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  4. The value of cos ( (π /7) ) cos ( (4π /7) ) cos ( (5π /7) ) is equal to

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Q. The value of $ \cos \left( \frac{\pi }{7} \right)\cos \left( \frac{4\pi }{7} \right)\cos \left( \frac{5\pi }{7} \right) $ is equal to

Jharkhand CECEJharkhand CECE 2010

Solution:

We have, $ \cos \frac{\pi }{7}\cos \frac{4\pi }{7}\cos \frac{5\pi }{7} $
$ =\cos \frac{\pi }{7}\cos \frac{4\pi }{7}\cos \left( \pi -\frac{2\pi }{7} \right) $
$ =-\cos \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7} $
$ [\because \,\,\cos (\pi -\theta )=-\cos \theta ] $
$ =\frac{1}{2\sin \frac{\pi }{7}}\left[ 2\sin \frac{\pi }{7}\cos \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7} \right] $
$ =\frac{-1}{2\sin \frac{\pi }{7}}\left[ \sin \frac{2\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7} \right] $
$ \left( \because \,\,2\sin \frac{\theta }{2}\cos \frac{\theta }{2}=\sin \theta \right) $
$ =\frac{-1}{{{2}^{2}}\sin \frac{\pi }{7}}\left[ 2\sin \frac{2\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7} \right] $
$ =\frac{-1}{{{2}^{2}}\sin \frac{\pi }{7}}\left[ \sin \frac{4\pi }{7}\cos \frac{4\pi }{7} \right] $
$ =\frac{-1}{{{2}^{3}}\sin \frac{\pi }{7}}\left[ 2\sin \frac{4\pi }{7}\cos \frac{4\pi }{7} \right] $
$ =\frac{-1}{8\sin \frac{\pi }{7}}\left[ \sin \frac{8\pi }{7} \right] $
$ =\frac{\sin \left( \pi +\frac{\pi }{7} \right)}{8\sin \frac{\pi }{7}} $
$ =\sin \frac{\frac{\pi }{7}}{8\sin \frac{\pi }{7}}=\frac{1}{8} $
$ [\because \,\,\sin (\pi +\theta )=-\sin \theta ] $