Let x∈R and let n∈N 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)sin⁡x+∑k=0|n/2|(−1)k(n2k+1)cos⁡x I have the intuition that sin⁡(x+y)=sin⁡xcos⁡y+sin⁡ycos⁡x should be used at some point.

Name: Let x∈R and let n∈N 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)sin⁡x+∑k=0|n/2|(−1)k(n2k+1)cos⁡x I have the intuition that sin⁡(x+y)=sin⁡xcos⁡y+sin⁡ycos⁡x should be used at some point.
Uploaded: 2022-02-23T17:22:57+00:00
Description: Let x∈R and let n∈N 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)sin⁡x+∑k=0|n/2|(−1)k(n2k+1)cos⁡x I have the intuition that sin⁡(x+y)=sin⁡xcos⁡y+sin⁡ycos⁡x should be used at some point.

Question

Damian Roberts · Accepted Answer

sin⁡(x+kπ2) is the imaginary part ofei(x+kπ2)=eix(eiπ2)k=eixikusing Euler&#039;s formula: eiφ=cos⁡(φ)+isin⁡(φ)⇒∑k=0nsin⁡(x+kπ2)Finally, 1+i=2(cos⁡π4+isin⁡π4)⇒(1+i)n=2n2(cos⁡nπ4+isin⁡nπ4) by Proof for de Moivre&#039;s Formula

search633504 · Accepted Answer

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