Question

The value of $\sin \left(\frac{\pi }{18}\right)\sin \left(\frac{5\pi }{18}\right)\sin \left(\frac{7\pi }{18}\right)$ is:

• A. $\frac{1}{16}$
• B. $\frac{1}{8}$
• C. $-\frac{1}{8}$
• D. $-1$

Answer 1

The correct option is B $\frac{1}{8}$

$\because \sin \left(\frac{\pi }{18}\right)\sin \left(\frac{5\pi }{18}\right)=\cos (\frac{\pi }{2}-\frac{\pi }{18})\cos (\frac{\pi }{2}-\frac{5\pi }{18})\cos (\frac{\pi }{2}-\frac{7\pi }{18})$
$=\cos \left(\frac{8\pi }{18}\right)\cos \left(\frac{4\pi }{18}\right)\cos \left(\frac{2\pi }{18}\right)=\cos \left(\frac{\pi }{9}\right)\cos n\left(\frac{2\pi }{9}\right)\cos \left(\frac{4\pi }{9}\right)=\frac{1}{2^3}=\frac{1}{8}(\alpha =\frac{\pi }{2^3+1})$.
option B