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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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The value of $\sqrt{2+\sqrt{2+\sqrt{2+2\cos 8\theta}}}$ where $\theta < \theta<\large\frac{\pi}{8}$ is equal to

$(a)\;2\cos\theta\qquad(b)\;\cos\theta\qquad(c)\;2\sin\theta\qquad(d)\;-2\cos\theta$

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1 Answer

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We have $\sqrt{2+\sqrt{2+\sqrt{2+2\cos 8\theta}}}$
$1+\cos 8\theta=2\cos^2\large\frac{8\theta}{2}$
$\Rightarrow \sqrt{2+\sqrt{2+\sqrt{4\cos ^24\theta}}}$
$\Rightarrow \sqrt{2+\sqrt{2+2\cos 4\theta}}$
$\Rightarrow \sqrt{2+\sqrt{2(1+\cos 4\theta)}}$
$1+\cos 4\theta=2\cos^22\theta$
$\Rightarrow \sqrt{2+\sqrt{2(2\cos^22\theta)}}$
$\Rightarrow \sqrt{2+2\cos 2\theta)}$
$\Rightarrow \sqrt{2(1+\cos 2\theta)}$
$\Rightarrow \sqrt{2(2\cos^2\theta)}$
$\Rightarrow 2\cos\theta$
Hence (a) is the correct answer.
answered Oct 14, 2013 by sreemathi.v
 

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