Sigma in Trigonometry

$\sum_{n=1}^{44} \cos n^\circ = \sum_{n=46}^{89} \sin n^\circ$ $= \sum_{n=1}^{44} \sin (n+45)^\circ$ $= \sum_{n=1}^{44} (\sin n^\circ \cos 45^\circ + \cos n^\circ \sin 45^\circ )$

We know that $\sin 45^\circ=\cos 45^\circ=\frac{1}{\sqrt{2}}$ so plugging that in we get $\sqrt{2}\sum_{n=1}^{44} \cos n^\circ = \sum_{n=1}^{44}( \sin n^\circ + \cos n^\circ )$

or $(\sqrt{2}-1)\sum_{n=1}^{44} \cos n^\circ=\sum_{n=1}^{44} \sin n^\circ$

It follows that the answer is $\sqrt{2}+1$

We make use of the sum to product identities, and group the terms whose arguments sum to $45^\circ$ ; e.g., $\sin(1^\circ)+\sin(44^\circ).$

Beautifully, all of the terms cancel out except for a factor of $\cot(22.5^\circ).$ We can use the double angle identities with tangent to find that $\cot(22.5^\circ) = \sqrt{2}+1,$ so our answer is $241.$

Graph the imaginary numbers $cis1^{\circ}, cis2^{\circ}, ... cis44^{\circ}$ on the complex plane. The real part of their sum is $\sum cos n^{\circ}$ , the imaginary part is $\sum sin n^{\circ}$ .

Add vectorially the pairs $(cis1^{\circ} + cis44^{\circ})$ , $(cis2^{\circ} + cis43^{\circ})$ ... $(cis22^{\circ} + cis23^{\circ})$ . The numbers in each pair are symmetric about $cis22.5^{\circ}$ , so the sum of each pair and hence the sum of all 22 pairs is symmetric about $cis22.5^{\circ}$ .

The desired ratio is $\frac{Re(cis22.5^{\circ})}{Im(cis22.5^{\circ})} = cot22.5^{\circ}$ . By double-angle formula, $tan45^{\circ}= \frac{2tan22.5^{\circ}}{1-tan^2 22.5^{\circ}}$ . Solving, $tan 22.5^{\circ} = \sqrt{2} - 1$ and $cot22.5^{\circ} = (\sqrt{2} - 1)^{-1} = \sqrt{2}+1$ .

I figured we could think of the nemerator as a Riemann sum approximation of the integral from 0 to pi/4 of cosx dx, using 44 subdivisions, and similarly the denominator as an aproximation of the integral of sin xdx from 0 to pi/4. The ratio of these two integrals is 1+sqrt2 : 1. Yeah, we're missing one rectangle in each sum, but as the width of the rectangles was only pi/176 I expected (correctly) that the difference would be negligible.

Interestingly, x is also identical to \int 0^45 cos(z) dz / \int 0^45 sin(z) dz = sqrt(2)/(2-sqrt(2)) = 1+ sqrt(2).