I got expressions that look like $$ \frac{1}{2} \mathrm e^{-\mathrm i r x} \left[\mathrm e^{2\mathrm i r x} \left(1 + \mathrm e^{-2\mathrm i x}\right)^r + \left(1 + \mathrm e^{2\mathrm i x}\right)^r\right] $$ and I want Mathematica to return $ 2^r\cos^rx $.
I tried Assumptions -> x > 0 && r > 0 and other things but nothing seems to give me this obvious answer. $x$ is a real number and $r$ is an integer.
A roundabout way at arriving at the result.
expr = 1/2 E^(-I r x) (E^(2 I r x) (1 + E^(-2 I x))^r + (1 + E^(2 I x))^r); Calculating a sequence for integer values of r
seq = Table[expr // FullSimplify, {r, 1, 5}] (* {2 Cos[x], 4 Cos[x]^2, 8 Cos[x]^3, 16 Cos[x]^4, 32 Cos[x]^5} *) Using FindSequenceFunction to generalize from the sequence
expr2 = FindSequenceFunction[seq, r] // Simplify (* 2^r Cos[x]^r *) Verifying that the expressions are equal for integer r
Assuming[Element[r, Integers], expr == expr2 // FullSimplify] (* True *) Another kind of workaround:
expr = 1/2 E^(-I r x) (E^(2I r x) (1+E^(-2I x))^r+(1+E^(2I x))^r); ExpToTrig[Simplify[Expand[expr] /. a_^(p_ q_) b_^p_ -> (a^q b)^p]] 