<s>[INST] Find the values of m and n(Trigononetry in series) $$\sin ^6(1)+\sin ^6(2)+... ...+\sin ^6(89)=\frac{m}{n}$$
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.
 [/INST] $$\begin{align} 1 &= (\sin^2 t + \cos^2 t)^3 \\
&= \sin^6 t + 3\sin^4t \cos^2 t + 3 \sin^2t+ \cos^4t + \cos^6 t \\
&= \sin^6 t + \cos^6 t +  3\sin^2t \cos^2 t \\
&=\sin^6 t + \cos^6 t +  \frac34\sin^2 (2t) \\
\end{align} $$
let $$ S= \sin^6 1^\circ + \sin^6 2\circ + \cdots + \sin^6 89^\circ. $$ then we have $$\begin{align}2S &= (\sin^6 1^\circ + \sin^6 89^\circ) + (\sin^6 2^\circ + \sin^6 87^\circ) + \cdots + (\sin^6 89^\circ + \sin^6 1^\circ)\\
&= (\sin^6 1^\circ + \cos^6 1^\circ) + (\sin^6 2^\circ + \cos^6 2^\circ) + \cdots + (\sin^6 89^\circ + \cos^6 89^\circ)\\
&= 89 - \frac34\left(\sin^2 2^\circ + \sin^2 4^\circ + \cdots+\sin^2178^\circ \right)\\
&=89 - \frac 34 - \frac32\left(\sin^2 2^\circ + \sin^2 4^\circ + \cdots+\sin^2 88^\circ \right)\\
&=89 - \frac 34 - \frac34\left((\sin^2 2^\circ + \sin^288^\circ) + (\sin^2 4^\circ + \sin^2 86) + \cdots+(\sin^2 88^\circ + \sin^2 2^\circ \right)\\
&=89 - \frac 34 - \frac34 44\\
\end{align}$$
so $$ S = \frac{221} 8$$
there is an easier way to this. thanks to claude. see the comments below. we can express $\sin^6 x $ as a $\cos$ series in the following way:
$$\sin^6 x = -\frac{15}{32}\cos 2x +\frac 3 {16} \cos 4x - \frac 1 {32} +\cos 6x + \frac 5{16} $$ the $\cos 2x,\cos 6x$ sum to zero and $\cos 4x$ terms sum to $-1$   therefore the
$$ S = -\frac{3}{16} + 89 \times \frac 5{16} = \frac{221}{8} .$$
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