Let x∈R and let n∈N then evaluate: ∑k=0n(nk)sin⁡(x+kπ2) I could only go up to breaking...

Aubrey Hendricks

Aubrey Hendricks

Answered

2022-01-26

Let xR and let nN then evaluate: k=0n(nk)sin(x+kπ2) I could only go up to breaking this sum in two parts of sinx and cosx in the following way: k=0n(nk)sin(x+kπ2)=k=0|n/2|(1)k(n2k)sinx+k=0|n/2|(1)k(n2k+1)cosx I have the intuition that sin(x+y)=sinxcosy+sinycosx should be used at some point.

Answer & Explanation

Damian Roberts

Damian Roberts

Expert

2022-01-27Added 14 answers

sin(x+kπ2) is the imaginary part of
ei(x+kπ2)=eix(eiπ2)k=eixik
using Euler's formula: eiφ=cos(φ)+isin(φ)
k=0nsin(x+kπ2)
Finally, 1+i=2(cosπ4+isinπ4)
(1+i)n=2n2(cosnπ4+isinnπ4) by Proof for de Moivre's Formula
search633504

search633504

Expert

2022-01-28Added 16 answers

k=0n(nk)sin(x+kπ2) is the imaginary part of k=0n(nk)ei(x+kπ2)=eixk=0n(nk)eikπ2 and k=0n(nk)eikπ2 is the binomial expansion of (1+eiπ2)n=(1+i)n

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