A Trigonometric Sum

It can be shown that the following sum:

$\displaystyle S=\sum_{k=1}^{89}\left(\sin^6\left(k^{\circ}\right)\right)$

is rational. Find the value of $S$.

I first used the co-function identity:

$\displaystyle \sin\left(90^{\circ}-x\right)=\cos(x)$

to express the sum as:

$\displaystyle S=\sum_{k=1}^{44} \sin^6\left(k^{\circ}\right)+\sin^6\left(45^{\circ}\right)+\sum_{k=1}^{44} \cos^6\left(k^{\circ}\right)$

Hence:

$\displaystyle S=\sum_{k=1}^{44}\left(\sin^6\left(k^{\circ}\right)+\cos^6\left(k^{\circ}\right)\right)+\frac{1}{8}$

Now consider the following (sum of 2 cubes and a Pythagorean identity):

$\displaystyle \sin^6(x)+\cos^6(x)=\sin^4(x)-\sin^2(x)\cos^2(x)+\cos^4(x)$

Now, if we write everything in terms of sine, we obtain:

$\displaystyle 3\sin^4(x)-3\sin^2(x)+1$

Factor and use a Pythagorean identity:

$\displaystyle 1-3\sin^2(x)\cos^2(x)$

Apply double-angle identity for cosine:

$\displaystyle \frac{4-3\left(1-\cos^2(2x)\right)}{4}$

Pythagorean identity:

$\displaystyle \frac{4-3\sin^2(2x)}{4}$

Double-angle identity for cosine:

$\displaystyle \frac{8-3\left(1-\cos(4x)\right)}{8}$

$\displaystyle \frac{3\cos(4x)+5}{8}$

Hence, we now have:

$\displaystyle S=\frac{1}{8}\sum_{k=1}^{44}\left(3\cos \left(4k^{\circ}\right)+5\right)+\frac{1}{8}$

$\displaystyle S=\frac{1}{8} \left(3\sum_{k=1}^{44} \left(3\cos \left(4k^{\circ} \right) \right)+44\cdot5+1 \right)$

$\displaystyle S=\frac{1}{8} \left(3\sum_{k=1}^{44} \left(\cos \left(4k^{\circ} \right) \right)+221 \right)$

Now, observe that:

$\displaystyle \cos\left(180^{\circ}-x\right)=-\cos(x)$

And we may write:

$\displaystyle S=\frac{1}{8} \left(3\sum_{k=1}^{22} \left(\cos \left(4k^{\circ} \right)-\cos \left(4k^{\circ} \right) \right)+221 \right)$

The sum goes to zero, and we are left with:

$\displaystyle S=\frac{221}{8}$