Summation and Trig met and made this (2)

Let $n-k = r$ . We have to evaluate :

$\displaystyle \sum_{r=0}^{n} {n \choose r} \sin^r \theta \cos^{n-r} \theta \cos\bigg(r \dfrac{\pi}{2}\bigg)$

Now, If $r$ is odd, then $\cos \frac{r \pi}{2}$ is $0$ . And if even, it would be $\cos \frac{r \pi}{2} = i^r$ .

Hence, we have to evaluate:

$\displaystyle \sum_{r \in even,0 \leq r \leq n} {n \choose r} \sin^r \theta \cos^{n-r} \theta i^r$

$\displaystyle \sum_{r \in even,0 \leq r \leq n} {n \choose r} (i\sin \theta)^r \cos^{n-r} \theta$

= $\dfrac{(\cos \theta + i \sin \theta)^n + (\cos \theta - i \sin \theta)^n}{2} = \boxed{\cos n \theta}$