Question

The value of $2{\cos }^3\frac{\pi }{7}-{\cos }^2\frac{\pi }{7}-\cos \frac{\pi }{7}$ is $=\frac{1}{b}$ Find b

Answer 1

Let $\frac{\pi }{7}=\theta$
Now,

$2{\cos }^3\theta -{\cos }^2\theta -\cos \theta$

$=\cos \theta [2{\cos }^2\theta -1]-{\cos }^2\theta$

$=\cos \theta \cos 2\theta -{\cos }^2\theta$

$=\cos \theta [\cos 2\theta -\cos \theta ]$

$=-2\cos \theta .\sin \frac{3\theta }{2}\sin \frac{\theta }{2}$

$=-2\cos \frac{2\pi }{14}\sin \frac{3\pi }{14}\sin \frac{\pi }{14}$

$=-2\cos \frac{2\pi }{14}\cos \frac{4\pi }{14}\cos \frac{6\pi }{14}$

$=-2\cos \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{3\pi }{7}$

$=2\cos \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}$

$=\frac{\sin 8\pi /7}{4\sin \pi /7}$

$=\frac{\sin (\pi +\left(\pi /7\right))}{4\sin \left(\pi /7\right)}=\frac{1}{4}$
Ans: b = 4