AngouriMath
SolveARootOfOnePlusASineBesideAnotherRootUnderTheSine(AngouriMath.Entity,AngouriMath.Entity.Variable)
Method (no overloads)
Summary
The root of k (1 +- f(y)) , or of it over something in f(y) , beside another
root of something inf(y) , for f a sine or a cosine, under
u = f(y) : with g the other function, R = sqrt(k (1 + s f)/D) is
K g/(sqrt(D) sqrt(1 - s f)) where K = R sqrt(D) sqrt(1 - s f)/g has the square
k , and so is constant between the zeros of g ; and g dy is
du for the sine and -du for the cosine. What is asked in u is the
rest oversqrt(D(u)) sqrt(1 - s u) .
root of something in
rest over
Remarks
The half angle at which 1 +- f is a square reads only roots of that kind, and
Rubi's 4.2.2.1sqrt(1 - cos(x))/sqrt(a - cos(x)) was declined and
sqrt((1 - cos(x))/(a - cos(x))) ran out of time; under u = cos(x) they are
1/(sqrt(1 + u) sqrt(a - u)) , answered in a tenth of a second. Only for a number
k , and where another root of something in the function with a symbol in it is
there: the rules after it answer the rest, and more shortly.
https://github.com/asc-community/AngouriMath/issues/233
Rubi's 4.2.2.1
there: the rules after it answer the rest, and more shortly.
https://github.com/asc-community/AngouriMath/issues/233
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