Timeout for ab-angle->ABCF A

Specification

Math

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

FPCore

(FPCore (a b angle)
  :precision binary64
  :pre TRUE
  (let* ((t_0 (* (/ angle 180.0) PI)))
  (+ (pow (* a (sin t_0)) 2.0) (pow (* b (cos t_0)) 2.0))))

C

double code(double a, double b, double angle) {
	double t_0 = (angle / 180.0) * ((double) M_PI);
	return pow((a * sin(t_0)), 2.0) + pow((b * cos(t_0)), 2.0);
}

Java

public static double code(double a, double b, double angle) {
	double t_0 = (angle / 180.0) * Math.PI;
	return Math.pow((a * Math.sin(t_0)), 2.0) + Math.pow((b * Math.cos(t_0)), 2.0);
}

Python

def code(a, b, angle):
	t_0 = (angle / 180.0) * math.pi
	return math.pow((a * math.sin(t_0)), 2.0) + math.pow((b * math.cos(t_0)), 2.0)

Julia

function code(a, b, angle)
	t_0 = Float64(Float64(angle / 180.0) * pi)
	return Float64((Float64(a * sin(t_0)) ^ 2.0) + (Float64(b * cos(t_0)) ^ 2.0))
end

MATLAB

function tmp = code(a, b, angle)
	t_0 = (angle / 180.0) * pi;
	tmp = ((a * sin(t_0)) ^ 2.0) + ((b * cos(t_0)) ^ 2.0);
end

Wolfram

code[a_, b_, angle_] := Block[{t$95$0 = N[(N[(angle / 180.0), $MachinePrecision] * Pi), $MachinePrecision]}, N[(N[Power[N[(a * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]

ReFlow

f(a, b, angle):
	a in [-inf, +inf],
	b in [-inf, +inf],
	angle in [-inf, +inf]
code: THEORY
BEGIN
f(a, b, angle: real): real =
	LET t_0 = ((angle / (180)) * (4 * atan(1))) IN
	((a * (sin(t_0))) ^ (2)) + ((b * (cos(t_0))) ^ (2))
END code