Aubrey Hendricks

2022-01-26

Let $x\in \mathbb{R}$ and let $n\in \mathbb{N}$ then evaluate: $\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right)\mathrm{sin}\left(x+\frac{k\pi }{2}\right)$ I could only go up to breaking this sum in two parts of sinx and cosx in the following way: $\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right)\mathrm{sin}\left(x+\frac{k\pi }{2}\right)=\sum _{k=0}^{|n/2|}\left(-1{\right)}^{k}\left(\genfrac{}{}{0}{}{n}{2k}\right)\mathrm{sin}x+\sum _{k=0}^{|n/2|}\left(-1{\right)}^{k}\left(\genfrac{}{}{0}{}{n}{2k+1}\right)\mathrm{cos}x$ I have the intuition that $\mathrm{sin}\left(x+y\right)=\mathrm{sin}x\mathrm{cos}y+\mathrm{sin}y\mathrm{cos}x$ should be used at some point.

Damian Roberts

Expert

$\mathrm{sin}\left(x+\frac{k\pi }{2}\right)$ is the imaginary part of
${e}^{i\left(x+\frac{k\pi }{2}\right)}={e}^{ix}{\left({e}^{i\frac{\pi }{2}}\right)}^{k}={e}^{ix}{i}^{k}$
using Euler's formula: ${e}^{i\phi }=\mathrm{cos}\left(\phi \right)+i\mathrm{sin}\left(\phi \right)$
$⇒\sum _{k=0}^{n}\mathrm{sin}\left(x+\frac{k\pi }{2}\right)$
Finally, $1+i=\sqrt{2}\left(\mathrm{cos}\frac{\pi }{4}+i\mathrm{sin}\frac{\pi }{4}\right)$
$⇒{\left(1+i\right)}^{n}={2}^{\frac{n}{2}}\left(\mathrm{cos}\frac{n\pi }{4}+i\mathrm{sin}\frac{n\pi }{4}\right)$ by Proof for de Moivre's Formula

search633504

Expert

$\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right)\mathrm{sin}\left(x+\frac{k\pi }{2}\right)$ is the imaginary part of $\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right){e}^{i\left(x+\frac{k\pi }{2}\right)}={e}^{ix}\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right){e}^{i\frac{k\pi }{2}}$ and $\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right){e}^{i\frac{k\pi }{2}}$ is the binomial expansion of $\left(1+{e}^{i\frac{\pi }{2}}{\right)}^{n}=\left(1+i{\right)}^{n}$