Question

Assertion :$\cos \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}=-\frac{1}{8}$ Reason: $\cos \theta \cos 2\theta \cos 4\theta ...\cos 2^{n-1}\theta =-\frac{1}{2^n}$ if $\theta =\frac{\pi }{2^n-1}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Assertion:

$\cos \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}=\frac{1}{2\sin \frac{\pi }{7}}\left(2\sin \frac{\pi }{7}\cos \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\right)$

$=\frac{1}{8}\frac{\sin \frac{8\pi }{7}}{\sin \frac{\pi }{7}}=\frac{1}{8}\frac{\sin (\pi +\frac{\pi }{7})}{\sin \frac{\pi }{7}}=-\frac{1}{8}$

Reason:

$\theta =\frac{\pi }{2^n-1}\Rightarrow 2^n\theta =\pi +\theta$

$\therefore \sin \left(2^n\theta \right)=-\sin \theta$

$\cos \theta \cos 2\theta ...\cos 2^{n-1}\theta =\frac{1}{2}\frac{(\sin \theta \cos \theta \cos 2\theta ...\cos 2^{n-1}\theta )}{\sin \theta }$

$=\frac{1}{2^n}\frac{\sin \left(2^n\theta \right)}{\sin \theta }=-\frac{1}{2^n}\frac{\sin \theta }{\sin \theta }=-\frac{1}{2^n}$