2
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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?

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