Math FPCore C Java Julia Wolfram TeX \[{\left(a \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} + {\left(b \cdot \cos \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2}
\]
↓
\[\begin{array}{l}
t_0 := \sqrt[3]{\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}}\\
t_1 := {t_0}^{4}\\
t_2 := \frac{angle}{180} \cdot \pi\\
{\left(a \cdot \sin t_2\right)}^{2} + {\left(b \cdot \cos \left(\left(\left(t_1 \cdot \sqrt[3]{t_1}\right) \cdot {\left(\sqrt[3]{t_0}\right)}^{2}\right) \cdot \sqrt[3]{t_2}\right)\right)}^{2}
\end{array}
\]
(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 (cbrt (cbrt (* PI (* angle 0.005555555555555556)))))
(t_1 (pow t_0 4.0))
(t_2 (* (/ angle 180.0) PI)))
(+
(pow (* a (sin t_2)) 2.0)
(pow
(* b (cos (* (* (* t_1 (cbrt t_1)) (pow (cbrt t_0) 2.0)) (cbrt t_2))))
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 = cbrt(cbrt((((double) M_PI) * (angle * 0.005555555555555556))));
double t_1 = pow(t_0, 4.0);
double t_2 = (angle / 180.0) * ((double) M_PI);
return pow((a * sin(t_2)), 2.0) + pow((b * cos((((t_1 * cbrt(t_1)) * pow(cbrt(t_0), 2.0)) * cbrt(t_2)))), 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.cbrt(Math.cbrt((Math.PI * (angle * 0.005555555555555556))));
double t_1 = Math.pow(t_0, 4.0);
double t_2 = (angle / 180.0) * Math.PI;
return Math.pow((a * Math.sin(t_2)), 2.0) + Math.pow((b * Math.cos((((t_1 * Math.cbrt(t_1)) * Math.pow(Math.cbrt(t_0), 2.0)) * Math.cbrt(t_2)))), 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 = cbrt(cbrt(Float64(pi * Float64(angle * 0.005555555555555556))))
t_1 = t_0 ^ 4.0
t_2 = Float64(Float64(angle / 180.0) * pi)
return Float64((Float64(a * sin(t_2)) ^ 2.0) + (Float64(b * cos(Float64(Float64(Float64(t_1 * cbrt(t_1)) * (cbrt(t_0) ^ 2.0)) * cbrt(t_2)))) ^ 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[Power[N[Power[N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision], 1/3], $MachinePrecision]}, Block[{t$95$1 = N[Power[t$95$0, 4.0], $MachinePrecision]}, Block[{t$95$2 = N[(N[(angle / 180.0), $MachinePrecision] * Pi), $MachinePrecision]}, N[(N[Power[N[(a * N[Sin[t$95$2], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Cos[N[(N[(N[(t$95$1 * N[Power[t$95$1, 1/3], $MachinePrecision]), $MachinePrecision] * N[Power[N[Power[t$95$0, 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] * N[Power[t$95$2, 1/3], $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 := \sqrt[3]{\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}}\\
t_1 := {t_0}^{4}\\
t_2 := \frac{angle}{180} \cdot \pi\\
{\left(a \cdot \sin t_2\right)}^{2} + {\left(b \cdot \cos \left(\left(\left(t_1 \cdot \sqrt[3]{t_1}\right) \cdot {\left(\sqrt[3]{t_0}\right)}^{2}\right) \cdot \sqrt[3]{t_2}\right)\right)}^{2}
\end{array}
Alternatives Alternative 1 Accuracy 79.8% Cost 136512
\[\begin{array}{l}
t_0 := \sqrt[3]{\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}}\\
t_1 := {t_0}^{4}\\
t_2 := \frac{angle}{180} \cdot \pi\\
{\left(a \cdot \sin t_2\right)}^{2} + {\left(b \cdot \cos \left(\left(\left(t_1 \cdot \sqrt[3]{t_1}\right) \cdot {\left(\sqrt[3]{t_0}\right)}^{2}\right) \cdot \sqrt[3]{t_2}\right)\right)}^{2}
\end{array}
\]
Alternative 2 Accuracy 79.8% Cost 91072
\[\begin{array}{l}
t_0 := \frac{angle}{180} \cdot \pi\\
{\left(a \cdot \sin t_0\right)}^{2} + {\left(b \cdot \cos \left(\sqrt[3]{t_0} \cdot {\left({\left(\sqrt[3]{\sqrt[3]{\frac{angle}{180} \cdot \sqrt[3]{{\pi}^{3}}}}\right)}^{2}\right)}^{3}\right)\right)}^{2}
\end{array}
\]
Alternative 3 Accuracy 79.8% Cost 78208
\[\begin{array}{l}
t_0 := \frac{angle}{180} \cdot \pi\\
t_1 := \sqrt[3]{t_0}\\
{\left(a \cdot \sin t_0\right)}^{2} + {\left(b \cdot \cos \left(t_1 \cdot {\left({\left(\sqrt[3]{t_1}\right)}^{2}\right)}^{3}\right)\right)}^{2}
\end{array}
\]
Alternative 4 Accuracy 79.8% Cost 78144
\[\begin{array}{l}
t_0 := \frac{angle}{180} \cdot \pi\\
{\left(a \cdot \sin t_0\right)}^{2} + {\left(b \cdot \cos \left(\sqrt[3]{t_0} \cdot e^{\log \left({\left(\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right)}^{2}\right)}\right)\right)}^{2}
\end{array}
\]
Alternative 5 Accuracy 79.8% Cost 52224
\[{\left(a \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} + {\left(b \cdot \cos \left({\left(\sqrt[3]{angle \cdot \left(\pi \cdot 0.005555555555555556\right)}\right)}^{3}\right)\right)}^{2}
\]
Alternative 6 Accuracy 79.8% Cost 52160
\[{\left(a \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} + {\left(b \cdot \log \left(e^{\cos \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)}\right)\right)}^{2}
\]
Alternative 7 Accuracy 79.9% Cost 52160
\[{\left(a \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} + {\left(b \cdot \mathsf{log1p}\left(\mathsf{expm1}\left(\cos \left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)\right)\right)\right)}^{2}
\]
Alternative 8 Accuracy 79.8% Cost 39360
\[\begin{array}{l}
t_0 := \frac{angle}{180} \cdot \pi\\
{\left(a \cdot \sin t_0\right)}^{2} + {\left(b \cdot \cos t_0\right)}^{2}
\end{array}
\]
Alternative 9 Accuracy 79.8% Cost 39360
\[{\left(a \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} + {\left(b \cdot \cos \left(\frac{\pi}{\frac{180}{angle}}\right)\right)}^{2}
\]
Alternative 10 Accuracy 79.7% Cost 26240
\[{b}^{2} + {\left(a \cdot \sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)}^{2}
\]
Alternative 11 Accuracy 79.8% Cost 26240
\[{\left(a \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} + {b}^{2}
\]
Alternative 12 Accuracy 77.1% Cost 20488
\[\begin{array}{l}
t_0 := \pi \cdot \left(a \cdot angle\right)\\
\mathbf{if}\;a \leq -3.8 \cdot 10^{-32}:\\
\;\;\;\;{b}^{2} + {\left(0.005555555555555556 \cdot \left(angle \cdot \left(a \cdot \pi\right)\right)\right)}^{2}\\
\mathbf{elif}\;a \leq 5.5 \cdot 10^{-87}:\\
\;\;\;\;{b}^{2} + {\left(a \cdot 0\right)}^{2}\\
\mathbf{else}:\\
\;\;\;\;{b}^{2} + t_0 \cdot \left(0.005555555555555556 \cdot \left(0.005555555555555556 \cdot t_0\right)\right)\\
\end{array}
\]
Alternative 13 Accuracy 77.0% Cost 20105
\[\begin{array}{l}
\mathbf{if}\;a \leq -2.45 \cdot 10^{-32} \lor \neg \left(a \leq 1.26 \cdot 10^{-86}\right):\\
\;\;\;\;{b}^{2} + {\left(\pi \cdot \left(a \cdot angle\right)\right)}^{2} \cdot 3.08641975308642 \cdot 10^{-5}\\
\mathbf{else}:\\
\;\;\;\;{b}^{2} + {\left(a \cdot 0\right)}^{2}\\
\end{array}
\]
Alternative 14 Accuracy 77.1% Cost 20104
\[\begin{array}{l}
\mathbf{if}\;a \leq -1.45 \cdot 10^{-32}:\\
\;\;\;\;{b}^{2} + {\left(0.005555555555555556 \cdot \left(angle \cdot \left(a \cdot \pi\right)\right)\right)}^{2}\\
\mathbf{elif}\;a \leq 2.9 \cdot 10^{-87}:\\
\;\;\;\;{b}^{2} + {\left(a \cdot 0\right)}^{2}\\
\mathbf{else}:\\
\;\;\;\;{b}^{2} + {\left(\pi \cdot \left(a \cdot angle\right)\right)}^{2} \cdot 3.08641975308642 \cdot 10^{-5}\\
\end{array}
\]
Alternative 15 Accuracy 77.1% Cost 20104
\[\begin{array}{l}
\mathbf{if}\;a \leq -8.5 \cdot 10^{-33}:\\
\;\;\;\;{b}^{2} + {\left(0.005555555555555556 \cdot \left(angle \cdot \left(a \cdot \pi\right)\right)\right)}^{2}\\
\mathbf{elif}\;a \leq 1.45 \cdot 10^{-86}:\\
\;\;\;\;{b}^{2} + {\left(a \cdot 0\right)}^{2}\\
\mathbf{else}:\\
\;\;\;\;{b}^{2} + {\left(a \cdot \frac{angle}{\frac{180}{\pi}}\right)}^{2}\\
\end{array}
\]
Alternative 16 Accuracy 56.1% Cost 13248
\[{b}^{2} + {\left(a \cdot 0\right)}^{2}
\]