# Sum of terms. Trigonometry involved

Given that $$-4 \sum_{k=1}^{1009} \frac{\sin (2k^\circ)\sin 1^\circ}{\cos (4k^\circ)+\cos 2^\circ}$$can be written in the form $\dfrac 1{\cos 1^\circ}-\dfrac 1{\cos n^\circ},$ where $n \in \mathbb Z^+,$ find the minimum value of $n$.

Source: Friend of mine

I thought of listing terms out, but that got me nowhere. There were too many terms. Additionally, the solution that he gave was very strange (there were a lot of spaces in his computation). Can someone guide me through the solution?

• Not sure this helps, but note that $\sum_{k=180n+1}^{180(n+1)}\frac{\sin (2k^\circ)\sin 1^\circ}{\cos (4k^\circ)+\cos 2^\circ} = 0$ for any integer $n$, so your sum is the same as $-4\sum_{k=901}^{1009} \frac{\sin (2k^\circ)\sin 1^\circ}{\cos (4k^\circ)+\cos 2^\circ}$. – rogerl Dec 6 '17 at 2:24
• – lab bhattacharjee Dec 6 '17 at 4:30
• @ lab bhattacharjee I was unable to find any use of the links. They showed me methods that I will keep in mind. @rogerl. Do I keep substituting values? – A Piercing Arrow Dec 6 '17 at 5:39
• @Skupp, Are you sure that the numerator is not $$\sin(4k^\circ)\sin(1^\circ)?$$ – lab bhattacharjee Dec 6 '17 at 9:26
• No, the numerator is $\sin{2k}\sin{1}$ – A Piercing Arrow Dec 6 '17 at 21:52