Computing $\sum_{k=0}^n{2n\choose 2k}(-1)^k\sin^{2k}\theta\cos^{2n-2k}\theta$ using Euler's formula

Question

Compute the following sum by using Euler's formula, $ e^{i \theta} = \cos \theta + i \sin \theta$, $$\cos^{2n}\theta-{2n\choose 2}\cos^{2n-2}\theta\sin^2\theta+...+(-1)^{n-1}{2n\choose 2n-2}\cos^2\theta\sin^{2n-2}\theta+(-1)^n\sin^{2n}\theta$$

I have tried to rewrite the expression as:

$$\sum_{k=0}^n{2n\choose 2k}(-1)^k\sin^{2k}\theta\cos^{2n-2k}\theta$$

But I have no certain idea about how to continue. Could you give me some hints? Thanks!

Answer 1

$(-1)^k\sin^{2k}t=(i\sin t)^{2k}$

$$(\cos t+i\sin t)^{2n}+(\cos t-i\sin t)^{2n}=?$$