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If $(cos\alpha+isin\alpha)(cos2\alpha+isin2\alpha)....(cosn\alpha+isinn\alpha)$ = 1, then $\alpha$ = ?

$\begin{array}{1 1}(A)4n \pi \\ (B) \large\frac{2n\pi}{n(n+1)} \\ (C) \large\frac{4n \pi}{n(n+1)} \\(D) \large\frac{n \pi}{n(n+1)} \end{array}$

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Toolbox:
  • $cos\;x=0\Rightarrow\:x=2n\pi$
  • $1+2+3+...........n=\large\frac{n(n+1)}{2}$
Given:$(cos\alpha+isin\alpha)(cos\;2\alpha+isin\;2\alpha).....(cos\;n\alpha+isin\;n\alpha)=1$
$\Rightarrow\:cos(\alpha+2\alpha+.......n\alpha)+isin(\alpha+2\alpha+....n\alpha)=1$
$\Rightarrow\:\alpha(1+2+........n)=2n\pi$
$\Rightarrow\:\alpha\:\large\frac{n(n+1)}{2}$$=2n\pi$
$\Rightarrow\:\alpha=\large\frac{4n\pi}{n(n+1)}$
answered Aug 2, 2013 by rvidyagovindarajan_1
 

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