Math FPCore C Julia Wolfram TeX \[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2}
\]
↓
\[\begin{array}{l}
t_0 := \mathsf{fma}\left(\pi \cdot angle, 0.005555555555555556, 1\right)\\
{\left(a \cdot \left(\cos t_0 \cdot \cos 1 + \sin t_0 \cdot \sin 1\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2}
\end{array}
\]
(FPCore (a b angle)
:precision binary64
(+
(pow (* a (cos (* PI (/ angle 180.0)))) 2.0)
(pow (* b (sin (* PI (/ angle 180.0)))) 2.0))) ↓
(FPCore (a b angle)
:precision binary64
(let* ((t_0 (fma (* PI angle) 0.005555555555555556 1.0)))
(+
(pow (* a (+ (* (cos t_0) (cos 1.0)) (* (sin t_0) (sin 1.0)))) 2.0)
(pow (* b (sin (* PI (/ angle 180.0)))) 2.0)))) double code(double a, double b, double angle) {
return pow((a * cos((((double) M_PI) * (angle / 180.0)))), 2.0) + pow((b * sin((((double) M_PI) * (angle / 180.0)))), 2.0);
}
↓
double code(double a, double b, double angle) {
double t_0 = fma((((double) M_PI) * angle), 0.005555555555555556, 1.0);
return pow((a * ((cos(t_0) * cos(1.0)) + (sin(t_0) * sin(1.0)))), 2.0) + pow((b * sin((((double) M_PI) * (angle / 180.0)))), 2.0);
}
function code(a, b, angle)
return Float64((Float64(a * cos(Float64(pi * Float64(angle / 180.0)))) ^ 2.0) + (Float64(b * sin(Float64(pi * Float64(angle / 180.0)))) ^ 2.0))
end
↓
function code(a, b, angle)
t_0 = fma(Float64(pi * angle), 0.005555555555555556, 1.0)
return Float64((Float64(a * Float64(Float64(cos(t_0) * cos(1.0)) + Float64(sin(t_0) * sin(1.0)))) ^ 2.0) + (Float64(b * sin(Float64(pi * Float64(angle / 180.0)))) ^ 2.0))
end
code[a_, b_, angle_] := N[(N[Power[N[(a * N[Cos[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Sin[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]
↓
code[a_, b_, angle_] := Block[{t$95$0 = N[(N[(Pi * angle), $MachinePrecision] * 0.005555555555555556 + 1.0), $MachinePrecision]}, N[(N[Power[N[(a * N[(N[(N[Cos[t$95$0], $MachinePrecision] * N[Cos[1.0], $MachinePrecision]), $MachinePrecision] + N[(N[Sin[t$95$0], $MachinePrecision] * N[Sin[1.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Sin[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]
{\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}
↓
\begin{array}{l}
t_0 := \mathsf{fma}\left(\pi \cdot angle, 0.005555555555555556, 1\right)\\
{\left(a \cdot \left(\cos t_0 \cdot \cos 1 + \sin t_0 \cdot \sin 1\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2}
\end{array}
Alternatives Alternative 1 Accuracy 80.4% Cost 78336
\[\begin{array}{l}
t_0 := \mathsf{fma}\left(\pi \cdot angle, 0.005555555555555556, 1\right)\\
{\left(a \cdot \left(\cos t_0 \cdot \cos 1 + \sin t_0 \cdot \sin 1\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2}
\end{array}
\]
Alternative 2 Accuracy 80.4% Cost 39488
\[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\frac{1}{\frac{180}{\pi \cdot angle}}\right)\right)}^{2}
\]
Alternative 3 Accuracy 80.4% Cost 39360
\[{\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(a \cdot \cos \left(\left(\pi \cdot angle\right) \cdot 0.005555555555555556\right)\right)}^{2}
\]
Alternative 4 Accuracy 80.4% Cost 39360
\[{\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(a \cdot \cos \left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)\right)}^{2}
\]
Alternative 5 Accuracy 80.5% Cost 39360
\[{\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(angle \cdot \frac{\pi}{180}\right)\right)}^{2}
\]
Alternative 6 Accuracy 80.4% Cost 26368
\[{\left(b \cdot \sin \left(\frac{1}{\frac{180}{\pi \cdot angle}}\right)\right)}^{2} + {a}^{2}
\]
Alternative 7 Accuracy 80.5% Cost 26240
\[{\left(b \cdot \sin \left(angle \cdot \frac{\pi}{180}\right)\right)}^{2} + {a}^{2}
\]
Alternative 8 Accuracy 75.5% Cost 19840
\[{a}^{2} + {\left(0.005555555555555556 \cdot \left(angle \cdot \left(\pi \cdot b\right)\right)\right)}^{2}
\]
Alternative 9 Accuracy 75.5% Cost 19840
\[{a}^{2} + {\left(angle \cdot \left(\pi \cdot \left(0.005555555555555556 \cdot b\right)\right)\right)}^{2}
\]
Alternative 10 Accuracy 75.5% Cost 19840
\[{a}^{2} + {\left(b \cdot \left(angle \cdot \frac{\pi}{180}\right)\right)}^{2}
\]
Alternative 11 Accuracy 57.8% Cost 13248
\[{a}^{2} + {\left(b \cdot 0\right)}^{2}
\]