ab-angle->ABCF A

Percentage Accurate: 80.2% → 80.2%

Time: 39.7s

Alternatives: 4

Speedup: 0.8×

• 37.1% 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: 80.2% 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

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

FPCore

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

C

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

Java

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

Julia

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

Wolfram

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

Derivation

1. Initial program 81.2%

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

   1. associate-*l/ 81.3%

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

   2. associate-*r/ 81.2%

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

   3. associate-*l/ 81.2%

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

   4. associate-*r/ 81.2%

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

3. Simplified 81.2%

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

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 inf 81.8%

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

   1. expm1-log1p-u 64.6%

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

7. Applied egg-rr 64.6%

   \[\leadsto {\left(a \cdot \sin \color{blue}{\left(\mathsf{expm1}\left(\mathsf{log1p}\left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\right)}\right)}^{2} + {\left(b \cdot 1\right)}^{2} \]
8. Step-by-step derivation

   1. expm1-log1p-u 64.3%

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

9. Applied egg-rr 64.3%

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

10. Final simplification 64.3%

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

Alternative 2: 80.2% accurate, 2.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 81.2%

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

   1. associate-*l/ 81.3%

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

   2. associate-*r/ 81.2%

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

   3. associate-*l/ 81.2%

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

   4. associate-*r/ 81.2%

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

3. Simplified 81.2%

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

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 inf 81.8%

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

   1. *-commutative 81.8%

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

   2. metadata-eval 81.8%

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

   3. div-inv 81.8%

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

   4. clear-num 81.8%

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

7. Applied egg-rr 81.8%

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

8. Taylor expanded in b around 0 81.8%

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

   1. unpow2 77.7%

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

10. Simplified 81.8%

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

11. Final simplification 81.8%

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

Alternative 3: 75.1% accurate, 2.0× speedup

Math

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

FPCore

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

C

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

Python

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

Julia

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

MATLAB

angle = abs(angle)
function tmp = code(a, b, angle)
	tmp = (b ^ 2.0) + (((angle * (a * pi)) ^ 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[(angle * N[(a * Pi), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * 3.08641975308642e-5), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 81.2%

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

   1. associate-*l/ 81.3%

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

   2. associate-*r/ 81.2%

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

   3. associate-*l/ 81.2%

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

   4. associate-*r/ 81.2%

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

3. Simplified 81.2%

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

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 77.6%

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

   1. *-commutative 77.6%

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

7. Simplified 77.6%

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

   1. *-commutative 77.6%

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

   2. unpow-prod-down 77.7%

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

   3. metadata-eval 77.7%

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

9. Applied egg-rr 77.7%

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

10. Final simplification 77.7%

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

Alternative 4: 75.2% accurate, 2.9× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 81.2%

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

   1. associate-*l/ 81.3%

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

   2. associate-*r/ 81.2%

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

   3. associate-*l/ 81.2%

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

   4. associate-*r/ 81.2%

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

3. Simplified 81.2%

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

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 inf 81.8%

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

6. Taylor expanded in angle around 0 66.7%

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

   1. associate-*r* 66.7%

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

   2. *-commutative 66.7%

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

   3. unpow2 66.7%

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

   4. unpow2 66.7%

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

   5. swap-sqr 66.7%

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

   6. unpow2 66.7%

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

   7. associate-*l* 66.7%

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

   8. unpow2 66.7%

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

   9. unpow2 66.7%

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

   10. swap-sqr 77.7%

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

   11. unpow2 77.7%

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

   12. associate-*r* 77.7%

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

   13. *-commutative 77.7%

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

   14. associate-*l* 77.7%

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

8. Simplified 77.7%

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

9. Taylor expanded in b around 0 77.7%

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

    1. unpow2 77.7%

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

11. Simplified 77.7%

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

12. Final simplification 77.7%

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

Reproduce

herbie shell --seed 2023217 
(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)))