Let $\mathrm{f}\left( r ight)=\frac{\sin \left[ (2r+1) heta ight]}{\cos heta },r\in {{\mathbb

Question

Let $\mathrm{f}\left( r ight)=\frac{\sin \left[ (2r+1)	heta ight]}{\cos 	heta },r\in {{\mathbb{Z}}^{+}}$. Show that $\mathrm{f}\left( r ight)-\mathrm{f}\left( r-1 ight)=A\cos \left( Br	heta ight)	an \left( 	heta ight)$, where $A$ and $B$ are to be determined. By using the result part (i), show that $\cos 2	heta +\cos 4	heta +...+\cos 2N	heta =\frac{1}{2}\left[ \frac{\sin \left[ \left( 2N+1 ight)	heta ight]}{\sin 	heta }-1 ight]$ .

Accepted Answer

$A = 2$, $B = 2$; $\sum_{r=1}^{N}\cos 2r\theta = \dfrac{1}{2}\left[\dfrac{\sin(2N+1)\theta}{\sin\theta} - 1\right]$

SRJC P1 Q7

SRJC P1 Q7

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