Let x∈R and let n∈N then evaluate: ∑k=0n(nk)sin(x+kπ2) I could only go up to breaking...
Let and let then evaluate:
I could only go up to breaking this sum in two parts of sinx and cosx in the following way:
I have the intuition that should be used at some point.
Answer & Explanation
is the imaginary part of
using Euler's formula:
by Proof for de Moivre's Formula
is the imaginary part of and
is the binomial expansion of