A Trigonometric Sum

It can be shown that the following sum:

[MATH]S=\sum_{k=1}^{89}\left(\sin^6\left(k^{\circ}\right)\right)[/MATH]

is rational. Find the value of $S$.

I first used the co-function identity:

[MATH]\sin\left(90^{\circ}-x\right)=\cos(x)[/MATH]

to express the sum as:

[MATH]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)[/MATH]

Hence:

[MATH]S=\sum_{k=1}^{44}\left(\sin^6\left(k^{\circ}\right)+\cos^6\left(k^{\circ}\right)\right)+\frac{1}{8}[/MATH]

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

[MATH]\sin^6(x)+\cos^6(x)=\sin^4(x)-\sin^2(x)\cos^2(x)+\cos^4(x)[/MATH]

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

[MATH]3\sin^4(x)-3\sin^2(x)+1[/MATH]

Factor and use a Pythagorean identity:

[MATH]1-3\sin^2(x)\cos^2(x)[/MATH]

Apply double-angle identity for cosine:

[MATH]\frac{4-3\left(1-\cos^2(2x)\right)}{4}[/MATH]

Pythagorean identity:

[MATH]\frac{4-3\sin^2(2x)}{4}[/MATH]

Double-angle identity for cosine:

[MATH]\frac{8-3\left(1-\cos(4x)\right)}{8}[/MATH]

[MATH]\frac{3\cos(4x)+5}{8}[/MATH]

Hence, we now have:

[MATH]S=\frac{1}{8}\sum_{k=1}^{44}\left(3\cos \left(4k^{\circ}\right)+5\right)+\frac{1}{8}[/MATH]

[MATH]S=\frac{1}{8} \left(3\sum_{k=1}^{44} \left(3\cos \left(4k^{\circ} \right) \right)+44\cdot5+1 \right)[/MATH]

[MATH]S=\frac{1}{8} \left(3\sum_{k=1}^{44} \left(\cos \left(4k^{\circ} \right) \right)+221 \right)[/MATH]

Now, observe that:

[MATH]\cos\left(180^{\circ}-x\right)=-\cos(x)[/MATH]

And we may write:

[MATH]S=\frac{1}{8} \left(3\sum_{k=1}^{22} \left(\cos \left(4k^{\circ} \right)-\cos \left(4k^{\circ} \right) \right)+221 \right)[/MATH]

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

[MATH]S=\frac{221}{8}[/MATH]