2

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?

A Piercing Arrow
  • 953
  • 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 06 '17 at 02:24
  • Use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html and https://math.stackexchange.com/questions/464031/find-the-sum-frac1-cos0-circ-cos1-circ-frac1-cos1-circ-cos2-cir and https://math.stackexchange.com/questions/425966/finite-series-reciprocals-of-sines – lab bhattacharjee Dec 06 '17 at 04:30
  • @ 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 06 '17 at 05:39
  • @Skupp, Are you sure that the numerator is not $$\sin(4k^\circ)\sin(1^\circ)?$$ – lab bhattacharjee Dec 06 '17 at 09:26
  • No, the numerator is $\sin{2k}\sin{1}$ – A Piercing Arrow Dec 06 '17 at 21:52

0 Answers0