AngouriMath
SolveByTakingARootOfAPerfectSquareInATrigonometricFunction(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A half-odd power of a perfect square in one trigonometric function of x is that power of the modulus: sqrt(b^2 + 2 a b sin(y) + a^2 sin(y)^2) is
sqrt(a^2) |sin(y) + b/a| , which is sqrt(a^2) sgn(sin(y) + b/a) (sin(y) + b/a) ,
the sign constant between the zeros of the linear in the sine.
the sign constant between the zeros of the linear in the sine.
Remarks
variable: Rubi's 4.7.7 has
were declined or past the budget where with the modulus written the rest is rational in
the function. At the top only, and only where each such root is a factor of the
integrand, so that its sign goes in front; a leading coefficient a number or one of
known sign for a real parameter, whose condition travels with the answer, as that
rule's does.
https://github.com/asc-community/AngouriMath/issues/718
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