De Moivre's Theorem: prove that $\sum_{j=0}^{50}(-1)^j\binom{101}{2j}\cos^{101-2j}(\frac {\pi}{101})\sin^{2j}(\frac {\pi}{101})=-1$

Question

(#):Suppose $\theta$ is a real number, and $z=\cos(\theta)+i\sin(\theta).$ Then for any $m \in \Bbb Z, z^m+\bar z^m=2\cos(m\theta)$ and $z^m-\bar z^m=2i\sin(m\theta).$

(b): Applying statement (#), prove that $\displaystyle\sum_{j=0}^{50}(-1)^j\binom{101}{2j}\cos^{101-2j}\frac\pi{101}\sin^{2j}\frac\pi{101}=-1$

I have done the proof for (#), but don't know how to do (b), can somebody help? thx:)

Answer 1

As$(-1)^j=i^{2j},$

we have, $$\sum_{j=0}^{50}\binom{101}{2j}\left(\cos\dfrac\pi{101}\right)^{101-2j}\left(i\sin\dfrac\pi{101}\right)^{2j}$$

Now use $\displaystyle(a+b)^n+(a-b)^n=2\sum_{r=0}^{2r\le n}\binom n{2r}a^{n-2r}b^{2r}$

Answer 2

Applying Trigonometric formula for ennuplication of circumference arcs. $\cos((2p+1)x)= \displaystyle\sum_{j=0}^{p}(-1)^j\binom{2p+1}{2j}\cos^{2p+1-2j}(x) \sin^{2j}(x)$

Setting $p=50$ and $x=\frac{\pi}{101}$, we get:

$\cos(2p+1)x=\cos \pi=-1$.