AngouriMath
SolveByTheHalfAngleWhereOnePlusAHyperbolicCosineIsASquare(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
Fractional powers of a ± a cosh(y) beside anything in the hyperbolic functions
ofy , by the half angle at which they are squares: 1 + cosh(y) is
2 cosh(y/2)^2 and 1 - cosh(y) is -2 sinh(y/2)^2 , so
sqrt(a + a cosh(y)) is sqrt(2a) cosh(y/2) -- no sign, the hyperbolic
cosine being positive -- and(a - a cosh(y))^(3/2) is
(-2a)^(3/2) sgn(sinh(y/2)) sinh(y/2)^3 , the sign a constant between the zeros
of the sine that comes out in front;cosh(y) and sinh(y) beside them are
2 cosh(y/2)^2 - 1 and 2 sinh(y/2) cosh(y/2) , and the question is asked
again inx . SolveByTheHalfAngleWhereOnePlusASineIsASquare(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)'s
identity for the hyperbolic cosine.
of
cosine being positive -- and
of the sine that comes out in front;
again in
identity for the hyperbolic cosine.
Remarks
Rubi's x^2 sqrt(a + a cosh(c + d x)) , x (a + a cosh(x))^(3/2) ,
cosh(x)/sqrt(a - a cosh(x)) and (A + B cosh(x))/(a - a cosh(x))^(5/2) were declined or searches past the budget: no substitution rationalises a root of a
hyperbolic function ofx beside a power of x , and beside the sine and
cosine ofy the root is of a quotient under every one. Exact, for any
a : cosh(y/2)^2 and sinh(y/2)^2 are not negative, so
(a q)^p = a^p q^p for the principal powers whatever a is, with
2a and -2a kept as the one constant they are; whole products 2p only. At the top only, as every rule that writes a sign for a function.
https://github.com/asc-community/AngouriMath/issues/718
hyperbolic function of
cosine of
https://github.com/asc-community/AngouriMath/issues/718
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