left( {1 + cos frac{pi }{8}} right),left( {1 | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(c) $\left( {1 + \cos \frac{\pi }{8}} \right)\,\left( {1 + \cos \frac{{3\pi }}{8}} \right)\,\left( {1 + \cos \frac{{5\pi }}{8}} \right)\,\left( {1 + \

Vedclass · Accepted Answer

$\frac{1}{8}$

Vedclass · Answer

(c) $\left( {1 + \cos \frac{\pi }{8}} \right)\,\left( {1 + \cos \frac{{3\pi }}{8}} \right)\,\left( {1 + \cos \frac{{5\pi }}{8}} \right)\,\left( {1 + \cos \frac{{7\pi }}{8}} \right)$

$ = \left( {1 + \cos \frac{\pi }{8} + \cos \frac{{7\pi }}{8} + \cos \frac{\pi }{8}\cos \frac{{7\pi }}{8}} \right)$

$\left( {1 + \cos \frac{{5\pi }}{8} + \cos \frac{{3\pi }}{8} + \cos \frac{{3\pi }}{8}\cos \frac{{5\pi }}{8}} \right)$

$ = \left( {1 + \cos \frac{\pi }{8} - \cos \frac{\pi }{8} + \cos \frac{\pi }{8}\cos \frac{{7\pi }}{8}} \right)$

$\left( {1 + \cos \frac{{5\pi }}{8} - \cos \frac{{5\pi }}{8} + \cos \frac{{3\pi }}{8}\cos \frac{{5\pi }}{8}} \right)$

$ = \left( {1 + \cos \frac{\pi }{8}\cos \frac{{7\pi }}{8}} \right)\,\left( {1 + \cos \frac{{3\pi }}{8}\cos \frac{{5\pi }}{8}} \right)$

$ = \frac{1}{4}\,\,\left( {2 + 2\cos \frac{\pi }{8}\cos \frac{{7\pi }}{8}} \right)\,\,\left( {2 + 2\cos \frac{{3\pi }}{8}\cos \frac{{5\pi }}{8}} \right)$
$ = \frac{1}{4}\left( {2 + \cos \frac{{3\pi }}{4} + \cos \pi } \right)\left( {2 + \cos \frac{\pi }{4} + \cos \pi } \right)$

$ = \frac{1}{4}\,\left( {1 + \cos \frac{{3\pi }}{4}} \right)\,\left( {1 + \cos \frac{\pi }{4}} \right) = \frac{1}{4}\left( {1 - \cos \frac{\pi }{4}} \right)\,\left( {1 + \cos \frac{\pi }{4}} \right)$

$ = \frac{1}{4}\left( {1 - {{\cos }^2}\frac{\pi }{4}} \right) = \frac{1}{4}\left( {1 - \frac{1}{2}} \right) = \frac{1}{8}$.

Aliter : $\left( {1 + \cos \frac{\pi }{8}} \right)\,\left( {1 + \cos \frac{{7\pi }}{8}} \right)\,\left( {1 + \cos \frac{{3\pi }}{8}} \right)\,\left( {1 + \cos \frac{{5\pi }}{8}} \right)$

$ = \left( {1 + \cos \frac{\pi }{8}} \right)\,\left( {1 - \cos \frac{\pi }{8}} \right)\,\left( {1 + \cos \frac{{3\pi }}{8}} \right)\,\left( {1 - \cos \frac{{3\pi }}{8}} \right)$

$ = \left( {1 - {{\cos }^2}\frac{\pi }{8}} \right){\rm{ }}\left( {1 - {{\cos }^2}\frac{{3\pi }}{8}} \right) = {\sin ^2}\frac{\pi }{8}{\sin ^2}\frac{{3\pi }}{8}$

$ = \frac{1}{4}{\left( {2\sin \frac{\pi }{8}.\sin \frac{{3\pi }}{8}} \right)^2}$$ = \frac{1}{4}{\left( {\cos \frac{\pi }{4} - \cos \frac{\pi }{2}} \right)^2} = \frac{1}{8}$.

$\left( {1 + \cos \frac{\pi }{8}} \right)\,\left( {1 + \cos \frac{{3\pi }}{8}} \right)\,\left( {1 + \cos \frac{{5\pi }}{8}} \right)\,\left( {1 + \cos \frac{{7\pi }}{8}} \right) = $

Similar Questions

Prove that $=\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x$

If $\frac{\sqrt{2} \sin \alpha}{\sqrt{1+\cos 2 \alpha}}=\frac{1}{7}$ and $\sqrt{\frac{1-\cos 2 \beta}{2}}=\frac{1}{\sqrt{10}}$ $\alpha, \beta \in\left(0, \frac{\pi}{2}\right),$ then $\tan (\alpha+2 \beta)$ is equal to

Let $0 < x < \frac{\pi }{4}.$ Then $\sec 2x - \tan 2x = $

$\frac{{\sin \theta + \sin 2\theta }}{{1 + \cos \theta + \cos 2\theta }} = $

$2 \sin \left(\frac{\pi}{22}\right) \sin \left(\frac{3 \pi}{22}\right) \sin \left(\frac{5 \pi}{22}\right) \sin \left(\frac{7 \pi}{22}\right) \sin \left(\frac{9 \pi}{22}\right)$ is