Our class has just started discussing complex numbers and I have a this as a homework.
I am trying to show that:
$$\cos^n\theta = \frac{1}{2^{n-1}}\sum_{r=0}^\frac{n-1}{2}\binom{n}{r}\cos[(n-2r)\theta]$$
for odd positive integers $n$.
We haven't discussed $e^{i\theta} = \cos\theta+i\sin\theta$ yet, so I assume that it is not necessary. I started my solution as follows:
$$\cos^n\theta=\frac{[(\cos\theta+i\sin\theta)+(\cos\theta-i\sin\theta)]^n}{2^n}$$
After using binomial expansion and power laws, I eventually reached this in my solution:
$$\cos^n\theta = \frac{1}{2^n}\sum_{r=0}^n\binom{n}{r}[\cos\{(n-2r)\theta\}+i\sin\{(n-2r)\theta\}]$$
As I plugged in some odd positive integers for $n$, I noticed that the $i\sin\{(n-2r)\theta\}$ terms cancel out, while the $\cos\{(n-2r)\theta\}$ terms are doubled.
Essentially I have found my proof however I am having difficulty expressing this explicitly. How should I go about finishing this proof?