Sinfull!

$\sum_{\theta=1^{\circ}}^{89^{\circ}}\sin^{6}\theta=\sum_{\theta=1^{\circ}}^{44^{\circ}}\sin^{6}\theta+\sin^{6}45^{\circ}+\sum_{\theta=46^{\circ}}^{89^{\circ}}\sin^{6}\theta$

Since $\sin\theta=\cos(90^{\circ}-\theta)$ and $\sin45^{\circ}=\frac{1}{\sqrt{2}}$ , therefore: $\sum_{\theta=1^{\circ}}^{44^{\circ}}\sin^{6}\theta+\sin^{6}45^{\circ}+\sum_{\theta=46^{\circ}}^{89^{\circ}}\sin^{6}\theta=\sum_{\theta=1^{\circ}}^{44^{\circ}}(\sin^{6}\theta+\cos^{6}\theta)+\frac{1}{2^{3}}$ Now, we need to simplify the summation above. Let $s=\sin^{2}\theta$ and $c=\cos^{2}\theta$ . Hence, by identity, $s+c=1$ and $4cs=\sin^{2}2\theta$ . So, $\sum_{\theta=1^{\circ}}^{44^{\circ}}(s^{3}+c^{3})=\sum_{\theta=1^{\circ}}^{44^{\circ}}((s+c)(s^{2}-sc+c^{2})$ $=\sum_{\theta=1^{\circ}}^{44^{\circ}}(1)((s+c)-3sc)$ $= \sum_{\theta=1^{\circ}}^{44^{\circ}}(1-3sc)$ $=\sum_{\theta=1^{\circ}}^{44^{\circ}}(1-\frac{3}{4}(4sc))$ $=\sum_{\theta=1^{\circ}}^{44^{\circ}}(1-\frac{3}{4}\sin^{2}2\theta) \Rightarrow 44-\frac{3}{4}\sum_{\theta=1^{\circ}}^{44^{\circ}}\sin^{2}2\theta$ Applying $\sin\theta=\cos(90^{\circ}-\theta)$ to the summation again yields: $\sum_{\theta=1^{\circ}}^{44^{\circ}}\sin^{2}2\theta=\sum_{\theta=1^{\circ}}^{22^{\circ}}\sin^{2}2\theta+\sum_{\theta=23^{\circ}}^{44^{\circ}}\sin^{2}2\theta$ $=\sum_{\theta=1^{\circ}}^{22^{\circ}}\sin^{2}2\theta+\sum_{\theta=1^{\circ}}^{22^{\circ}}\cos^{2}2\theta \Rightarrow \sum_{\theta=1^{\circ}}^{22^{\circ}}(\sin^{2}2\theta+\cos^{2}2\theta)=22$ With that, we can compute $\displaystyle\sum_{\theta=1^{\circ}}^{89^{\circ}}\sin^{6}\theta$ to be: $\sum_{\theta=1^{\circ}}^{89^{\circ}}\sin^{6}\theta=44-\frac{3}{4}(22)+\frac{1}{2^{3}}=\frac{221}{8}$ The desired answer is $221+8=\boxed{229}$ .