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

 Method (no overloads)

Summary

A sum of a cosine and a sine of one argument turned into one cosine:
a cos(y) + b sin(y) is R cos(u) for R = sqrt(a^2 + b^2) and
u = y - phi, where cos(y) = (a cos(u) - b sin(u))/R and
sin(y) = (b cos(u) + a sin(u))/R. 1/(a cos(x) + b sin(x))^3 is
sec(u)^3/R^3.

Remarks

This is the case SolveByHomogeneousTrigonometricSubstitution(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) cannot take:
a quotient whose degrees in the sine and cosine differ by an odd number, which the
tangent leaves with a root of 1 + t^2. After the rotation the denominator is one
power of one cosine, and what is above it a polynomial in cos(u) and
sin(u) with a/R and b/R for coefficients -- shapes the rules for
powers of the sine and cosine answer in a fraction of a second, where the original was
declined after the substitution search had spent its budget.
 The way back needs no phi, whose quadrant would depend on the signs of
a and b: cos(u) and sin(u) are
(a cos(y) + b sin(y))/R and (a sin(y) - b cos(y))/R, and a bare
u is y less a constant, which belongs to the constant of integration.
An answer holding u in any other way -- tan(u/2), sin(2u) -- is
declined rather than rewritten. R is the real root for real a and
b, not both zero; where a^2 + b^2 is zero -- a cos(y) + i a sin(y),
which is a e^(i y) -- there is nothing to rotate.
https://github.com/asc-community/AngouriMath/issues/718

























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