ab-angle->ABCF A

Percentage Accurate: 80.3% → 80.2%

Time: 37.4s

Alternatives: 10

Speedup: N/A×

• 36.6% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Initial Program: 80.3% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Alternative 1: 80.2% accurate, 0.8× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

Wolfram

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

Derivation

1. Initial program 82.5%

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

   1. associate-*l/ 82.5%

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

   2. associate-*l/ 82.6%

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

3. Simplified 82.6%

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

   1. expm1-log1p-u 66.3%

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

5. Applied egg-rr 66.3%

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

6. Final simplification 66.3%

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

Alternative 2: 80.3% accurate, 0.8× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

Wolfram

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

Derivation

1. Initial program 82.5%

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

   1. associate-*l/ 82.5%

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

   2. associate-*r/ 82.5%

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

   3. expm1-log1p-u 67.0%

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

   4. div-inv 67.0%

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

   5. metadata-eval 67.0%

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

3. Applied egg-rr 67.0%

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

4. Final simplification 67.0%

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

Alternative 3: 80.3% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 82.5%

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

   1. associate-*l/ 82.5%

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

   2. associate-*r/ 82.5%

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

   3. associate-*l/ 82.6%

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

   4. associate-*r/ 82.6%

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

3. Simplified 82.6%

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

4. Taylor expanded in angle around inf 82.6%

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

5. Final simplification 82.6%

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

Alternative 4: 80.2% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 82.5%

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

   1. associate-*l/ 82.5%

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

   2. associate-*l/ 82.6%

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

3. Simplified 82.6%

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

4. Final simplification 82.6%

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

Alternative 5: 80.1% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

Wolfram

NOTE: angle should be positive before calling this function
code[a_, b_, angle_] := N[(N[Power[b, 2.0], $MachinePrecision] + N[Power[N[(a * N[Sin[N[(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 82.5%

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

   1. associate-*l/ 82.5%

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

   2. associate-*r/ 82.5%

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

   3. associate-*l/ 82.6%

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

   4. associate-*r/ 82.6%

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

3. Simplified 82.6%

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

4. Taylor expanded in angle around 0 81.8%

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

   1. div-inv 81.8%

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

   2. metadata-eval 81.8%

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

   3. expm1-log1p-u 66.3%

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

   4. associate-*r* 66.3%

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

   5. *-commutative 66.3%

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

   6. associate-*r* 66.3%

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

6. Applied egg-rr 66.3%

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

7. Final simplification 66.3%

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

Alternative 6: 80.0% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

Wolfram

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

Derivation

1. Initial program 82.5%

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

   1. associate-*l/ 82.5%

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

   2. associate-*l/ 82.6%

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

3. Simplified 82.6%

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

   1. expm1-log1p-u 66.3%

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

5. Applied egg-rr 66.3%

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

6. Taylor expanded in angle around 0 65.7%

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

7. Final simplification 65.7%

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

Alternative 7: 80.2% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 82.5%

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

   1. associate-*l/ 82.5%

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

   2. associate-*r/ 82.5%

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

   3. associate-*l/ 82.6%

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

   4. associate-*r/ 82.6%

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

3. Simplified 82.6%

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

4. Taylor expanded in angle around 0 81.8%

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

5. Final simplification 81.8%

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

Alternative 8: 80.2% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 82.5%

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

   1. associate-*l/ 82.5%

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

   2. associate-*r/ 82.5%

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

   3. associate-*l/ 82.6%

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

   4. associate-*r/ 82.6%

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

3. Simplified 82.6%

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

4. Taylor expanded in angle around 0 81.8%

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

   1. associate-*r/ 81.8%

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

   2. *-commutative 81.8%

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

   3. associate-/l* 81.8%

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

6. Applied egg-rr 81.8%

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

7. Final simplification 81.8%

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

Alternative 9: 75.3% accurate, 2.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 82.5%

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

   1. associate-*l/ 82.5%

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

   2. associate-*r/ 82.5%

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

   3. associate-*l/ 82.6%

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

   4. associate-*r/ 82.6%

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

3. Simplified 82.6%

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

4. Taylor expanded in angle around 0 81.8%

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

5. Taylor expanded in angle around 0 77.1%

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

   1. *-commutative 77.1%

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

7. Simplified 77.1%

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

8. Taylor expanded in angle around 0 64.1%

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

   1. *-commutative 64.1%

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

   2. *-commutative 64.1%

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

   3. unpow2 64.1%

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

   4. unpow2 64.1%

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

   5. swap-sqr 64.1%

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

   6. unpow2 64.1%

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

   7. swap-sqr 77.1%

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

   8. associate-*r* 77.2%

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

   9. associate-*r* 77.1%

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

   10. unpow2 77.1%

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

   11. associate-*r* 77.1%

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

   12. *-commutative 77.1%

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

10. Simplified 77.1%

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

11. Final simplification 77.1%

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

Alternative 10: 75.3% accurate, 2.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 82.5%

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

   1. associate-*l/ 82.5%

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

   2. associate-*r/ 82.5%

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

   3. associate-*l/ 82.6%

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

   4. associate-*r/ 82.6%

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

3. Simplified 82.6%

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

4. Taylor expanded in angle around 0 81.8%

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

5. Taylor expanded in angle around 0 77.1%

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

   1. associate-*r* 77.1%

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

   2. metadata-eval 77.1%

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

   3. associate-/r/ 77.1%

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

   4. associate-*l/ 77.1%

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

   5. *-lft-identity 77.1%

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

   6. associate-/r/ 77.1%

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

   7. *-commutative 77.1%

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

7. Simplified 77.1%

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

8. Final simplification 77.1%

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

Reproduce

herbie shell --seed 2023255 
(FPCore (a b angle)
  :name "ab-angle->ABCF A"
  :precision binary64
  (+ (pow (* a (sin (* (/ angle 180.0) PI))) 2.0) (pow (* b (cos (* (/ angle 180.0) PI))) 2.0)))