Discussion Forum

$\cos \left(\frac{\pi}{7}\right)\cos \left(\frac{2\pi}{7}\right)\cos \left(\frac{4\pi}{7}\right)=$

Answer:

Explanation:

We have

$\cos \left(\frac{\pi}{7}\right)\cos \left(\frac{2\pi}{7}\right)\cos \left(\frac{4\pi}{7}\right)$

 Let A =$\frac{\pi}{7}\Rightarrow \pi=7A$

  = $\cos A \cos 2A \cos 4A= \cos A \cos 2A \cos 2^{2} A$

 Now, we will use the formula

  $\cos A \cos 2A \cos 2^{2} A.......\cos ^{n-1} A= \frac{\sin  2^{n} A}{2^{n} \sin A}$

$\therefore$    $\cos A \cos2A\cos 2^{2}A=\frac{\sin  2^{3} A}{2^{3} \sin A}$

=$\frac{ \sin 8 A}{8 \sin A}= \frac{ \sin(7A+A)}{8 \sin A}= \frac{\sin(\pi+A)}{8 \sin A}$

                                                                      [$\because 7A=\pi]$

 =$\frac{-\sin A}{8 \sin A}$            $[\because \sin(\pi+\theta)=-\sin \theta)$

   = $-\frac{1}{8}$