trigonometry - Computing $sum_{k=0}^n{2nchoose 2k}(-1)^ksin^{2k}theta cos^{2n-2k}theta$ using Euler's formula

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

I have tried to rewrite the expression as:

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

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