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The general value of the real angle $\theta$, which satisfies the equation. $(\cos \theta+i \sin \theta)(\cos 2 \theta+i \sin 2 \theta) \ldots$$(\cos n \theta+t \sin n \theta)=1$ is given by, (assuming $k$ is an integer)
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The correct answer is:
$\frac{4 k \pi}{n(n+1)}$
We have,
$(\cos\theta+i\sin\theta)(cos2\theta+isin2\theta)...(cosn\theta+isinn\theta) = 1$
$\Rightarrow e^{i\theta}.e^{i(2\theta)}.e^{i(3\theta)}. ...e^{i(n\theta)} = 1$
$\Rightarrow e^{i\theta(1+2+3+...+n}=1$
$\Rightarrow e^{\frac{in(n+1)\theta}{2}}=1$
$\Rightarrow cos(\frac{n(n+1)}{2}\theta)+isin(\frac{n(n+1)}{2}\theta = 1 + 0i$
$\Rightarrow cos(\frac{n(n+1)}{2}\theta)=1$
$\Rightarrow \frac{n(n+1)}{2}\theta = 2k\pi$
$\Rightarrow \theta=\frac{4k}{n(n+1)}\pi$
$(\cos\theta+i\sin\theta)(cos2\theta+isin2\theta)...(cosn\theta+isinn\theta) = 1$
$\Rightarrow e^{i\theta}.e^{i(2\theta)}.e^{i(3\theta)}. ...e^{i(n\theta)} = 1$
$\Rightarrow e^{i\theta(1+2+3+...+n}=1$
$\Rightarrow e^{\frac{in(n+1)\theta}{2}}=1$
$\Rightarrow cos(\frac{n(n+1)}{2}\theta)+isin(\frac{n(n+1)}{2}\theta = 1 + 0i$
$\Rightarrow cos(\frac{n(n+1)}{2}\theta)=1$
$\Rightarrow \frac{n(n+1)}{2}\theta = 2k\pi$
$\Rightarrow \theta=\frac{4k}{n(n+1)}\pi$
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