# Trigonmetric sum of inverses

Prove that: $$\sum^{45}_{k=1}\frac{1}{\cos1^\circ-\cos(87+4k)^\circ}=\frac{1}{2\sin 1^\circ}$$

Numerically, this is accurate comparing the lhs and rhs. Some ideas: We can transform the question into proving that:

$$\sum^{45}_{k=1}\frac{1}{\sin(43+2k)^\circ\sin(44+2k)^\circ}=\frac{1}{\sin 1^\circ}$$ Since clearly every term is positive, there is no hope in trying to cancel terms. I have also tried to transform this into a sum of inverses of roots of an equation so that I can use Vieta's formulae, but I have had no success.

• Where did the "$n$" come from? – Edward Jiang Nov 21 '14 at 4:20
• @EdwardJiang Sorry, it was a $1$.Edited! – chubakueno Nov 21 '14 at 4:23

$$\sin1^\circ=\sin[(44+2k)^\circ-(43+2k)^\circ]$$ $$=\sin(44+2k)^\circ\cos(43+2k)^\circ-\cos(44+2k)^\circ\sin(43+2k)^\circ$$
$$\implies\frac{\sin1^\circ}{\sin(44+2k)^\circ\sin(43+2k)^\circ}=\cot(43+2k)^\circ-\cot(44+2k)^\circ$$