Sixth power trig series!

$S=\sin^61^\circ+\sin^62^\circ+\sin^63^\circ+\cdots+\sin^689^\circ$

Using the fact that $\sin^6\theta=\cos^6(90-\theta)$ , $\begin{aligned}S &=\sin^61^\circ+\sin^62^\circ+\sin^63^\circ+\cdots+\sin^645^\circ+\cos^644^\circ+\cos^643^\circ+\cos^642^\circ+\cdots+\cos^61^\circ \\ \\ &=(\sin^61^\circ+\cos^61^\circ)+(\sin^62^\circ+\cos^62^\circ)+(\sin^63^\circ+\cos^63^\circ)+\cdots+(\sin^644^\circ+\cos^644^\circ)+\sin^645^\circ\end{aligned}$

Using the factorization $a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)((a+b)^2-3ab)$ and substituting $a=\sin^2\theta, b=\cos^2\theta$ we get, $\sin^6\theta+\cos^6\theta=(\sin^2\theta+\cos^2\theta)\left((\sin^2\theta+\cos^2\theta)^2-3(\sin^2\theta)(\cos^2\theta)\right)$ Since $\sin^2\theta+\cos^2\theta=1$ , $\sin^6\theta+\cos^6\theta=1-3(\sin^2\theta)(\cos^2\theta)$ Therefore, $\begin{aligned}S &=(\sin^61^\circ+\cos^61^\circ)+(\sin^62^\circ+\cos^62^\circ)+(\sin^63^\circ+\cos^63^\circ)+\cdots +(\sin^644^\circ+\cos^644^\circ)+\sin^645^\circ \\ \\ &=(1-3(\sin^21)(\cos^21))+(1-3(\sin^22)(\cos^22))+(1-3(\sin^23)(\cos^23))+\cdots +(1-3(\sin^244)(\cos^244))+\frac{1}{8} \\ \\ &= 44-\left(3(\sin^21)(\cos^21)+3(\sin^22)(\cos^22)+3(\sin^23)(\cos^23)+\cdots +3(\sin^244)(\cos^244)\right )+\frac{1}{8} \end{aligned}$

Also, $\sin(2\theta)=2\sin\theta \cos\theta$ $\implies \sin^2(2\theta) =4\sin^2\theta \cos^2\theta$ , which gives us

$\begin{aligned} S &=\frac{353}{8}-\frac{3}{4}(\sin^22+\sin^24+\sin^26+\cdots+\sin^288)\\ \\ &=\frac{353}{8}-\frac{3}{4}(\sin^22+\sin^24+\cdots+\sin^244+\cos^244+\cos^242+\cdots+\cos^24+\cos^22) \\ \\ &=\frac{353}{8}-\frac{3}{4}((\sin^22+\cos^22)+(\sin^24+\cos^24)+\cdots+(\sin^244+\cos^244)) \end{aligned}$

Again using the identity $\sin^2\theta+\cos^2\theta=1$ we get,

$S=\frac{353}{8}-\frac{3}{4}(22)=\frac{353}{8}-\frac{132}{8}=\frac{221}{8}$

Therefore, $m=221, n=8$ and $m+n=\boxed{229}$

$\begin{aligned} S & = \sin^6 1^\circ + \sin^6 2^\circ + \sin^6 3^\circ + \cdots + \sin^6 89^\circ \\ & = \sum_{k=1}^{89} \sin^6 k^\circ \\ & = \sum_{k=1}^{44} \sin^6 k^\circ + \sin^6 45^\circ + \sum_{k=46}^{89} \sin^6 k^\circ \\ & = \sum_{k=1}^{44} \sin^6 k^\circ + \frac 18 + \sum_{k=46}^{89} \cos^6 (90-k)^\circ \\ & = \sum_{k=1}^{44} \sin^6 k^\circ + \frac 18 + \sum_{k=1}^{44} \cos^6 k^\circ \\ & = \sum_{k=1}^{44} \sin^6 k^\circ + \frac 18 + \sum_{k=1}^{44} \left(1-\sin^2 k^\circ\right)^3 \\ & = \sum_{k=1}^{44} \sin^6 k^\circ + \frac 18 + \sum_{k=1}^{44} \left(1-3\sin^2 k^\circ + 3\sin^4 k^\circ - \sin^6 k^\circ \right) \\ & = \frac 18 + \sum_{k=1}^{44} 1 - 3 \sum_{k=1}^{44} \sin^2 k^\circ \left(1 - \sin^2 k^\circ \right) \\ & = \frac 18 + 44 - 3 \sum_{k=1}^{44} \sin^2 k^\circ \cos^2 k^\circ \\ & = \frac {353}8 - \frac 34 \sum_{k=1}^{44} \sin^2 2k^\circ \\ & = \frac {353}8 - \frac 34 \left( \sum_{k=1}^{22} \sin^2 2k^\circ + \sum_{k=23}^{44} \cos^2 (90-2k)^\circ \right) \\ & = \frac {353}8 - \frac 34 \left( \sum_{k=1}^{22} \sin^2 2k^\circ + \sum_{k=1}^{22} \cos^2 2k^\circ \right) \\ & = \frac {353}8 - \frac 34 \sum_{k=1}^{22} 1 \\ & = \frac {353}8 - \frac {66}4 \\ & = \frac {221}8 \end{aligned}$

$\implies m+n = 221+8 = \boxed{229}$