ab-angle->ABCF C

Percentage Accurate: 79.7% → 79.7%
Time: 22.9s
Alternatives: 10
Speedup: 0.8×

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 10 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: 79.7% 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: 79.7% accurate, 0.7× speedup?

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

\\
{\left(a \cdot \cos \left(\sqrt[3]{{\pi}^{3}} \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}
\end{array}
Derivation
  1. Initial program 81.1%

    \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
  2. Step-by-step derivation
    1. Simplified81.2%

      \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
    2. Add Preprocessing
    3. Step-by-step derivation
      1. add-cbrt-cube81.2%

        \[\leadsto {\left(a \cdot \cos \left(\color{blue}{\sqrt[3]{\left(\pi \cdot \pi\right) \cdot \pi}} \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
      2. pow381.2%

        \[\leadsto {\left(a \cdot \cos \left(\sqrt[3]{\color{blue}{{\pi}^{3}}} \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
    4. Applied egg-rr81.2%

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

    Alternative 2: 79.7% accurate, 0.8× speedup?

    \[\begin{array}{l} \\ \begin{array}{l} t_0 := \pi \cdot \left(angle \cdot 0.005555555555555556\right)\\ {\left(b \cdot \sin t\_0\right)}^{2} + {\cos t\_0}^{2} \cdot {a}^{2} \end{array} \end{array} \]
    (FPCore (a b angle)
     :precision binary64
     (let* ((t_0 (* PI (* angle 0.005555555555555556))))
       (+ (pow (* b (sin t_0)) 2.0) (* (pow (cos t_0) 2.0) (pow a 2.0)))))
    double code(double a, double b, double angle) {
    	double t_0 = ((double) M_PI) * (angle * 0.005555555555555556);
    	return pow((b * sin(t_0)), 2.0) + (pow(cos(t_0), 2.0) * pow(a, 2.0));
    }
    
    public static double code(double a, double b, double angle) {
    	double t_0 = Math.PI * (angle * 0.005555555555555556);
    	return Math.pow((b * Math.sin(t_0)), 2.0) + (Math.pow(Math.cos(t_0), 2.0) * Math.pow(a, 2.0));
    }
    
    def code(a, b, angle):
    	t_0 = math.pi * (angle * 0.005555555555555556)
    	return math.pow((b * math.sin(t_0)), 2.0) + (math.pow(math.cos(t_0), 2.0) * math.pow(a, 2.0))
    
    function code(a, b, angle)
    	t_0 = Float64(pi * Float64(angle * 0.005555555555555556))
    	return Float64((Float64(b * sin(t_0)) ^ 2.0) + Float64((cos(t_0) ^ 2.0) * (a ^ 2.0)))
    end
    
    function tmp = code(a, b, angle)
    	t_0 = pi * (angle * 0.005555555555555556);
    	tmp = ((b * sin(t_0)) ^ 2.0) + ((cos(t_0) ^ 2.0) * (a ^ 2.0));
    end
    
    code[a_, b_, angle_] := Block[{t$95$0 = N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]}, N[(N[Power[N[(b * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[(N[Power[N[Cos[t$95$0], $MachinePrecision], 2.0], $MachinePrecision] * N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]
    
    \begin{array}{l}
    
    \\
    \begin{array}{l}
    t_0 := \pi \cdot \left(angle \cdot 0.005555555555555556\right)\\
    {\left(b \cdot \sin t\_0\right)}^{2} + {\cos t\_0}^{2} \cdot {a}^{2}
    \end{array}
    \end{array}
    
    Derivation
    1. Initial program 81.1%

      \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
    2. Step-by-step derivation
      1. Simplified81.2%

        \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
      2. Add Preprocessing
      3. Step-by-step derivation
        1. *-commutative81.2%

          \[\leadsto {\color{blue}{\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a\right)}}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
        2. unpow-prod-down81.2%

          \[\leadsto \color{blue}{{\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{2} \cdot {a}^{2}} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
      4. Applied egg-rr81.2%

        \[\leadsto \color{blue}{{\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{2} \cdot {a}^{2}} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
      5. Final simplification81.2%

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

      Alternative 3: 79.7% accurate, 1.0× speedup?

      \[\begin{array}{l} \\ \begin{array}{l} t_0 := \pi \cdot \left(angle \cdot 0.005555555555555556\right)\\ {\left(\mathsf{hypot}\left(a \cdot \cos t\_0, b \cdot \sin t\_0\right)\right)}^{2} \end{array} \end{array} \]
      (FPCore (a b angle)
       :precision binary64
       (let* ((t_0 (* PI (* angle 0.005555555555555556))))
         (pow (hypot (* a (cos t_0)) (* b (sin t_0))) 2.0)))
      double code(double a, double b, double angle) {
      	double t_0 = ((double) M_PI) * (angle * 0.005555555555555556);
      	return pow(hypot((a * cos(t_0)), (b * sin(t_0))), 2.0);
      }
      
      public static double code(double a, double b, double angle) {
      	double t_0 = Math.PI * (angle * 0.005555555555555556);
      	return Math.pow(Math.hypot((a * Math.cos(t_0)), (b * Math.sin(t_0))), 2.0);
      }
      
      def code(a, b, angle):
      	t_0 = math.pi * (angle * 0.005555555555555556)
      	return math.pow(math.hypot((a * math.cos(t_0)), (b * math.sin(t_0))), 2.0)
      
      function code(a, b, angle)
      	t_0 = Float64(pi * Float64(angle * 0.005555555555555556))
      	return hypot(Float64(a * cos(t_0)), Float64(b * sin(t_0))) ^ 2.0
      end
      
      function tmp = code(a, b, angle)
      	t_0 = pi * (angle * 0.005555555555555556);
      	tmp = hypot((a * cos(t_0)), (b * sin(t_0))) ^ 2.0;
      end
      
      code[a_, b_, angle_] := Block[{t$95$0 = N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]}, N[Power[N[Sqrt[N[(a * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision] ^ 2 + N[(b * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision] ^ 2], $MachinePrecision], 2.0], $MachinePrecision]]
      
      \begin{array}{l}
      
      \\
      \begin{array}{l}
      t_0 := \pi \cdot \left(angle \cdot 0.005555555555555556\right)\\
      {\left(\mathsf{hypot}\left(a \cdot \cos t\_0, b \cdot \sin t\_0\right)\right)}^{2}
      \end{array}
      \end{array}
      
      Derivation
      1. Initial program 81.1%

        \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
      2. Step-by-step derivation
        1. Simplified81.2%

          \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
        2. Add Preprocessing
        3. Step-by-step derivation
          1. add-sqr-sqrt81.2%

            \[\leadsto \color{blue}{\sqrt{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \cdot \sqrt{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}}} \]
          2. pow281.2%

            \[\leadsto \color{blue}{{\left(\sqrt{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}}\right)}^{2}} \]
          3. unpow281.2%

            \[\leadsto {\left(\sqrt{\color{blue}{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}}\right)}^{2} \]
          4. unpow281.2%

            \[\leadsto {\left(\sqrt{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) + \color{blue}{\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}}\right)}^{2} \]
          5. hypot-define81.2%

            \[\leadsto {\color{blue}{\left(\mathsf{hypot}\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right), b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right)}}^{2} \]
          6. *-commutative81.2%

            \[\leadsto {\left(\mathsf{hypot}\left(\color{blue}{\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a}, b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right)}^{2} \]
          7. *-commutative81.2%

            \[\leadsto {\left(\mathsf{hypot}\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a, \color{blue}{\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b}\right)\right)}^{2} \]
        4. Applied egg-rr81.2%

          \[\leadsto \color{blue}{{\left(\mathsf{hypot}\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a, \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)\right)}^{2}} \]
        5. Final simplification81.2%

          \[\leadsto {\left(\mathsf{hypot}\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right), b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)\right)}^{2} \]
        6. Add Preprocessing

        Alternative 4: 62.4% accurate, 1.3× speedup?

        \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;b \leq 1.8 \cdot 10^{+142}:\\ \;\;\;\;{\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{2} \cdot {a}^{2}\\ \mathbf{else}:\\ \;\;\;\;{\left(b \cdot \sin \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\right)}^{2}\\ \end{array} \end{array} \]
        (FPCore (a b angle)
         :precision binary64
         (if (<= b 1.8e+142)
           (* (pow (cos (* PI (* angle 0.005555555555555556))) 2.0) (pow a 2.0))
           (pow (* b (sin (* angle (* PI 0.005555555555555556)))) 2.0)))
        double code(double a, double b, double angle) {
        	double tmp;
        	if (b <= 1.8e+142) {
        		tmp = pow(cos((((double) M_PI) * (angle * 0.005555555555555556))), 2.0) * pow(a, 2.0);
        	} else {
        		tmp = pow((b * sin((angle * (((double) M_PI) * 0.005555555555555556)))), 2.0);
        	}
        	return tmp;
        }
        
        public static double code(double a, double b, double angle) {
        	double tmp;
        	if (b <= 1.8e+142) {
        		tmp = Math.pow(Math.cos((Math.PI * (angle * 0.005555555555555556))), 2.0) * Math.pow(a, 2.0);
        	} else {
        		tmp = Math.pow((b * Math.sin((angle * (Math.PI * 0.005555555555555556)))), 2.0);
        	}
        	return tmp;
        }
        
        def code(a, b, angle):
        	tmp = 0
        	if b <= 1.8e+142:
        		tmp = math.pow(math.cos((math.pi * (angle * 0.005555555555555556))), 2.0) * math.pow(a, 2.0)
        	else:
        		tmp = math.pow((b * math.sin((angle * (math.pi * 0.005555555555555556)))), 2.0)
        	return tmp
        
        function code(a, b, angle)
        	tmp = 0.0
        	if (b <= 1.8e+142)
        		tmp = Float64((cos(Float64(pi * Float64(angle * 0.005555555555555556))) ^ 2.0) * (a ^ 2.0));
        	else
        		tmp = Float64(b * sin(Float64(angle * Float64(pi * 0.005555555555555556)))) ^ 2.0;
        	end
        	return tmp
        end
        
        function tmp_2 = code(a, b, angle)
        	tmp = 0.0;
        	if (b <= 1.8e+142)
        		tmp = (cos((pi * (angle * 0.005555555555555556))) ^ 2.0) * (a ^ 2.0);
        	else
        		tmp = (b * sin((angle * (pi * 0.005555555555555556)))) ^ 2.0;
        	end
        	tmp_2 = tmp;
        end
        
        code[a_, b_, angle_] := If[LessEqual[b, 1.8e+142], N[(N[Power[N[Cos[N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision], 2.0], $MachinePrecision] * N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision], N[Power[N[(b * N[Sin[N[(angle * N[(Pi * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]]
        
        \begin{array}{l}
        
        \\
        \begin{array}{l}
        \mathbf{if}\;b \leq 1.8 \cdot 10^{+142}:\\
        \;\;\;\;{\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{2} \cdot {a}^{2}\\
        
        \mathbf{else}:\\
        \;\;\;\;{\left(b \cdot \sin \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\right)}^{2}\\
        
        
        \end{array}
        \end{array}
        
        Derivation
        1. Split input into 2 regimes
        2. if b < 1.8000000000000001e142

          1. Initial program 77.9%

            \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
          2. Step-by-step derivation
            1. Simplified78.0%

              \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
            2. Add Preprocessing
            3. Step-by-step derivation
              1. add-cbrt-cube57.8%

                \[\leadsto \color{blue}{\sqrt[3]{\left(\left({\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\right) \cdot \left({\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\right)\right) \cdot \left({\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\right)}} \]
              2. pow1/357.1%

                \[\leadsto \color{blue}{{\left(\left(\left({\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\right) \cdot \left({\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\right)\right) \cdot \left({\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\right)\right)}^{0.3333333333333333}} \]
            4. Applied egg-rr57.1%

              \[\leadsto \color{blue}{{\left({\left({\left(\mathsf{hypot}\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a, \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)\right)}^{2}\right)}^{3}\right)}^{0.3333333333333333}} \]
            5. Taylor expanded in a around inf 61.3%

              \[\leadsto \color{blue}{{a}^{2} \cdot {\cos \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2}} \]
            6. Step-by-step derivation
              1. *-commutative61.3%

                \[\leadsto {a}^{2} \cdot {\cos \color{blue}{\left(\left(angle \cdot \pi\right) \cdot 0.005555555555555556\right)}}^{2} \]
              2. *-commutative61.3%

                \[\leadsto {a}^{2} \cdot {\cos \left(\color{blue}{\left(\pi \cdot angle\right)} \cdot 0.005555555555555556\right)}^{2} \]
              3. associate-*r*61.4%

                \[\leadsto {a}^{2} \cdot {\cos \color{blue}{\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}}^{2} \]
            7. Simplified61.4%

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

            if 1.8000000000000001e142 < b

            1. Initial program 97.7%

              \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
            2. Step-by-step derivation
              1. Simplified97.7%

                \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
              2. Add Preprocessing
              3. Taylor expanded in a around 0 65.4%

                \[\leadsto \color{blue}{{b}^{2} \cdot {\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2}} \]
              4. Step-by-step derivation
                1. *-commutative65.4%

                  \[\leadsto \color{blue}{{\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2} \cdot {b}^{2}} \]
                2. *-commutative65.4%

                  \[\leadsto {\sin \color{blue}{\left(\left(angle \cdot \pi\right) \cdot 0.005555555555555556\right)}}^{2} \cdot {b}^{2} \]
                3. *-commutative65.4%

                  \[\leadsto {\sin \left(\color{blue}{\left(\pi \cdot angle\right)} \cdot 0.005555555555555556\right)}^{2} \cdot {b}^{2} \]
                4. associate-*r*65.4%

                  \[\leadsto {\sin \color{blue}{\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}}^{2} \cdot {b}^{2} \]
                5. unpow265.4%

                  \[\leadsto \color{blue}{\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)} \cdot {b}^{2} \]
                6. unpow265.4%

                  \[\leadsto \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \color{blue}{\left(b \cdot b\right)} \]
                7. swap-sqr86.2%

                  \[\leadsto \color{blue}{\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right) \cdot \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)} \]
                8. unpow286.2%

                  \[\leadsto \color{blue}{{\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)}^{2}} \]
                9. *-commutative86.2%

                  \[\leadsto {\color{blue}{\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}}^{2} \]
                10. associate-*r*86.2%

                  \[\leadsto {\left(b \cdot \sin \color{blue}{\left(\left(\pi \cdot angle\right) \cdot 0.005555555555555556\right)}\right)}^{2} \]
                11. *-commutative86.2%

                  \[\leadsto {\left(b \cdot \sin \left(\color{blue}{\left(angle \cdot \pi\right)} \cdot 0.005555555555555556\right)\right)}^{2} \]
                12. associate-*r*86.3%

                  \[\leadsto {\left(b \cdot \sin \color{blue}{\left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)}\right)}^{2} \]
              5. Simplified86.3%

                \[\leadsto \color{blue}{{\left(b \cdot \sin \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
            3. Recombined 2 regimes into one program.
            4. Final simplification65.4%

              \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq 1.8 \cdot 10^{+142}:\\ \;\;\;\;{\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{2} \cdot {a}^{2}\\ \mathbf{else}:\\ \;\;\;\;{\left(b \cdot \sin \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\right)}^{2}\\ \end{array} \]
            5. Add Preprocessing

            Alternative 5: 79.6% accurate, 1.3× speedup?

            \[\begin{array}{l} \\ {\left(\mathsf{hypot}\left(a, b \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)\right)}^{2} \end{array} \]
            (FPCore (a b angle)
             :precision binary64
             (pow (hypot a (* b (sin (* 0.005555555555555556 (* PI angle))))) 2.0))
            double code(double a, double b, double angle) {
            	return pow(hypot(a, (b * sin((0.005555555555555556 * (((double) M_PI) * angle))))), 2.0);
            }
            
            public static double code(double a, double b, double angle) {
            	return Math.pow(Math.hypot(a, (b * Math.sin((0.005555555555555556 * (Math.PI * angle))))), 2.0);
            }
            
            def code(a, b, angle):
            	return math.pow(math.hypot(a, (b * math.sin((0.005555555555555556 * (math.pi * angle))))), 2.0)
            
            function code(a, b, angle)
            	return hypot(a, Float64(b * sin(Float64(0.005555555555555556 * Float64(pi * angle))))) ^ 2.0
            end
            
            function tmp = code(a, b, angle)
            	tmp = hypot(a, (b * sin((0.005555555555555556 * (pi * angle))))) ^ 2.0;
            end
            
            code[a_, b_, angle_] := N[Power[N[Sqrt[a ^ 2 + N[(b * N[Sin[N[(0.005555555555555556 * N[(Pi * angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] ^ 2], $MachinePrecision], 2.0], $MachinePrecision]
            
            \begin{array}{l}
            
            \\
            {\left(\mathsf{hypot}\left(a, b \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)\right)}^{2}
            \end{array}
            
            Derivation
            1. Initial program 81.1%

              \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
            2. Step-by-step derivation
              1. Simplified81.2%

                \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
              2. Add Preprocessing
              3. Step-by-step derivation
                1. add-cbrt-cube62.2%

                  \[\leadsto \color{blue}{\sqrt[3]{\left(\left({\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\right) \cdot \left({\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\right)\right) \cdot \left({\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\right)}} \]
                2. pow1/361.4%

                  \[\leadsto \color{blue}{{\left(\left(\left({\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\right) \cdot \left({\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\right)\right) \cdot \left({\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\right)\right)}^{0.3333333333333333}} \]
              4. Applied egg-rr61.4%

                \[\leadsto \color{blue}{{\left({\left({\left(\mathsf{hypot}\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a, \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)\right)}^{2}\right)}^{3}\right)}^{0.3333333333333333}} \]
              5. Taylor expanded in angle around 0 61.5%

                \[\leadsto {\left({\left({\left(\mathsf{hypot}\left(\color{blue}{1} \cdot a, \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)\right)}^{2}\right)}^{3}\right)}^{0.3333333333333333} \]
              6. Taylor expanded in a around 0 71.5%

                \[\leadsto \color{blue}{{b}^{2} \cdot {\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2} + {a}^{2}} \]
              7. Step-by-step derivation
                1. +-commutative71.5%

                  \[\leadsto \color{blue}{{a}^{2} + {b}^{2} \cdot {\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2}} \]
                2. unpow271.5%

                  \[\leadsto \color{blue}{a \cdot a} + {b}^{2} \cdot {\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2} \]
                3. *-commutative71.5%

                  \[\leadsto a \cdot a + \color{blue}{{\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2} \cdot {b}^{2}} \]
                4. *-commutative71.5%

                  \[\leadsto a \cdot a + {\sin \color{blue}{\left(\left(angle \cdot \pi\right) \cdot 0.005555555555555556\right)}}^{2} \cdot {b}^{2} \]
                5. *-commutative71.5%

                  \[\leadsto a \cdot a + {\sin \left(\color{blue}{\left(\pi \cdot angle\right)} \cdot 0.005555555555555556\right)}^{2} \cdot {b}^{2} \]
                6. associate-*r*71.5%

                  \[\leadsto a \cdot a + {\sin \color{blue}{\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}}^{2} \cdot {b}^{2} \]
                7. unpow271.5%

                  \[\leadsto a \cdot a + \color{blue}{\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)} \cdot {b}^{2} \]
                8. unpow271.5%

                  \[\leadsto a \cdot a + \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \color{blue}{\left(b \cdot b\right)} \]
                9. swap-sqr81.0%

                  \[\leadsto a \cdot a + \color{blue}{\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right) \cdot \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)} \]
                10. rem-square-sqrt81.0%

                  \[\leadsto \color{blue}{\sqrt{a \cdot a + \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right) \cdot \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)} \cdot \sqrt{a \cdot a + \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right) \cdot \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)}} \]
                11. hypot-undefine81.0%

                  \[\leadsto \color{blue}{\mathsf{hypot}\left(a, \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)} \cdot \sqrt{a \cdot a + \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right) \cdot \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)} \]
              8. Simplified81.0%

                \[\leadsto \color{blue}{{\left(\mathsf{hypot}\left(a, b \cdot \sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)\right)}^{2}} \]
              9. Final simplification81.0%

                \[\leadsto {\left(\mathsf{hypot}\left(a, b \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)\right)}^{2} \]
              10. Add Preprocessing

              Alternative 6: 62.4% accurate, 2.0× speedup?

              \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;b \leq 5.6 \cdot 10^{+142}:\\ \;\;\;\;{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\\ \mathbf{else}:\\ \;\;\;\;{\left(b \cdot \sin \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\right)}^{2}\\ \end{array} \end{array} \]
              (FPCore (a b angle)
               :precision binary64
               (if (<= b 5.6e+142)
                 (pow (* a (cos (* PI (* angle 0.005555555555555556)))) 2.0)
                 (pow (* b (sin (* angle (* PI 0.005555555555555556)))) 2.0)))
              double code(double a, double b, double angle) {
              	double tmp;
              	if (b <= 5.6e+142) {
              		tmp = pow((a * cos((((double) M_PI) * (angle * 0.005555555555555556)))), 2.0);
              	} else {
              		tmp = pow((b * sin((angle * (((double) M_PI) * 0.005555555555555556)))), 2.0);
              	}
              	return tmp;
              }
              
              public static double code(double a, double b, double angle) {
              	double tmp;
              	if (b <= 5.6e+142) {
              		tmp = Math.pow((a * Math.cos((Math.PI * (angle * 0.005555555555555556)))), 2.0);
              	} else {
              		tmp = Math.pow((b * Math.sin((angle * (Math.PI * 0.005555555555555556)))), 2.0);
              	}
              	return tmp;
              }
              
              def code(a, b, angle):
              	tmp = 0
              	if b <= 5.6e+142:
              		tmp = math.pow((a * math.cos((math.pi * (angle * 0.005555555555555556)))), 2.0)
              	else:
              		tmp = math.pow((b * math.sin((angle * (math.pi * 0.005555555555555556)))), 2.0)
              	return tmp
              
              function code(a, b, angle)
              	tmp = 0.0
              	if (b <= 5.6e+142)
              		tmp = Float64(a * cos(Float64(pi * Float64(angle * 0.005555555555555556)))) ^ 2.0;
              	else
              		tmp = Float64(b * sin(Float64(angle * Float64(pi * 0.005555555555555556)))) ^ 2.0;
              	end
              	return tmp
              end
              
              function tmp_2 = code(a, b, angle)
              	tmp = 0.0;
              	if (b <= 5.6e+142)
              		tmp = (a * cos((pi * (angle * 0.005555555555555556)))) ^ 2.0;
              	else
              		tmp = (b * sin((angle * (pi * 0.005555555555555556)))) ^ 2.0;
              	end
              	tmp_2 = tmp;
              end
              
              code[a_, b_, angle_] := If[LessEqual[b, 5.6e+142], N[Power[N[(a * N[Cos[N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision], N[Power[N[(b * N[Sin[N[(angle * N[(Pi * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]]
              
              \begin{array}{l}
              
              \\
              \begin{array}{l}
              \mathbf{if}\;b \leq 5.6 \cdot 10^{+142}:\\
              \;\;\;\;{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\\
              
              \mathbf{else}:\\
              \;\;\;\;{\left(b \cdot \sin \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\right)}^{2}\\
              
              
              \end{array}
              \end{array}
              
              Derivation
              1. Split input into 2 regimes
              2. if b < 5.6e142

                1. Initial program 77.9%

                  \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
                2. Step-by-step derivation
                  1. Simplified78.0%

                    \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
                  2. Add Preprocessing
                  3. Taylor expanded in a around inf 61.3%

                    \[\leadsto \color{blue}{{a}^{2} \cdot {\cos \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2}} \]
                  4. Step-by-step derivation
                    1. *-commutative61.3%

                      \[\leadsto \color{blue}{{\cos \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2} \cdot {a}^{2}} \]
                    2. *-commutative61.3%

                      \[\leadsto {\cos \color{blue}{\left(\left(angle \cdot \pi\right) \cdot 0.005555555555555556\right)}}^{2} \cdot {a}^{2} \]
                    3. *-commutative61.3%

                      \[\leadsto {\cos \left(\color{blue}{\left(\pi \cdot angle\right)} \cdot 0.005555555555555556\right)}^{2} \cdot {a}^{2} \]
                    4. associate-*r*61.4%

                      \[\leadsto {\cos \color{blue}{\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}}^{2} \cdot {a}^{2} \]
                    5. unpow261.4%

                      \[\leadsto \color{blue}{\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)} \cdot {a}^{2} \]
                    6. unpow261.4%

                      \[\leadsto \left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \color{blue}{\left(a \cdot a\right)} \]
                    7. swap-sqr61.4%

                      \[\leadsto \color{blue}{\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a\right) \cdot \left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a\right)} \]
                    8. unpow261.4%

                      \[\leadsto \color{blue}{{\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a\right)}^{2}} \]
                    9. *-commutative61.4%

                      \[\leadsto {\color{blue}{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}}^{2} \]
                  5. Simplified61.4%

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

                  if 5.6e142 < b

                  1. Initial program 97.7%

                    \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
                  2. Step-by-step derivation
                    1. Simplified97.7%

                      \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
                    2. Add Preprocessing
                    3. Taylor expanded in a around 0 65.4%

                      \[\leadsto \color{blue}{{b}^{2} \cdot {\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2}} \]
                    4. Step-by-step derivation
                      1. *-commutative65.4%

                        \[\leadsto \color{blue}{{\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2} \cdot {b}^{2}} \]
                      2. *-commutative65.4%

                        \[\leadsto {\sin \color{blue}{\left(\left(angle \cdot \pi\right) \cdot 0.005555555555555556\right)}}^{2} \cdot {b}^{2} \]
                      3. *-commutative65.4%

                        \[\leadsto {\sin \left(\color{blue}{\left(\pi \cdot angle\right)} \cdot 0.005555555555555556\right)}^{2} \cdot {b}^{2} \]
                      4. associate-*r*65.4%

                        \[\leadsto {\sin \color{blue}{\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}}^{2} \cdot {b}^{2} \]
                      5. unpow265.4%

                        \[\leadsto \color{blue}{\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)} \cdot {b}^{2} \]
                      6. unpow265.4%

                        \[\leadsto \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \color{blue}{\left(b \cdot b\right)} \]
                      7. swap-sqr86.2%

                        \[\leadsto \color{blue}{\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right) \cdot \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)} \]
                      8. unpow286.2%

                        \[\leadsto \color{blue}{{\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)}^{2}} \]
                      9. *-commutative86.2%

                        \[\leadsto {\color{blue}{\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}}^{2} \]
                      10. associate-*r*86.2%

                        \[\leadsto {\left(b \cdot \sin \color{blue}{\left(\left(\pi \cdot angle\right) \cdot 0.005555555555555556\right)}\right)}^{2} \]
                      11. *-commutative86.2%

                        \[\leadsto {\left(b \cdot \sin \left(\color{blue}{\left(angle \cdot \pi\right)} \cdot 0.005555555555555556\right)\right)}^{2} \]
                      12. associate-*r*86.3%

                        \[\leadsto {\left(b \cdot \sin \color{blue}{\left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)}\right)}^{2} \]
                    5. Simplified86.3%

                      \[\leadsto \color{blue}{{\left(b \cdot \sin \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
                  3. Recombined 2 regimes into one program.
                  4. Add Preprocessing

                  Alternative 7: 62.4% accurate, 2.0× speedup?

                  \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;b \leq 5.5 \cdot 10^{+142}:\\ \;\;\;\;{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\\ \mathbf{else}:\\ \;\;\;\;{\left(b \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)}^{2}\\ \end{array} \end{array} \]
                  (FPCore (a b angle)
                   :precision binary64
                   (if (<= b 5.5e+142)
                     (pow (* a (cos (* PI (* angle 0.005555555555555556)))) 2.0)
                     (pow (* b (sin (* 0.005555555555555556 (* PI angle)))) 2.0)))
                  double code(double a, double b, double angle) {
                  	double tmp;
                  	if (b <= 5.5e+142) {
                  		tmp = pow((a * cos((((double) M_PI) * (angle * 0.005555555555555556)))), 2.0);
                  	} else {
                  		tmp = pow((b * sin((0.005555555555555556 * (((double) M_PI) * angle)))), 2.0);
                  	}
                  	return tmp;
                  }
                  
                  public static double code(double a, double b, double angle) {
                  	double tmp;
                  	if (b <= 5.5e+142) {
                  		tmp = Math.pow((a * Math.cos((Math.PI * (angle * 0.005555555555555556)))), 2.0);
                  	} else {
                  		tmp = Math.pow((b * Math.sin((0.005555555555555556 * (Math.PI * angle)))), 2.0);
                  	}
                  	return tmp;
                  }
                  
                  def code(a, b, angle):
                  	tmp = 0
                  	if b <= 5.5e+142:
                  		tmp = math.pow((a * math.cos((math.pi * (angle * 0.005555555555555556)))), 2.0)
                  	else:
                  		tmp = math.pow((b * math.sin((0.005555555555555556 * (math.pi * angle)))), 2.0)
                  	return tmp
                  
                  function code(a, b, angle)
                  	tmp = 0.0
                  	if (b <= 5.5e+142)
                  		tmp = Float64(a * cos(Float64(pi * Float64(angle * 0.005555555555555556)))) ^ 2.0;
                  	else
                  		tmp = Float64(b * sin(Float64(0.005555555555555556 * Float64(pi * angle)))) ^ 2.0;
                  	end
                  	return tmp
                  end
                  
                  function tmp_2 = code(a, b, angle)
                  	tmp = 0.0;
                  	if (b <= 5.5e+142)
                  		tmp = (a * cos((pi * (angle * 0.005555555555555556)))) ^ 2.0;
                  	else
                  		tmp = (b * sin((0.005555555555555556 * (pi * angle)))) ^ 2.0;
                  	end
                  	tmp_2 = tmp;
                  end
                  
                  code[a_, b_, angle_] := If[LessEqual[b, 5.5e+142], N[Power[N[(a * N[Cos[N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision], N[Power[N[(b * N[Sin[N[(0.005555555555555556 * N[(Pi * angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]]
                  
                  \begin{array}{l}
                  
                  \\
                  \begin{array}{l}
                  \mathbf{if}\;b \leq 5.5 \cdot 10^{+142}:\\
                  \;\;\;\;{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\\
                  
                  \mathbf{else}:\\
                  \;\;\;\;{\left(b \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)}^{2}\\
                  
                  
                  \end{array}
                  \end{array}
                  
                  Derivation
                  1. Split input into 2 regimes
                  2. if b < 5.50000000000000035e142

                    1. Initial program 77.9%

                      \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
                    2. Step-by-step derivation
                      1. Simplified78.0%

                        \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
                      2. Add Preprocessing
                      3. Taylor expanded in a around inf 61.3%

                        \[\leadsto \color{blue}{{a}^{2} \cdot {\cos \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2}} \]
                      4. Step-by-step derivation
                        1. *-commutative61.3%

                          \[\leadsto \color{blue}{{\cos \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2} \cdot {a}^{2}} \]
                        2. *-commutative61.3%

                          \[\leadsto {\cos \color{blue}{\left(\left(angle \cdot \pi\right) \cdot 0.005555555555555556\right)}}^{2} \cdot {a}^{2} \]
                        3. *-commutative61.3%

                          \[\leadsto {\cos \left(\color{blue}{\left(\pi \cdot angle\right)} \cdot 0.005555555555555556\right)}^{2} \cdot {a}^{2} \]
                        4. associate-*r*61.4%

                          \[\leadsto {\cos \color{blue}{\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}}^{2} \cdot {a}^{2} \]
                        5. unpow261.4%

                          \[\leadsto \color{blue}{\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)} \cdot {a}^{2} \]
                        6. unpow261.4%

                          \[\leadsto \left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \color{blue}{\left(a \cdot a\right)} \]
                        7. swap-sqr61.4%

                          \[\leadsto \color{blue}{\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a\right) \cdot \left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a\right)} \]
                        8. unpow261.4%

                          \[\leadsto \color{blue}{{\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a\right)}^{2}} \]
                        9. *-commutative61.4%

                          \[\leadsto {\color{blue}{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}}^{2} \]
                      5. Simplified61.4%

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

                      if 5.50000000000000035e142 < b

                      1. Initial program 97.7%

                        \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
                      2. Step-by-step derivation
                        1. Simplified97.7%

                          \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
                        2. Add Preprocessing
                        3. Step-by-step derivation
                          1. add-cbrt-cube84.5%

                            \[\leadsto \color{blue}{\sqrt[3]{\left(\left({\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\right) \cdot \left({\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\right)\right) \cdot \left({\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\right)}} \]
                          2. pow1/383.4%

                            \[\leadsto \color{blue}{{\left(\left(\left({\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\right) \cdot \left({\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\right)\right) \cdot \left({\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\right)\right)}^{0.3333333333333333}} \]
                        4. Applied egg-rr83.4%

                          \[\leadsto \color{blue}{{\left({\left({\left(\mathsf{hypot}\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a, \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)\right)}^{2}\right)}^{3}\right)}^{0.3333333333333333}} \]
                        5. Taylor expanded in angle around 0 83.4%

                          \[\leadsto {\left({\left({\left(\mathsf{hypot}\left(\color{blue}{1} \cdot a, \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)\right)}^{2}\right)}^{3}\right)}^{0.3333333333333333} \]
                        6. Taylor expanded in a around 0 65.4%

                          \[\leadsto \color{blue}{{b}^{2} \cdot {\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2}} \]
                        7. Step-by-step derivation
                          1. *-commutative65.4%

                            \[\leadsto \color{blue}{{\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2} \cdot {b}^{2}} \]
                          2. *-commutative65.4%

                            \[\leadsto {\sin \color{blue}{\left(\left(angle \cdot \pi\right) \cdot 0.005555555555555556\right)}}^{2} \cdot {b}^{2} \]
                          3. *-commutative65.4%

                            \[\leadsto {\sin \left(\color{blue}{\left(\pi \cdot angle\right)} \cdot 0.005555555555555556\right)}^{2} \cdot {b}^{2} \]
                          4. associate-*r*65.4%

                            \[\leadsto {\sin \color{blue}{\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}}^{2} \cdot {b}^{2} \]
                          5. unpow265.4%

                            \[\leadsto \color{blue}{\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)} \cdot {b}^{2} \]
                          6. unpow265.4%

                            \[\leadsto \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \color{blue}{\left(b \cdot b\right)} \]
                          7. swap-sqr86.2%

                            \[\leadsto \color{blue}{\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right) \cdot \left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)} \]
                          8. unpow286.2%

                            \[\leadsto \color{blue}{{\left(\sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot b\right)}^{2}} \]
                          9. *-commutative86.2%

                            \[\leadsto {\color{blue}{\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}}^{2} \]
                          10. associate-*r*86.2%

                            \[\leadsto {\left(b \cdot \sin \color{blue}{\left(\left(\pi \cdot angle\right) \cdot 0.005555555555555556\right)}\right)}^{2} \]
                          11. *-commutative86.2%

                            \[\leadsto {\left(b \cdot \sin \left(\color{blue}{\left(angle \cdot \pi\right)} \cdot 0.005555555555555556\right)\right)}^{2} \]
                          12. *-commutative86.2%

                            \[\leadsto {\left(b \cdot \sin \color{blue}{\left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}\right)}^{2} \]
                        8. Simplified86.2%

                          \[\leadsto \color{blue}{{\left(b \cdot \sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)}^{2}} \]
                      3. Recombined 2 regimes into one program.
                      4. Final simplification65.4%

                        \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq 5.5 \cdot 10^{+142}:\\ \;\;\;\;{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}\\ \mathbf{else}:\\ \;\;\;\;{\left(b \cdot \sin \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)}^{2}\\ \end{array} \]
                      5. Add Preprocessing

                      Alternative 8: 56.9% accurate, 2.0× speedup?

                      \[\begin{array}{l} \\ {\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \end{array} \]
                      (FPCore (a b angle)
                       :precision binary64
                       (pow (* a (cos (* PI (* angle 0.005555555555555556)))) 2.0))
                      double code(double a, double b, double angle) {
                      	return pow((a * cos((((double) M_PI) * (angle * 0.005555555555555556)))), 2.0);
                      }
                      
                      public static double code(double a, double b, double angle) {
                      	return Math.pow((a * Math.cos((Math.PI * (angle * 0.005555555555555556)))), 2.0);
                      }
                      
                      def code(a, b, angle):
                      	return math.pow((a * math.cos((math.pi * (angle * 0.005555555555555556)))), 2.0)
                      
                      function code(a, b, angle)
                      	return Float64(a * cos(Float64(pi * Float64(angle * 0.005555555555555556)))) ^ 2.0
                      end
                      
                      function tmp = code(a, b, angle)
                      	tmp = (a * cos((pi * (angle * 0.005555555555555556)))) ^ 2.0;
                      end
                      
                      code[a_, b_, angle_] := N[Power[N[(a * N[Cos[N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]
                      
                      \begin{array}{l}
                      
                      \\
                      {\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}
                      \end{array}
                      
                      Derivation
                      1. Initial program 81.1%

                        \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
                      2. Step-by-step derivation
                        1. Simplified81.2%

                          \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
                        2. Add Preprocessing
                        3. Taylor expanded in a around inf 56.8%

                          \[\leadsto \color{blue}{{a}^{2} \cdot {\cos \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2}} \]
                        4. Step-by-step derivation
                          1. *-commutative56.8%

                            \[\leadsto \color{blue}{{\cos \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2} \cdot {a}^{2}} \]
                          2. *-commutative56.8%

                            \[\leadsto {\cos \color{blue}{\left(\left(angle \cdot \pi\right) \cdot 0.005555555555555556\right)}}^{2} \cdot {a}^{2} \]
                          3. *-commutative56.8%

                            \[\leadsto {\cos \left(\color{blue}{\left(\pi \cdot angle\right)} \cdot 0.005555555555555556\right)}^{2} \cdot {a}^{2} \]
                          4. associate-*r*56.8%

                            \[\leadsto {\cos \color{blue}{\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}}^{2} \cdot {a}^{2} \]
                          5. unpow256.8%

                            \[\leadsto \color{blue}{\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)} \cdot {a}^{2} \]
                          6. unpow256.8%

                            \[\leadsto \left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \color{blue}{\left(a \cdot a\right)} \]
                          7. swap-sqr56.8%

                            \[\leadsto \color{blue}{\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a\right) \cdot \left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a\right)} \]
                          8. unpow256.8%

                            \[\leadsto \color{blue}{{\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a\right)}^{2}} \]
                          9. *-commutative56.8%

                            \[\leadsto {\color{blue}{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}}^{2} \]
                        5. Simplified56.8%

                          \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
                        6. Add Preprocessing

                        Alternative 9: 56.9% accurate, 2.0× speedup?

                        \[\begin{array}{l} \\ {\left(a \cdot \cos \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)}^{2} \end{array} \]
                        (FPCore (a b angle)
                         :precision binary64
                         (pow (* a (cos (* 0.005555555555555556 (* PI angle)))) 2.0))
                        double code(double a, double b, double angle) {
                        	return pow((a * cos((0.005555555555555556 * (((double) M_PI) * angle)))), 2.0);
                        }
                        
                        public static double code(double a, double b, double angle) {
                        	return Math.pow((a * Math.cos((0.005555555555555556 * (Math.PI * angle)))), 2.0);
                        }
                        
                        def code(a, b, angle):
                        	return math.pow((a * math.cos((0.005555555555555556 * (math.pi * angle)))), 2.0)
                        
                        function code(a, b, angle)
                        	return Float64(a * cos(Float64(0.005555555555555556 * Float64(pi * angle)))) ^ 2.0
                        end
                        
                        function tmp = code(a, b, angle)
                        	tmp = (a * cos((0.005555555555555556 * (pi * angle)))) ^ 2.0;
                        end
                        
                        code[a_, b_, angle_] := N[Power[N[(a * N[Cos[N[(0.005555555555555556 * N[(Pi * angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]
                        
                        \begin{array}{l}
                        
                        \\
                        {\left(a \cdot \cos \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)}^{2}
                        \end{array}
                        
                        Derivation
                        1. Initial program 81.1%

                          \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
                        2. Step-by-step derivation
                          1. Simplified81.2%

                            \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
                          2. Add Preprocessing
                          3. Step-by-step derivation
                            1. unpow281.2%

                              \[\leadsto \color{blue}{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                            2. *-commutative81.2%

                              \[\leadsto \left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \color{blue}{\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a\right)} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                            3. associate-*r*81.2%

                              \[\leadsto \color{blue}{\left(\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot a} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                            4. *-commutative81.2%

                              \[\leadsto \left(\color{blue}{\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a\right)} \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot a + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                          4. Applied egg-rr81.2%

                            \[\leadsto \color{blue}{\left(\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot a} + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                          5. Step-by-step derivation
                            1. add-sqr-sqrt75.7%

                              \[\leadsto \left(\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a\right) \cdot \cos \left(\color{blue}{\left(\sqrt{\pi} \cdot \sqrt{\pi}\right)} \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot a + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                            2. pow275.7%

                              \[\leadsto \left(\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a\right) \cdot \cos \left(\color{blue}{{\left(\sqrt{\pi}\right)}^{2}} \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot a + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                          6. Applied egg-rr75.7%

                            \[\leadsto \left(\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot a\right) \cdot \cos \left(\color{blue}{{\left(\sqrt{\pi}\right)}^{2}} \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot a + {\left(b \cdot \sin \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2} \]
                          7. Taylor expanded in a around inf 56.8%

                            \[\leadsto \color{blue}{{a}^{2} \cdot {\cos \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2}} \]
                          8. Step-by-step derivation
                            1. unpow256.8%

                              \[\leadsto \color{blue}{\left(a \cdot a\right)} \cdot {\cos \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}^{2} \]
                            2. *-commutative56.8%

                              \[\leadsto \left(a \cdot a\right) \cdot {\cos \color{blue}{\left(\left(angle \cdot \pi\right) \cdot 0.005555555555555556\right)}}^{2} \]
                            3. *-commutative56.8%

                              \[\leadsto \left(a \cdot a\right) \cdot {\cos \left(\color{blue}{\left(\pi \cdot angle\right)} \cdot 0.005555555555555556\right)}^{2} \]
                            4. associate-*r*56.8%

                              \[\leadsto \left(a \cdot a\right) \cdot {\cos \color{blue}{\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}}^{2} \]
                            5. unpow256.8%

                              \[\leadsto \left(a \cdot a\right) \cdot \color{blue}{\left(\cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right) \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)} \]
                            6. swap-sqr56.8%

                              \[\leadsto \color{blue}{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right) \cdot \left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)} \]
                            7. unpow256.8%

                              \[\leadsto \color{blue}{{\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}} \]
                            8. associate-*r*56.7%

                              \[\leadsto {\left(a \cdot \cos \color{blue}{\left(\left(\pi \cdot angle\right) \cdot 0.005555555555555556\right)}\right)}^{2} \]
                            9. *-commutative56.7%

                              \[\leadsto {\left(a \cdot \cos \left(\color{blue}{\left(angle \cdot \pi\right)} \cdot 0.005555555555555556\right)\right)}^{2} \]
                            10. *-commutative56.7%

                              \[\leadsto {\left(a \cdot \cos \color{blue}{\left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}\right)}^{2} \]
                          9. Simplified56.7%

                            \[\leadsto \color{blue}{{\left(a \cdot \cos \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)}^{2}} \]
                          10. Final simplification56.7%

                            \[\leadsto {\left(a \cdot \cos \left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)}^{2} \]
                          11. Add Preprocessing

                          Alternative 10: 57.1% accurate, 4.1× speedup?

                          \[\begin{array}{l} \\ {a}^{2} \end{array} \]
                          (FPCore (a b angle) :precision binary64 (pow a 2.0))
                          double code(double a, double b, double angle) {
                          	return pow(a, 2.0);
                          }
                          
                          real(8) function code(a, b, angle)
                              real(8), intent (in) :: a
                              real(8), intent (in) :: b
                              real(8), intent (in) :: angle
                              code = a ** 2.0d0
                          end function
                          
                          public static double code(double a, double b, double angle) {
                          	return Math.pow(a, 2.0);
                          }
                          
                          def code(a, b, angle):
                          	return math.pow(a, 2.0)
                          
                          function code(a, b, angle)
                          	return a ^ 2.0
                          end
                          
                          function tmp = code(a, b, angle)
                          	tmp = a ^ 2.0;
                          end
                          
                          code[a_, b_, angle_] := N[Power[a, 2.0], $MachinePrecision]
                          
                          \begin{array}{l}
                          
                          \\
                          {a}^{2}
                          \end{array}
                          
                          Derivation
                          1. Initial program 81.1%

                            \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
                          2. Step-by-step derivation
                            1. Simplified81.2%

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

                              \[\leadsto \color{blue}{{a}^{2}} \]
                            4. Add Preprocessing

                            Reproduce

                            ?
                            herbie shell --seed 2024111 
                            (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)))