Using the formula for $\sin P-\sin Q$, show that $\sin \left[ \left( 2r+1 ight) heta ight]-\sin

Question

Using the formula for $\sin P-\sin Q$, show that $\sin \left[ \left( 2r+1 ight)	heta ight]-\sin \left[ \left( 2r-1 ight)	heta ight]\equiv 2\cos \left( 2r	heta ight)\sin 	heta $. [1] Given that $\sin 	heta 
e 0$, using the method of differences, show that $\sum\limits_{r=1}^{n}{\cos \left( 2r	heta ight)}=\frac{\sin \left[ \left( 2n+1 ight)	heta ight]-\sin 	heta }{2\sin 	heta }$ . [2] Hence find $\sum\limits_{r=1}^{n}{{{\cos }^{2}}\left( \frac{r\pi }{5} ight)}$ in terms of $n$. Explain why the infinite series ${{\cos }^{2}}\left( \frac{\pi }{5} ight)+{{\cos }^{2}}\left( \frac{2\pi }{5} ight)+{{\cos }^{2}}\left( \frac{3\pi }{5} ight)+...$ is divergent. [3]

Accepted Answer

$\displaystyle\sum_{r=1}^{n}\cos(r\theta) = \frac{\sin[(2n+1)\theta] - \sin\theta}{2\sin\theta}$; Series is divergent\)

2017 JJC P1 Q3

2017 JJC P1 Q3

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