555555555556 * (Math.PI * angle)))), 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((0.005555555555555556 * (math.pi * angle)))), 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(0.005555555555555556 * Float64(pi * angle)))) ^ 2.0))
end

MATLAB

function tmp = code(a, b, angle)
	tmp = ((a * cos((pi * (angle / 180.0)))) ^ 2.0) + ((b * sin((0.005555555555555556 * (pi * angle)))) ^ 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[(0.005555555555555556 * N[(Pi * angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 78.3%

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

2. Taylor expanded in angle around inf 78.2%

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

3. Final simplification 78.2%

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

Alternative 5: 79.9% 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 78.3%

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

2. Final simplification 78.3%

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

Alternative 6: 79.9% accurate, 1.0× speedup

Math

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

FPCore

(FPCore (a b angle)
 :precision binary64
 (+
  (pow (* b (sin (* PI (/ angle 180.0)))) 2.0)
  (pow (* a (cos (/ PI (/ 180.0 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((((double) M_PI) / (180.0 / 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.PI / (180.0 / 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.pi / (180.0 / 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(pi / Float64(180.0 / angle)))) ^ 2.0))
end

MATLAB

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

Derivation

1. Initial program 78.3%

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

   1. clear-num 78.3%

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

   2. un-div-inv 78.4%

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

3. Applied egg-rr 78.4%

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

4. Final simplification 78.4%

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

Alternative 7: 64.9% accurate, 1.0× speedup

Math

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

FPCore

(FPCore (a b angle)
 :precision binary64
 (+
  (pow a 2.0)
  (pow (* b (sin (expm1 (log1p (* PI (* angle 0.005555555555555556)))))) 2.0)))

C

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

Java

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

Python

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

Julia

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

Wolfram

code[a_, b_, angle_] := N[(N[Power[a, 2.0], $MachinePrecision] + N[Power[N[(b * N[Sin[N[(Exp[N[Log[1 + N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 78.3%

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

2. Taylor expanded in angle around 0 78.1%

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

   1. expm1-log1p-u 65.2%

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

   2. div-inv 65.2%

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

   3. metadata-eval 65.2%

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

4. Applied egg-rr 65.2%

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

5. Final simplification 65.2%

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

Alternative 8: 79.7% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 78.3%

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

2. Taylor expanded in angle around 0 78.1%

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

3. Taylor expanded in b around 0 78.0%

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

4. Final simplification 78.0%

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

Alternative 9: 79.7% accurate, 1.5× speedup

Math

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

FPCore

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

Python

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

Julia

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

MATLAB

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

Derivation

1. Initial program 78.3%

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

2. Taylor expanded in angle around 0 78.1%

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

3. Final simplification 78.1%

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

Alternative 10: 74.9% accurate, 2.0× speedup

Math

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

FPCore

(FPCore (a b angle)
 :precision binary64
 (+
  (pow a 2.0)
  (* 3.08641975308642e-5 (* (* angle b) (* PI (* PI (* angle b)))))))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 78.3%

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

2. Taylor expanded in angle around 0 78.1%

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

3. Taylor expanded in angle around 0 71.7%

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

4. Taylor expanded in angle around 0 62.3%

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

   1. associate-*r* 62.3%

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

   2. unpow2 62.3%

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

   3. unpow2 62.3%

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

   4. unswap-sqr 71.7%

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

   5. *-commutative 71.7%

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

   6. unpow2 71.7%

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

   7. swap-sqr 71.7%

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

   8. unpow2 71.7%

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

   9. associate-*r* 71.7%

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

   10. *-commutative 71.7%

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

   11. *-commutative 71.7%

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

6. Simplified 71.7%

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

   1. unpow2 71.7%

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

   2. associate-*r* 71.7%

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

   3. associate-*l* 71.7%

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

   4. *-commutative 71.7%

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

   5. *-commutative 71.7%

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

   6. associate-*l* 71.8%

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

8. Applied egg-rr 71.8%

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

9. Final simplification 71.8%

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

Alternative 11: 74.9% accurate, 2.0× speedup

Math

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

FPCore

(FPCore (a b angle)
 :precision binary64
 (+ (pow a 2.0) (* 3.08641975308642e-5 (pow (* b (* PI angle)) 2.0))))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 78.3%

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

2. Taylor expanded in angle around 0 78.1%

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

3. Taylor expanded in angle around 0 71.7%

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

4. Taylor expanded in angle around 0 62.3%

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

   1. associate-*r* 62.3%

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

   2. unpow2 62.3%

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

   3. unpow2 62.3%

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

   4. unswap-sqr 71.7%

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

   5. *-commutative 71.7%

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

   6. unpow2 71.7%

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

   7. swap-sqr 71.7%

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

   8. unpow2 71.7%

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

   9. associate-*r* 71.7%

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

   10. *-commutative 71.7%

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

   11. *-commutative 71.7%

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

6. Simplified 71.7%

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

7. Final simplification 71.7%

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

Alternative 12: 74.9% accurate, 2.0× speedup

Math

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

FPCore

(FPCore (a b angle)
 :precision binary64
 (+ (pow a 2.0) (* 3.08641975308642e-5 (pow (* PI (* angle b)) 2.0))))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 78.3%

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

2. Taylor expanded in angle around 0 78.1%

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

3. Taylor expanded in angle around 0 71.7%

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

   1. *-commutative 71.7%

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

   2. unpow-prod-down 71.7%

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

   3. associate-*r* 71.7%

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

   4. *-commutative 71.7%

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

   5. metadata-eval 71.7%

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

5. Applied egg-rr 71.7%

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

6. Final simplification 71.7%

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

Alternative 13: 74.9% accurate, 2.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 78.3%

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

2. Taylor expanded in angle around 0 78.1%

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

3. Taylor expanded in angle around 0 71.7%

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

4. Final simplification 71.7%

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

Reproduce

herbie shell --seed 2023257 
(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)))