ab-angle->ABCF A

Percentage Accurate: 80.0% → 80.0%

Time: 44.0s

Alternatives: 8

Speedup: 1.5×

• 37.5% 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: 80.0% 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: 80.0% accurate, 0.8× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 76.6%

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

   1. *-commutative 76.6%

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

   2. add-sqr-sqrt 33.4%

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

   3. associate-/l* 33.4%

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

   4. associate-*r/ 33.4%

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

3. Applied egg-rr 33.4%

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

4. Final simplification 33.4%

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

Alternative 2: 80.0% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 76.6%

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

   1. associate-*l/ 76.5%

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

   2. clear-num 76.6%

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

   3. associate-/r* 76.7%

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

3. Applied egg-rr 76.7%

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

4. Final simplification 76.7%

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

Alternative 3: 80.1% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 76.6%

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

   1. unpow2 76.6%

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

   2. associate-*l/ 76.6%

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

   3. associate-*r/ 76.6%

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

   4. unpow2 76.6%

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

   5. associate-*l/ 76.6%

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

   6. associate-*r/ 76.6%

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

3. Simplified 76.6%

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

4. Final simplification 76.6%

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

Alternative 4: 80.0% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 76.6%

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

   1. *-commutative 76.6%

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

   2. associate-*r/ 76.5%

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

   3. associate-/l* 76.7%

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

3. Applied egg-rr 76.7%

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

4. Final simplification 76.7%

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

Alternative 5: 79.9% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 76.6%

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

   1. unpow2 76.6%

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

   2. associate-*l/ 76.6%

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

   3. associate-*r/ 76.6%

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

   4. unpow2 76.6%

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

   5. associate-*l/ 76.6%

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

   6. associate-*r/ 76.6%

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

3. Simplified 76.6%

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

4. Taylor expanded in angle around 0 76.4%

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

5. Taylor expanded in angle around inf 76.4%

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

6. Final simplification 76.4%

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

Alternative 6: 80.0% 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 76.6%

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

   1. unpow2 76.6%

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

   2. associate-*l/ 76.6%

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

   3. associate-*r/ 76.6%

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

   4. unpow2 76.6%

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

   5. associate-*l/ 76.6%

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

   6. associate-*r/ 76.6%

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

3. Simplified 76.6%

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

4. Taylor expanded in angle around 0 76.4%

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

5. Final simplification 76.4%

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

Alternative 7: 75.2% 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 76.6%

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

   1. unpow2 76.6%

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

   2. associate-*l/ 76.6%

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

   3. associate-*r/ 76.6%

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

   4. unpow2 76.6%

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

   5. associate-*l/ 76.6%

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

   6. associate-*r/ 76.6%

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

3. Simplified 76.6%

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

4. Taylor expanded in angle around 0 76.4%

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

5. Taylor expanded in angle around 0 71.4%

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

   1. *-commutative 71.4%

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

7. Simplified 71.4%

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

   1. unpow-prod-down 71.4%

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

   2. *-commutative 71.4%

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

   3. associate-*l* 71.4%

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

   4. metadata-eval 71.4%

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

9. Applied egg-rr 71.4%

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

10. Final simplification 71.4%

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

Alternative 8: 75.2% accurate, 2.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 76.6%

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

   1. unpow2 76.6%

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

   2. associate-*l/ 76.6%

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

   3. associate-*r/ 76.6%

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

   4. unpow2 76.6%

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

   5. associate-*l/ 76.6%

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

   6. associate-*r/ 76.6%

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

3. Simplified 76.6%

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

4. Taylor expanded in angle around 0 76.4%

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

5. Taylor expanded in angle around 0 71.4%

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

   1. *-commutative 71.4%

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

7. Simplified 71.4%

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

   1. unpow-prod-down 71.4%

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

   2. *-commutative 71.4%

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

   3. associate-*l* 71.4%

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

   4. metadata-eval 71.4%

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

9. Applied egg-rr 71.4%

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

10. Taylor expanded in angle around 0 71.4%

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

    1. associate-*r* 71.4%

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

    2. *-commutative 71.4%

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

    3. *-commutative 71.4%

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

12. Simplified 71.4%

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

13. Final simplification 71.4%

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

Reproduce

herbie shell --seed 2023305 
(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)))