ab-angle->ABCF C

Percentage Accurate: 79.4% → 79.4%
Time: 15.8s
Alternatives: 7
Speedup: 2.0×

Specification

?
\[\begin{array}{l} \\ \begin{array}{l} t_0 := \pi \cdot \frac{angle}{180}\\ {\left(a \cdot \cos t\_0\right)}^{2} + {\left(b \cdot \sin t\_0\right)}^{2} \end{array} \end{array} \]
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* PI (/ angle 180.0))))
   (+ (pow (* a (cos t_0)) 2.0) (pow (* b (sin t_0)) 2.0))))
double code(double a, double b, double angle) {
	double t_0 = ((double) M_PI) * (angle / 180.0);
	return pow((a * cos(t_0)), 2.0) + pow((b * sin(t_0)), 2.0);
}
public static double code(double a, double b, double angle) {
	double t_0 = Math.PI * (angle / 180.0);
	return Math.pow((a * Math.cos(t_0)), 2.0) + Math.pow((b * Math.sin(t_0)), 2.0);
}
def code(a, b, angle):
	t_0 = math.pi * (angle / 180.0)
	return math.pow((a * math.cos(t_0)), 2.0) + math.pow((b * math.sin(t_0)), 2.0)
function code(a, b, angle)
	t_0 = Float64(pi * Float64(angle / 180.0))
	return Float64((Float64(a * cos(t_0)) ^ 2.0) + (Float64(b * sin(t_0)) ^ 2.0))
end
function tmp = code(a, b, angle)
	t_0 = pi * (angle / 180.0);
	tmp = ((a * cos(t_0)) ^ 2.0) + ((b * sin(t_0)) ^ 2.0);
end
code[a_, b_, angle_] := Block[{t$95$0 = N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]}, N[(N[Power[N[(a * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := \pi \cdot \frac{angle}{180}\\
{\left(a \cdot \cos t\_0\right)}^{2} + {\left(b \cdot \sin t\_0\right)}^{2}
\end{array}
\end{array}

Sampling outcomes in binary64 precision:

Local Percentage Accuracy vs ?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Accuracy vs Speed?

Herbie found 7 alternatives:

AlternativeAccuracySpeedup
The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

Initial Program: 79.4% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \pi \cdot \frac{angle}{180}\\ {\left(a \cdot \cos t\_0\right)}^{2} + {\left(b \cdot \sin t\_0\right)}^{2} \end{array} \end{array} \]
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* PI (/ angle 180.0))))
   (+ (pow (* a (cos t_0)) 2.0) (pow (* b (sin t_0)) 2.0))))
double code(double a, double b, double angle) {
	double t_0 = ((double) M_PI) * (angle / 180.0);
	return pow((a * cos(t_0)), 2.0) + pow((b * sin(t_0)), 2.0);
}
public static double code(double a, double b, double angle) {
	double t_0 = Math.PI * (angle / 180.0);
	return Math.pow((a * Math.cos(t_0)), 2.0) + Math.pow((b * Math.sin(t_0)), 2.0);
}
def code(a, b, angle):
	t_0 = math.pi * (angle / 180.0)
	return math.pow((a * math.cos(t_0)), 2.0) + math.pow((b * math.sin(t_0)), 2.0)
function code(a, b, angle)
	t_0 = Float64(pi * Float64(angle / 180.0))
	return Float64((Float64(a * cos(t_0)) ^ 2.0) + (Float64(b * sin(t_0)) ^ 2.0))
end
function tmp = code(a, b, angle)
	t_0 = pi * (angle / 180.0);
	tmp = ((a * cos(t_0)) ^ 2.0) + ((b * sin(t_0)) ^ 2.0);
end
code[a_, b_, angle_] := Block[{t$95$0 = N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]}, N[(N[Power[N[(a * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := \pi \cdot \frac{angle}{180}\\
{\left(a \cdot \cos t\_0\right)}^{2} + {\left(b \cdot \sin t\_0\right)}^{2}
\end{array}
\end{array}

Alternative 1: 79.4% accurate, 0.8× speedup?

\[\begin{array}{l} \\ {a}^{2} + {\left(b \cdot \sin \left(\sqrt[3]{{\pi}^{3}} \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \end{array} \]
(FPCore (a b angle)
 :precision binary64
 (+
  (pow a 2.0)
  (pow
   (* b (sin (* (cbrt (pow PI 3.0)) (* angle 0.005555555555555556))))
   2.0)))
double code(double a, double b, double angle) {
	return pow(a, 2.0) + pow((b * sin((cbrt(pow(((double) M_PI), 3.0)) * (angle * 0.005555555555555556)))), 2.0);
}
public static double code(double a, double b, double angle) {
	return Math.pow(a, 2.0) + Math.pow((b * Math.sin((Math.cbrt(Math.pow(Math.PI, 3.0)) * (angle * 0.005555555555555556)))), 2.0);
}
function code(a, b, angle)
	return Float64((a ^ 2.0) + (Float64(b * sin(Float64(cbrt((pi ^ 3.0)) * Float64(angle * 0.005555555555555556)))) ^ 2.0))
end
code[a_, b_, angle_] := N[(N[Power[a, 2.0], $MachinePrecision] + N[Power[N[(b * N[Sin[N[(N[Power[N[Power[Pi, 3.0], $MachinePrecision], 1/3], $MachinePrecision] * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]
\begin{array}{l}

\\
{a}^{2} + {\left(b \cdot \sin \left(\sqrt[3]{{\pi}^{3}} \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}
\end{array}
Derivation
  1. Initial program 77.3%

    \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
  2. Step-by-step derivation
    1. Simplified77.2%

      \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
    2. Add Preprocessing
    3. Step-by-step derivation
      1. add-cbrt-cube77.3%

        \[\leadsto {\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\color{blue}{\sqrt[3]{\left(\pi \cdot \pi\right) \cdot \pi}} \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
      2. pow377.3%

        \[\leadsto {\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\sqrt[3]{\color{blue}{{\pi}^{3}}} \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
    4. Applied egg-rr77.3%

      \[\leadsto {\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\color{blue}{\sqrt[3]{{\pi}^{3}}} \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
    5. Taylor expanded in angle around 0 77.3%

      \[\leadsto {\left(a \cdot \color{blue}{1}\right)}^{2} + {\left(b \cdot \sin \left(\sqrt[3]{{\pi}^{3}} \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
    6. Final simplification77.3%

      \[\leadsto {a}^{2} + {\left(b \cdot \sin \left(\sqrt[3]{{\pi}^{3}} \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
    7. Add Preprocessing

    Alternative 2: 79.4% accurate, 1.0× speedup?

    \[\begin{array}{l} \\ \begin{array}{l} t_0 := \pi \cdot \frac{angle}{180}\\ {\left(a \cdot \cos t\_0\right)}^{2} + {\left(b \cdot \sin t\_0\right)}^{2} \end{array} \end{array} \]
    (FPCore (a b angle)
     :precision binary64
     (let* ((t_0 (* PI (/ angle 180.0))))
       (+ (pow (* a (cos t_0)) 2.0) (pow (* b (sin t_0)) 2.0))))
    double code(double a, double b, double angle) {
    	double t_0 = ((double) M_PI) * (angle / 180.0);
    	return pow((a * cos(t_0)), 2.0) + pow((b * sin(t_0)), 2.0);
    }
    
    public static double code(double a, double b, double angle) {
    	double t_0 = Math.PI * (angle / 180.0);
    	return Math.pow((a * Math.cos(t_0)), 2.0) + Math.pow((b * Math.sin(t_0)), 2.0);
    }
    
    def code(a, b, angle):
    	t_0 = math.pi * (angle / 180.0)
    	return math.pow((a * math.cos(t_0)), 2.0) + math.pow((b * math.sin(t_0)), 2.0)
    
    function code(a, b, angle)
    	t_0 = Float64(pi * Float64(angle / 180.0))
    	return Float64((Float64(a * cos(t_0)) ^ 2.0) + (Float64(b * sin(t_0)) ^ 2.0))
    end
    
    function tmp = code(a, b, angle)
    	t_0 = pi * (angle / 180.0);
    	tmp = ((a * cos(t_0)) ^ 2.0) + ((b * sin(t_0)) ^ 2.0);
    end
    
    code[a_, b_, angle_] := Block[{t$95$0 = N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]}, N[(N[Power[N[(a * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]
    
    \begin{array}{l}
    
    \\
    \begin{array}{l}
    t_0 := \pi \cdot \frac{angle}{180}\\
    {\left(a \cdot \cos t\_0\right)}^{2} + {\left(b \cdot \sin t\_0\right)}^{2}
    \end{array}
    \end{array}
    
    Derivation
    1. Initial program 77.3%

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

    Alternative 3: 79.4% accurate, 1.3× speedup?

    \[\begin{array}{l} \\ {a}^{2} + {\left(b \cdot \sin \left(0.005555555555555556 \cdot \frac{\pi}{\frac{1}{angle}}\right)\right)}^{2} \end{array} \]
    (FPCore (a b angle)
     :precision binary64
     (+
      (pow a 2.0)
      (pow (* b (sin (* 0.005555555555555556 (/ PI (/ 1.0 angle))))) 2.0)))
    double code(double a, double b, double angle) {
    	return pow(a, 2.0) + pow((b * sin((0.005555555555555556 * (((double) M_PI) / (1.0 / angle))))), 2.0);
    }
    
    public static double code(double a, double b, double angle) {
    	return Math.pow(a, 2.0) + Math.pow((b * Math.sin((0.005555555555555556 * (Math.PI / (1.0 / angle))))), 2.0);
    }
    
    def code(a, b, angle):
    	return math.pow(a, 2.0) + math.pow((b * math.sin((0.005555555555555556 * (math.pi / (1.0 / angle))))), 2.0)
    
    function code(a, b, angle)
    	return Float64((a ^ 2.0) + (Float64(b * sin(Float64(0.005555555555555556 * Float64(pi / Float64(1.0 / angle))))) ^ 2.0))
    end
    
    function tmp = code(a, b, angle)
    	tmp = (a ^ 2.0) + ((b * sin((0.005555555555555556 * (pi / (1.0 / angle))))) ^ 2.0);
    end
    
    code[a_, b_, angle_] := N[(N[Power[a, 2.0], $MachinePrecision] + N[Power[N[(b * N[Sin[N[(0.005555555555555556 * N[(Pi / N[(1.0 / angle), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]
    
    \begin{array}{l}
    
    \\
    {a}^{2} + {\left(b \cdot \sin \left(0.005555555555555556 \cdot \frac{\pi}{\frac{1}{angle}}\right)\right)}^{2}
    \end{array}
    
    Derivation
    1. Initial program 77.3%

      \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
    2. Step-by-step derivation
      1. Simplified77.2%

        \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
      2. Add Preprocessing
      3. Step-by-step derivation
        1. metadata-eval77.2%

          \[\leadsto {\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot \color{blue}{\frac{1}{180}}\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
        2. div-inv77.3%

          \[\leadsto {\left(a \cdot \cos \left(\pi \cdot \color{blue}{\frac{angle}{180}}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
        3. clear-num77.3%

          \[\leadsto {\left(a \cdot \cos \left(\pi \cdot \color{blue}{\frac{1}{\frac{180}{angle}}}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
        4. un-div-inv77.2%

          \[\leadsto {\left(a \cdot \cos \color{blue}{\left(\frac{\pi}{\frac{180}{angle}}\right)}\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
      4. Applied egg-rr77.2%

        \[\leadsto {\left(a \cdot \cos \color{blue}{\left(\frac{\pi}{\frac{180}{angle}}\right)}\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
      5. Step-by-step derivation
        1. add-exp-log41.4%

          \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} + {\left(b \cdot \sin \color{blue}{\left(e^{\log \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}\right)}\right)}^{2} \]
      6. Applied egg-rr41.4%

        \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} + {\left(b \cdot \sin \color{blue}{\left(e^{\log \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}\right)}\right)}^{2} \]
      7. Step-by-step derivation
        1. rem-exp-log77.2%

          \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} + {\left(b \cdot \sin \color{blue}{\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}\right)}^{2} \]
        2. metadata-eval77.2%

          \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot \color{blue}{\frac{1}{180}}\right)\right)\right)}^{2} \]
        3. div-inv77.3%

          \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \color{blue}{\frac{angle}{180}}\right)\right)}^{2} \]
        4. clear-num76.9%

          \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \color{blue}{\frac{1}{\frac{180}{angle}}}\right)\right)}^{2} \]
        5. *-un-lft-identity76.9%

          \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} + {\left(b \cdot \sin \color{blue}{\left(1 \cdot \left(\pi \cdot \frac{1}{\frac{180}{angle}}\right)\right)}\right)}^{2} \]
        6. div-inv77.0%

          \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} + {\left(b \cdot \sin \left(1 \cdot \color{blue}{\frac{\pi}{\frac{180}{angle}}}\right)\right)}^{2} \]
        7. metadata-eval77.0%

          \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} + {\left(b \cdot \sin \left(\color{blue}{\frac{1}{1}} \cdot \frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} \]
        8. times-frac77.0%

          \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} + {\left(b \cdot \sin \color{blue}{\left(\frac{1 \cdot \pi}{1 \cdot \frac{180}{angle}}\right)}\right)}^{2} \]
        9. *-un-lft-identity77.0%

          \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} + {\left(b \cdot \sin \left(\frac{1 \cdot \pi}{\color{blue}{\frac{180}{angle}}}\right)\right)}^{2} \]
        10. div-inv76.9%

          \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} + {\left(b \cdot \sin \left(\frac{1 \cdot \pi}{\color{blue}{180 \cdot \frac{1}{angle}}}\right)\right)}^{2} \]
        11. times-frac77.3%

          \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} + {\left(b \cdot \sin \color{blue}{\left(\frac{1}{180} \cdot \frac{\pi}{\frac{1}{angle}}\right)}\right)}^{2} \]
        12. metadata-eval77.3%

          \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} + {\left(b \cdot \sin \left(\color{blue}{0.005555555555555556} \cdot \frac{\pi}{\frac{1}{angle}}\right)\right)}^{2} \]
      8. Applied egg-rr77.3%

        \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2} + {\left(b \cdot \sin \color{blue}{\left(0.005555555555555556 \cdot \frac{\pi}{\frac{1}{angle}}\right)}\right)}^{2} \]
      9. Taylor expanded in angle around 0 77.3%

        \[\leadsto {\left(a \cdot \color{blue}{1}\right)}^{2} + {\left(b \cdot \sin \left(0.005555555555555556 \cdot \frac{\pi}{\frac{1}{angle}}\right)\right)}^{2} \]
      10. Final simplification77.3%

        \[\leadsto {a}^{2} + {\left(b \cdot \sin \left(0.005555555555555556 \cdot \frac{\pi}{\frac{1}{angle}}\right)\right)}^{2} \]
      11. Add Preprocessing

      Alternative 4: 53.6% accurate, 2.0× speedup?

      \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;a \leq 4.1 \cdot 10^{-129}:\\ \;\;\;\;{\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \end{array} \]
      (FPCore (a b angle)
       :precision binary64
       (if (<= a 4.1e-129)
         (pow (* b (sin (* PI (* angle 0.005555555555555556)))) 2.0)
         (* a a)))
      double code(double a, double b, double angle) {
      	double tmp;
      	if (a <= 4.1e-129) {
      		tmp = pow((b * sin((((double) M_PI) * (angle * 0.005555555555555556)))), 2.0);
      	} else {
      		tmp = a * a;
      	}
      	return tmp;
      }
      
      public static double code(double a, double b, double angle) {
      	double tmp;
      	if (a <= 4.1e-129) {
      		tmp = Math.pow((b * Math.sin((Math.PI * (angle * 0.005555555555555556)))), 2.0);
      	} else {
      		tmp = a * a;
      	}
      	return tmp;
      }
      
      def code(a, b, angle):
      	tmp = 0
      	if a <= 4.1e-129:
      		tmp = math.pow((b * math.sin((math.pi * (angle * 0.005555555555555556)))), 2.0)
      	else:
      		tmp = a * a
      	return tmp
      
      function code(a, b, angle)
      	tmp = 0.0
      	if (a <= 4.1e-129)
      		tmp = Float64(b * sin(Float64(pi * Float64(angle * 0.005555555555555556)))) ^ 2.0;
      	else
      		tmp = Float64(a * a);
      	end
      	return tmp
      end
      
      function tmp_2 = code(a, b, angle)
      	tmp = 0.0;
      	if (a <= 4.1e-129)
      		tmp = (b * sin((pi * (angle * 0.005555555555555556)))) ^ 2.0;
      	else
      		tmp = a * a;
      	end
      	tmp_2 = tmp;
      end
      
      code[a_, b_, angle_] := If[LessEqual[a, 4.1e-129], N[Power[N[(b * N[Sin[N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision], N[(a * a), $MachinePrecision]]
      
      \begin{array}{l}
      
      \\
      \begin{array}{l}
      \mathbf{if}\;a \leq 4.1 \cdot 10^{-129}:\\
      \;\;\;\;{\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\\
      
      \mathbf{else}:\\
      \;\;\;\;a \cdot a\\
      
      
      \end{array}
      \end{array}
      
      Derivation
      1. Split input into 2 regimes
      2. if a < 4.1e-129

        1. Initial program 77.5%

          \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
        2. Step-by-step derivation
          1. Simplified77.5%

            \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
          2. Add Preprocessing
          3. Taylor expanded in a around 0 35.8%

            \[\leadsto \color{blue}{{b}^{2} \cdot {\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2}} \]
          4. Step-by-step derivation
            1. *-commutative35.8%

              \[\leadsto \color{blue}{{\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2} \cdot {b}^{2}} \]
            2. associate-*r*35.8%

              \[\leadsto {\sin \color{blue}{\left(\left(0.005555555555555556 \cdot angle\right) \cdot \pi\right)}}^{2} \cdot {b}^{2} \]
            3. *-commutative35.8%

              \[\leadsto {\sin \left(\color{blue}{\left(angle \cdot 0.005555555555555556\right)} \cdot \pi\right)}^{2} \cdot {b}^{2} \]
            4. *-commutative35.8%

              \[\leadsto {\sin \color{blue}{\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}}^{2} \cdot {b}^{2} \]
            5. unpow235.8%

              \[\leadsto \color{blue}{\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)} \cdot {b}^{2} \]
            6. unpow235.8%

              \[\leadsto \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \color{blue}{\left(b \cdot b\right)} \]
            7. swap-sqr44.6%

              \[\leadsto \color{blue}{\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right) \cdot \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)} \]
            8. unpow244.6%

              \[\leadsto \color{blue}{{\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)}^{2}} \]
            9. *-commutative44.6%

              \[\leadsto {\color{blue}{\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}}^{2} \]
            10. *-commutative44.6%

              \[\leadsto {\left(b \cdot \sin \left(\pi \cdot \color{blue}{\left(0.005555555555555556 \cdot angle\right)}\right)\right)}^{2} \]
          5. Simplified44.6%

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

          if 4.1e-129 < a

          1. Initial program 76.9%

            \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
          2. Step-by-step derivation
            1. Simplified76.9%

              \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
            2. Add Preprocessing
            3. Taylor expanded in angle around 0 67.9%

              \[\leadsto \color{blue}{{a}^{2}} \]
            4. Step-by-step derivation
              1. unpow267.9%

                \[\leadsto \color{blue}{a \cdot a} \]
            5. Applied egg-rr67.9%

              \[\leadsto \color{blue}{a \cdot a} \]
          3. Recombined 2 regimes into one program.
          4. Final simplification53.3%

            \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 4.1 \cdot 10^{-129}:\\ \;\;\;\;{\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \]
          5. Add Preprocessing

          Alternative 5: 53.6% accurate, 2.0× speedup?

          \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;a \leq 7.5 \cdot 10^{-128}:\\ \;\;\;\;{\left(b \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)}^{2}\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \end{array} \]
          (FPCore (a b angle)
           :precision binary64
           (if (<= a 7.5e-128)
             (pow (* b (sin (* 0.005555555555555556 (* PI angle)))) 2.0)
             (* a a)))
          double code(double a, double b, double angle) {
          	double tmp;
          	if (a <= 7.5e-128) {
          		tmp = pow((b * sin((0.005555555555555556 * (((double) M_PI) * angle)))), 2.0);
          	} else {
          		tmp = a * a;
          	}
          	return tmp;
          }
          
          public static double code(double a, double b, double angle) {
          	double tmp;
          	if (a <= 7.5e-128) {
          		tmp = Math.pow((b * Math.sin((0.005555555555555556 * (Math.PI * angle)))), 2.0);
          	} else {
          		tmp = a * a;
          	}
          	return tmp;
          }
          
          def code(a, b, angle):
          	tmp = 0
          	if a <= 7.5e-128:
          		tmp = math.pow((b * math.sin((0.005555555555555556 * (math.pi * angle)))), 2.0)
          	else:
          		tmp = a * a
          	return tmp
          
          function code(a, b, angle)
          	tmp = 0.0
          	if (a <= 7.5e-128)
          		tmp = Float64(b * sin(Float64(0.005555555555555556 * Float64(pi * angle)))) ^ 2.0;
          	else
          		tmp = Float64(a * a);
          	end
          	return tmp
          end
          
          function tmp_2 = code(a, b, angle)
          	tmp = 0.0;
          	if (a <= 7.5e-128)
          		tmp = (b * sin((0.005555555555555556 * (pi * angle)))) ^ 2.0;
          	else
          		tmp = a * a;
          	end
          	tmp_2 = tmp;
          end
          
          code[a_, b_, angle_] := If[LessEqual[a, 7.5e-128], N[Power[N[(b * N[Sin[N[(0.005555555555555556 * N[(Pi * angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision], N[(a * a), $MachinePrecision]]
          
          \begin{array}{l}
          
          \\
          \begin{array}{l}
          \mathbf{if}\;a \leq 7.5 \cdot 10^{-128}:\\
          \;\;\;\;{\left(b \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)}^{2}\\
          
          \mathbf{else}:\\
          \;\;\;\;a \cdot a\\
          
          
          \end{array}
          \end{array}
          
          Derivation
          1. Split input into 2 regimes
          2. if a < 7.50000000000000021e-128

            1. Initial program 77.5%

              \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
            2. Step-by-step derivation
              1. Simplified77.5%

                \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
              2. Add Preprocessing
              3. Taylor expanded in a around 0 35.8%

                \[\leadsto \color{blue}{{b}^{2} \cdot {\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2}} \]
              4. Step-by-step derivation
                1. *-commutative35.8%

                  \[\leadsto \color{blue}{{\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2} \cdot {b}^{2}} \]
                2. associate-*r*35.8%

                  \[\leadsto {\sin \color{blue}{\left(\left(0.005555555555555556 \cdot angle\right) \cdot \pi\right)}}^{2} \cdot {b}^{2} \]
                3. *-commutative35.8%

                  \[\leadsto {\sin \left(\color{blue}{\left(angle \cdot 0.005555555555555556\right)} \cdot \pi\right)}^{2} \cdot {b}^{2} \]
                4. *-commutative35.8%

                  \[\leadsto {\sin \color{blue}{\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}}^{2} \cdot {b}^{2} \]
                5. unpow235.8%

                  \[\leadsto \color{blue}{\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)} \cdot {b}^{2} \]
                6. unpow235.8%

                  \[\leadsto \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \color{blue}{\left(b \cdot b\right)} \]
                7. swap-sqr44.6%

                  \[\leadsto \color{blue}{\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right) \cdot \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)} \]
                8. unpow244.6%

                  \[\leadsto \color{blue}{{\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)}^{2}} \]
                9. *-commutative44.6%

                  \[\leadsto {\color{blue}{\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}}^{2} \]
                10. *-commutative44.6%

                  \[\leadsto {\left(b \cdot \sin \color{blue}{\left(\left(angle \cdot 0.005555555555555556\right) \cdot \pi\right)}\right)}^{2} \]
                11. *-commutative44.6%

                  \[\leadsto {\left(b \cdot \sin \left(\color{blue}{\left(0.005555555555555556 \cdot angle\right)} \cdot \pi\right)\right)}^{2} \]
                12. associate-*r*44.6%

                  \[\leadsto {\left(b \cdot \sin \color{blue}{\left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}\right)}^{2} \]
              5. Simplified44.6%

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

              if 7.50000000000000021e-128 < a

              1. Initial program 76.9%

                \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
              2. Step-by-step derivation
                1. Simplified76.9%

                  \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
                2. Add Preprocessing
                3. Taylor expanded in angle around 0 67.9%

                  \[\leadsto \color{blue}{{a}^{2}} \]
                4. Step-by-step derivation
                  1. unpow267.9%

                    \[\leadsto \color{blue}{a \cdot a} \]
                5. Applied egg-rr67.9%

                  \[\leadsto \color{blue}{a \cdot a} \]
              3. Recombined 2 regimes into one program.
              4. Final simplification53.3%

                \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 7.5 \cdot 10^{-128}:\\ \;\;\;\;{\left(b \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)}^{2}\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \]
              5. Add Preprocessing

              Alternative 6: 79.5% accurate, 2.0× speedup?

              \[\begin{array}{l} \\ a \cdot a + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \end{array} \]
              (FPCore (a b angle)
               :precision binary64
               (+ (* a a) (pow (* b (sin (* PI (* angle 0.005555555555555556)))) 2.0)))
              double code(double a, double b, double angle) {
              	return (a * a) + pow((b * sin((((double) M_PI) * (angle * 0.005555555555555556)))), 2.0);
              }
              
              public static double code(double a, double b, double angle) {
              	return (a * a) + Math.pow((b * Math.sin((Math.PI * (angle * 0.005555555555555556)))), 2.0);
              }
              
              def code(a, b, angle):
              	return (a * a) + math.pow((b * math.sin((math.pi * (angle * 0.005555555555555556)))), 2.0)
              
              function code(a, b, angle)
              	return Float64(Float64(a * a) + (Float64(b * sin(Float64(pi * Float64(angle * 0.005555555555555556)))) ^ 2.0))
              end
              
              function tmp = code(a, b, angle)
              	tmp = (a * a) + ((b * sin((pi * (angle * 0.005555555555555556)))) ^ 2.0);
              end
              
              code[a_, b_, angle_] := N[(N[(a * a), $MachinePrecision] + N[Power[N[(b * N[Sin[N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]
              
              \begin{array}{l}
              
              \\
              a \cdot a + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}
              \end{array}
              
              Derivation
              1. Initial program 77.3%

                \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
              2. Step-by-step derivation
                1. Simplified77.2%

                  \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
                2. Add Preprocessing
                3. Step-by-step derivation
                  1. metadata-eval77.2%

                    \[\leadsto {\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot \color{blue}{\frac{1}{180}}\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                  2. div-inv77.3%

                    \[\leadsto {\left(a \cdot \cos \left(\pi \cdot \color{blue}{\frac{angle}{180}}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                  3. clear-num77.3%

                    \[\leadsto {\left(a \cdot \cos \left(\pi \cdot \color{blue}{\frac{1}{\frac{180}{angle}}}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                  4. un-div-inv77.2%

                    \[\leadsto {\left(a \cdot \cos \color{blue}{\left(\frac{\pi}{\frac{180}{angle}}\right)}\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                4. Applied egg-rr77.2%

                  \[\leadsto {\left(a \cdot \cos \color{blue}{\left(\frac{\pi}{\frac{180}{angle}}\right)}\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                5. Step-by-step derivation
                  1. *-commutative77.2%

                    \[\leadsto {\color{blue}{\left(\cos \left(\frac{\pi}{\frac{180}{angle}}\right) \cdot a\right)}}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                  2. unpow-prod-down77.2%

                    \[\leadsto \color{blue}{{\cos \left(\frac{\pi}{\frac{180}{angle}}\right)}^{2} \cdot {a}^{2}} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                  3. pow277.2%

                    \[\leadsto \color{blue}{\left(\cos \left(\frac{\pi}{\frac{180}{angle}}\right) \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\right)} \cdot {a}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                  4. sqr-cos-a77.2%

                    \[\leadsto \color{blue}{\left(0.5 + 0.5 \cdot \cos \left(2 \cdot \frac{\pi}{\frac{180}{angle}}\right)\right)} \cdot {a}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                  5. div-inv77.3%

                    \[\leadsto \left(0.5 + 0.5 \cdot \cos \left(2 \cdot \color{blue}{\left(\pi \cdot \frac{1}{\frac{180}{angle}}\right)}\right)\right) \cdot {a}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                  6. clear-num77.3%

                    \[\leadsto \left(0.5 + 0.5 \cdot \cos \left(2 \cdot \left(\pi \cdot \color{blue}{\frac{angle}{180}}\right)\right)\right) \cdot {a}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                  7. div-inv77.2%

                    \[\leadsto \left(0.5 + 0.5 \cdot \cos \left(2 \cdot \left(\pi \cdot \color{blue}{\left(angle \cdot \frac{1}{180}\right)}\right)\right)\right) \cdot {a}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                  8. metadata-eval77.2%

                    \[\leadsto \left(0.5 + 0.5 \cdot \cos \left(2 \cdot \left(\pi \cdot \left(angle \cdot \color{blue}{0.005555555555555556}\right)\right)\right)\right) \cdot {a}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                  9. sqr-cos-a77.2%

                    \[\leadsto \color{blue}{\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)} \cdot {a}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                  10. pow277.2%

                    \[\leadsto \color{blue}{{\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{2}} \cdot {a}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                  11. unpow277.2%

                    \[\leadsto {\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{2} \cdot \color{blue}{\left(a \cdot a\right)} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                  12. associate-*r*77.2%

                    \[\leadsto \color{blue}{\left({\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{2} \cdot a\right) \cdot a} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                6. Applied egg-rr77.2%

                  \[\leadsto \color{blue}{\left({\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{2} \cdot a\right) \cdot a} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                7. Taylor expanded in angle around 0 77.3%

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

                Alternative 7: 56.0% accurate, 139.0× speedup?

                \[\begin{array}{l} \\ a \cdot a \end{array} \]
                (FPCore (a b angle) :precision binary64 (* a a))
                double code(double a, double b, double angle) {
                	return a * a;
                }
                
                real(8) function code(a, b, angle)
                    real(8), intent (in) :: a
                    real(8), intent (in) :: b
                    real(8), intent (in) :: angle
                    code = a * a
                end function
                
                public static double code(double a, double b, double angle) {
                	return a * a;
                }
                
                def code(a, b, angle):
                	return a * a
                
                function code(a, b, angle)
                	return Float64(a * a)
                end
                
                function tmp = code(a, b, angle)
                	tmp = a * a;
                end
                
                code[a_, b_, angle_] := N[(a * a), $MachinePrecision]
                
                \begin{array}{l}
                
                \\
                a \cdot a
                \end{array}
                
                Derivation
                1. Initial program 77.3%

                  \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
                2. Step-by-step derivation
                  1. Simplified77.2%

                    \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
                  2. Add Preprocessing
                  3. Taylor expanded in angle around 0 56.5%

                    \[\leadsto \color{blue}{{a}^{2}} \]
                  4. Step-by-step derivation
                    1. unpow256.5%

                      \[\leadsto \color{blue}{a \cdot a} \]
                  5. Applied egg-rr56.5%

                    \[\leadsto \color{blue}{a \cdot a} \]
                  6. Add Preprocessing

                  Reproduce

                  ?
                  herbie shell --seed 2024137 
                  (FPCore (a b angle)
                    :name "ab-angle->ABCF C"
                    :precision binary64
                    (+ (pow (* a (cos (* PI (/ angle 180.0)))) 2.0) (pow (* b (sin (* PI (/ angle 180.0)))) 2.0)))