# Simplification of a product of trigonometric functions

I have this expression

2 Csc[2 y[x2]] Sech[x[x2]]^2 (Cos[y[x2]]^4 Cosh[x[x2]]^8 Sin[y[x2]]^4)^(1/4)


which is 1. But Mathematica doesn't do the simplification. I know that Mathematica doesn't simplify square roots very well. What would be the conditions I must give?

• Maybe TrigExpand@PowerExpand[expr] – user1066 Mar 1 '18 at 18:22
• or TrigReduce@PowerExpand[expr]. – anderstood Mar 1 '18 at 18:37
• Your assumption is not correct. For example, the expression is -1 for {y[x2]->-π/4, x[x2]->0}. – Carl Woll Mar 1 '18 at 18:40
• The problem is that it is not 1. Sometimes it is -1. This can be easily checked by plotting it. – Alexei Boulbitch May 31 '18 at 8:10

The following works.

FullSimplify[2 Csc[2 y[x2]] Sech[x[x2]]^2 (Cos[y[x2]]^4 Cosh[x[x2]]^8 Sin[y[x2]]^4)^(1/4),
Assumptions -> y[x2] \[Element] Reals && x[x2] \[Element] Reals]

1/Sign[Cos[y[x2]] Sin[y[x2]]]


This gives 1.

Define the expression.

expr = 2 Csc[
2 y[x2]] Sech[
x[x2]]^2 (Cos[y[x2]]^4 Cosh[x[x2]]^8 Sin[y[x2]]^4)^(1/4)


And then

expr // Factor // TrigExpand // PowerExpand

FullSimplify[2 Csc[2 y[x2]] Sech[x[x2]]^2 (Cos[y[x2]]^4 Cosh[x[x2]]^8 Sin[y[x2]]^4)^(1/4),
Element[_,Reals]]
PiecewiseExpand[%,Reals]

1/Sign[Cos[y[x2]] Sin[y[x2]]]

Piecewise[{{-1, Cos[y[x2]]*Sin[y[x2]] < 0}, {1, Cos[y[x2]]*Sin[y[x2]] > 0}}, ComplexInfinity]