Sum of powers of sine 
Find $\displaystyle \sum_{n=1}^{89} \sin^6(n) = \frac{m}{n}$

Let $x = \sin(n)$ and let $y = \cos(n)$. 
Since $\cos(n) = \sin(90 - n)$ it follows that
$= \sin^6(1) +\sin^6(1) + ... + \sin^6(45) + \cos^6(1) + \cos^6(2) + ... + \cos^6(44)$. 
EDIT:
$x^6 + y^6 = (x^2 + y^2)(x^4 - (xy)^2 + y^4)$. 
$(x^2 + y^2)^2 = x^4 + y^4 + 2(xy)^2 \implies x^4 + y^4 - (xy)^2 = (x^2 + y^2)^2 - 3(xy)^2$
Hence,
$x^6 + y^6 = (x^2 + y^2)\bigg((x^2 + y^2)^2 - 3(xy)^2 \bigg)$
Since $x = \sin(n)$ and $y= \cos(n)$, it really follows that:
$x^6 + y^6 = (1)(1 - 3(xy)^2)) = 1 - 3(xy)^2$.
Since $\sin(n)\cos(n) = \frac{\sin(2n)}{2}$, let $t = \frac{\sin(2n)}{2}$
So then,
$x^6 + y^6 = 1 - \frac{3}{4} t^2$, also $\sin^6(45) = \frac{1}{8}$. 
Hence,
$S = \frac{1}{8} + (1 - 3[\sin^2(1)\cos^2(1) + ... + \sin^2(44)\cos^2(44)])$. 
$ = \frac{1}{8} + 1 - \frac{3}{4}[\sin^2(2) + \sin^2(4) + .. + \sin^2(88)]$
$ = \frac{1}{8} + 1 - \frac{3}{4} [ \sin^2(2) + \cos^2(2) + \sin^2(4) + \cos^2(4)+ ... + \sin^2(44) + \cos^2(44)]$
$ = \frac{1}{8} + 1 - \frac{3}{4} [22 ] = \frac{1}{8} + 1 - \frac{33}{2} = \frac{1 + 8 - 132}{8} = \frac{-123}{8}$.
But that is still not right?

Let $$S = \sum^{89}_{n=1}\sin^6(n)\;,$$ as you have used $\sin(90^\circ-n) = \cos (n)$.
So we get $$S = \sum^{89}_{n=1}\sin^6(90^{\circ}-n) = \sum^{89}_{n=1}\cos^6(n)\;,$$ as you have used $\sin(90-n) = \cos (n)$.
Now Add these two equation, We get $$2S = \sum^{89}_{n=1}\left[\sin^6(n)+\cos^{6}(n)\right] = \sum^{89}_{n=1}\left[1-3\sin^2 (n)\cdot \cos^2(n)\right]$$
Above we have uesd the formula If $\bullet\; \sin^2 x+\cos^2 x+(-1) =0\;,$ Then $$\sin^6 x+\cos^6 x-1 = 3\cdot \sin^2 x\cdot \cos^2 x\cdot -1$$
So we get $$\sin^6 x+\cos^6 x= 1-3\sin^2 x \cdot \cos^2 x.$$
So we get $$2S = \sum^{89}_{n=1}1-\frac{3}{4}\sum^{89}_{n=1}\sin^2(2n) = 89-\frac{3}{4}T$$
Where $$T = \sum^{89}_{n=1}\sin^2(2n)=2\sum^{44}_{n=1}\sin^2(n)+1$$
