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SolveARationalFunctionOfASineOverASymbolicQuadratic​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable,​System.​Boolean)

 Method (no overloads)

Summary

A rational function of a sine, or of a cosine, over a quadratic in it with a symbol
among the coefficients: sin(x)/(a + b sin(x) + c sin(x)^2), and the same with
the other function to even powers, a cosecant or a secant beside. Written in
s = sin(x) it is a rational function of s, divided down and split over
the written factors as any is, and the piece over the quadratic is taken by the two
roots: with q = sqrt(b^2 - 4 a c) and r = (-b ± q)/(2c),
(d + e s)/(a + b s + c s^2) is (d + e r_1)/(q (s - r_1)) - (d + e r_2)/(q (s - r_2)),
and 1/(sin(x) - r) is -2 atan((r tan(x/2) - 1)/sqrt(r^2 - 1))/sqrt(r^2 - 1) for any complex r but ±1; 1/(cos(x) - r) is
-2 atan(tan(x/2)/σ)/((1 + r) σ) with σ = sqrt((r - 1)/(r + 1)).

Remarks

Under the half-angle the quadratic in the sine is a quartic in t with a symbol
in every coefficient, which nothing factors, and the substitution u = sin(x) wants a cosine above the bar that an even power does not give; Rubi's
trig^m (a + b sin^n + c sin^(2n))^p files lost every even numerator to that.
The two closed forms are each checked by differentiation in the summary of the
biquadratic rule's kind: d/dx atan(u/s)/s is u'/(s^2 + u^2) whatever
s is, so neither asks the sign of a root, where a piecewise on it, as the
quadratic rule writes for a symbolic root, has no value for the conjugate pair the
ordinary case gives. On each interval between the poles of tan(x/2), the
standing property of the half-angle.
Exact wherever the two roots differ, the generic case, with a root at ±1 taken as the square it makes under the half-angle; a discriminant that is zero as
written declines.
https://github.com/asc-community/AngouriMath/issues/718

























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