Six Degrees of Separation

Note first that $\cos(k^{\circ}) = \sin(90^{\circ} - k^{\circ})$ , so

$\displaystyle\sum_{k=1}^{89} \cos^{6}(k^{\circ}) = \sum_{k=1}^{89} \sin^{6}(k^{\circ}) \Longrightarrow \sum_{k= 1}^{89} \cos^{6}(k^{\circ}) = \dfrac{1}{2} \sum_{k=1}^{89} (\sin^{6}(k^{\circ}) + \cos^{6}(k^{\circ})).$

Now in general, (in any measure), we have that $\sin^{6}(x) + \cos^{6}(x) =$

$(\sin^{2}(x) + \cos^{2}(x))(\sin^{4}(x) - \sin^{2}(x)\cos^{2}(x) + \cos^{4}(x)) =$

$(\sin^{2}(x) + \cos^{2}(x))^{2} - 3\sin^{2}(x)\cos^{2}(x) = 1 - \dfrac{3}{4}\sin^{2}(2x).$

Thus $\displaystyle\sum_{k=1}^{89} \cos^{6}(k^{\circ}) = \dfrac{1}{2} \sum_{k=1}^{89} (1 - \dfrac{3}{4}\sin^{2}(2k^{\circ})) = \dfrac{89}{2} - \dfrac{3}{8}\displaystyle\sum_{k=1}^{89} \sin^{2}(2k^{\circ}).$

Now in this last sum we can "pair up" terms such as $\sin^{2}(2^{\circ}) + \sin^{2}(88^{\circ}) = 1$ and $\sin^{2}(92^{\circ}) + \sin^{2}(188^{\circ}) = 1$ . There will be $44$ such pairings, which along with $\sin^{2}(90^{\circ}) = 1$ gives us a value of $45$ . Thus the desired sum is

$\dfrac{89}{2} - \dfrac{3}{8}*45 = \dfrac{221}{8},$ and so $a + b = 221 + 8 = \boxed{229}.$