TITLE: Find the values of m and n(Trigononetry in series)
QUESTION [5 upvotes]: $$\sin ^6(1)+\sin ^6(2)+... ...+\sin ^6(89)=\frac{m}{n}$$
Find $\frac{m}{n}$ in its simplest form , and hence find both values. (All angles are in degree)
I've no idea how to start to solve this questions. Need some guidance for it. Thanks in advance.

REPLY [2 votes]: Found a way to solve from another site.
$$S=\frac{1}{8}\left(3\sum_{k=1}^{44}\left(\cos\left(4k\frac{\pi}{180}\right)\right)+221\right)$$
$$\sum_{k=1}^{44} \left( \cos \left(4k \frac{ \pi}{180} \right) \right)= \sum_{k=1}^{22} \left( \cos \left(4k \frac{ \pi}{180} \right) \right)- \sum_{k=1}^{22} \left( \cos \left(4k \frac{ \pi}{180} \right) \right)=0$$
Hence:
$$S=\frac{1}{8}\left(3\cdot0+221\right)=\frac{221}{8}$$