Find the sum of $\sin^2(n)$

Question

I have no clue how to solve this a detailed solution would be great$$\sum_{n=1}^N \sin^2(n)=? $$

Answer 1

NOTE If this is a school problem i'll just give a hint, i'll post a full solution later.
You can try to consider that $\sin(n) = Im(\cos(n)+i*\sin(n))$, with $Im(.)$ the function that takes the imaginary part of a complex number.

If your question is for the sum of $\sin^2(n)$ (as written in the title) remember that:
$\sin^2(x) = \frac{1-\cos(2x)}{2}$ and then you can use the same method as above

Solution below (for $\sin^2$)

$\sum_{n=1}^N \sin^2(n) = \sum_{n=1}^N \frac{1-\cos(2n)}{2}$
And $\sum_{n=1}^N \cos(2n) = Re\Bigl(\sum_{n=1}^N \cos(2n)+i\sin(2n)\Bigl)$ $=\sum_{n=1}^N e^{2in}=\sum_{n=1}^N (e^{2i})^n=e^{2i}\sum_{n=0}^{N-1} (e^{2i})^n$ $=e^{2i}*\frac{1-(e^{2i})^{N+1}}{1-e^{2i}} = e^{2i}*\frac{e^{i(N+1)}(e^{-i(N+1)}-e^{i(N+1)})}{e^i(e^{-i}-e^i)}$ $=e^{iN+2}\frac{\sin(N+1)}{sin(1)}$
Thus $Re\Bigl(\sum_{n=1}^N e^{2in}\Bigl) = \frac{\cos(N+2)\sin(n+1)}{\sin(1)}$
Finally $\sum_{n=1}^N \sin^2(n) = \frac N2-\frac{\cos(N+2)\sin(n+1)}{2\sin(1)}$

Hope i didn't make any mistakes :)