Math FPCore C Java Julia Wolfram TeX \[{\left(a \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} + {\left(b \cdot \cos \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2}
\]
↓
\[\begin{array}{l}
t_0 := \sin \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\\
a \cdot \left(t_0 \cdot \left(t_0 \cdot a\right)\right) + {\left(b \cdot \cos \left(angle \cdot \frac{\sqrt[3]{\pi} \cdot {\left(\sqrt[3]{\pi}\right)}^{2}}{180}\right)\right)}^{2}
\end{array}
\]
(FPCore (a b angle)
:precision binary64
(+
(pow (* a (sin (* (/ angle 180.0) PI))) 2.0)
(pow (* b (cos (* (/ angle 180.0) PI))) 2.0))) ↓
(FPCore (a b angle)
:precision binary64
(let* ((t_0 (sin (* angle (* PI 0.005555555555555556)))))
(+
(* a (* t_0 (* t_0 a)))
(pow
(* b (cos (* angle (/ (* (cbrt PI) (pow (cbrt PI) 2.0)) 180.0))))
2.0)))) double code(double a, double b, double angle) {
return pow((a * sin(((angle / 180.0) * ((double) M_PI)))), 2.0) + pow((b * cos(((angle / 180.0) * ((double) M_PI)))), 2.0);
}
↓
double code(double a, double b, double angle) {
double t_0 = sin((angle * (((double) M_PI) * 0.005555555555555556)));
return (a * (t_0 * (t_0 * a))) + pow((b * cos((angle * ((cbrt(((double) M_PI)) * pow(cbrt(((double) M_PI)), 2.0)) / 180.0)))), 2.0);
}
public static double code(double a, double b, double angle) {
return Math.pow((a * Math.sin(((angle / 180.0) * Math.PI))), 2.0) + Math.pow((b * Math.cos(((angle / 180.0) * Math.PI))), 2.0);
}
↓
public static double code(double a, double b, double angle) {
double t_0 = Math.sin((angle * (Math.PI * 0.005555555555555556)));
return (a * (t_0 * (t_0 * a))) + Math.pow((b * Math.cos((angle * ((Math.cbrt(Math.PI) * Math.pow(Math.cbrt(Math.PI), 2.0)) / 180.0)))), 2.0);
}
function code(a, b, angle)
return Float64((Float64(a * sin(Float64(Float64(angle / 180.0) * pi))) ^ 2.0) + (Float64(b * cos(Float64(Float64(angle / 180.0) * pi))) ^ 2.0))
end
↓
function code(a, b, angle)
t_0 = sin(Float64(angle * Float64(pi * 0.005555555555555556)))
return Float64(Float64(a * Float64(t_0 * Float64(t_0 * a))) + (Float64(b * cos(Float64(angle * Float64(Float64(cbrt(pi) * (cbrt(pi) ^ 2.0)) / 180.0)))) ^ 2.0))
end
code[a_, b_, angle_] := N[(N[Power[N[(a * N[Sin[N[(N[(angle / 180.0), $MachinePrecision] * Pi), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Cos[N[(N[(angle / 180.0), $MachinePrecision] * Pi), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]
↓
code[a_, b_, angle_] := Block[{t$95$0 = N[Sin[N[(angle * N[(Pi * 0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]}, N[(N[(a * N[(t$95$0 * N[(t$95$0 * a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[Power[N[(b * N[Cos[N[(angle * N[(N[(N[Power[Pi, 1/3], $MachinePrecision] * N[Power[N[Power[Pi, 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]
{\left(a \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} + {\left(b \cdot \cos \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2}
↓
\begin{array}{l}
t_0 := \sin \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\\
a \cdot \left(t_0 \cdot \left(t_0 \cdot a\right)\right) + {\left(b \cdot \cos \left(angle \cdot \frac{\sqrt[3]{\pi} \cdot {\left(\sqrt[3]{\pi}\right)}^{2}}{180}\right)\right)}^{2}
\end{array}
Alternatives Alternative 1 Accuracy 78.9% Cost 71872
\[\begin{array}{l}
t_0 := \sin \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\\
a \cdot \left(t_0 \cdot \left(t_0 \cdot a\right)\right) + {\left(b \cdot \cos \left(angle \cdot \frac{\sqrt[3]{\pi} \cdot {\left(\sqrt[3]{\pi}\right)}^{2}}{180}\right)\right)}^{2}
\end{array}
\]
Alternative 2 Accuracy 78.9% Cost 33024
\[\begin{array}{l}
t_0 := \sin \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\\
a \cdot \left(t_0 \cdot \left(t_0 \cdot a\right)\right) + {b}^{2}
\end{array}
\]
Alternative 3 Accuracy 79.8% Cost 26240
\[{b}^{2} + {\left(a \cdot \sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)}^{2}
\]
Alternative 4 Accuracy 79.8% Cost 26240
\[{b}^{2} + {\left(a \cdot \sin \left(angle \cdot \frac{\pi}{180}\right)\right)}^{2}
\]
Alternative 5 Accuracy 74.5% Cost 19840
\[{b}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\left(angle \cdot \left(\pi \cdot a\right)\right)}^{2}
\]
Alternative 6 Accuracy 74.6% Cost 19840
\[{b}^{2} + {\left(\pi \cdot \left(angle \cdot a\right)\right)}^{2} \cdot 3.08641975308642 \cdot 10^{-5}
\]
Alternative 7 Accuracy 74.6% Cost 19840
\[{b}^{2} + {\left(0.005555555555555556 \cdot \left(angle \cdot \left(\pi \cdot a\right)\right)\right)}^{2}
\]
Alternative 8 Accuracy 74.6% Cost 19840
\[{b}^{2} + {\left(0.005555555555555556 \cdot \left(\pi \cdot \left(angle \cdot a\right)\right)\right)}^{2}
\]