# Finite Series - reciprocals of sines

Find the sum of the finite series $$\sum _{k=1}^{k=89} \frac{1}{\sin(k^{\circ})\sin((k+1)^{\circ})}$$ This problem was asked in a test in my school. The answer seems to be $\dfrac{\cos1^{\circ}}{\sin^21^{\circ}}$ but I do not know how. I have tried reducing it using sum to product formulae and found out the actual value and it agrees well. Haven't been successful in telescoping it.

$$\frac{1}{\sin k^\circ\sin(k+1)^\circ}=\frac1{\sin1^\circ}\frac{\sin (k+1-k)^\circ}{\sin k^\circ\sin(k+1)^\circ}$$ $$=\frac1{\sin1^\circ}\cdot\frac{\cos k^\circ\sin(k+1)^\circ-\sin k^\circ\cos(k+1)^\circ}{\sin k^\circ\sin(k+1)^\circ}=\frac1{\sin1^\circ}\left(\cot k^\circ-\cot(k+1)^\circ\right)$$
• @cmtappu96, observe that only $\cot1^\circ$ surviving the summation to validate the answer – lab bhattacharjee Jun 21 '13 at 5:59
• Thank you. My bad. I was trying to somehow get a factor of $\sin(k+1)-\sin(k)$ in the numerator. – Chaitanya Tappu Jun 21 '13 at 6:16