Question

$K=\sin \left(\frac{\pi }{18}\right)\sin \left(\frac{5\pi }{18}\right)\sin \left(\frac{7\pi }{18}\right)$. Find the value of K.

• A. $\frac{1}{4}$
• B. $\frac{1}{6}$
• C. $\frac{1}{8}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

$\frac{1}{8}$

$K=\sin \left(\frac{\pi }{18}\right)\sin \left(\frac{5\pi }{18}\right)\sin \left(\frac{7\pi }{18}\right)$
$\Rightarrow K=\frac{\left[2\sin \left(\frac{\pi }{18}\right)\cos \left(\frac{\pi }{18}\right)\right]\left[2\sin \left(\frac{5\pi }{18}\right)\cos \left(\frac{5\pi }{18}\right)\right]\left[2\sin \left(\frac{7\pi }{18}\right)\cos \left(\frac{7\pi }{18}\right)\right]}{8\cos \left(\frac{\pi }{18}\right)\cos \left(\frac{5\pi }{18}\right)\cos \left(\frac{7\pi }{18}\right)}$
$\Rightarrow K=\frac{\sin \left(\frac{\pi }{9}\right)\sin \left(\frac{5\pi }{9}\right)\sin \left(\frac{7\pi }{9}\right)}{8\cos \left(\frac{\pi }{18}\right)\cos \left(\frac{5\pi }{18}\right)\cos \left(\frac{7\pi }{18}\right)}$ ...$\{\because 2\sin \theta \cos \theta =\sin 2\theta \}$
$\Rightarrow K=\frac{\cos (\frac{\pi }{2}-\frac{\pi }{9})\cos (\frac{\pi }{2}-\frac{5\pi }{9})\cos (\frac{\pi }{2}-\frac{7\pi }{9})}{8\cos \left(\frac{\pi }{18}\right)\cos \left(\frac{5\pi }{18}\right)\cos \left(\frac{7\pi }{18}\right)}$
$\Rightarrow K=\frac{\cos \left(\frac{7\pi }{18}\right)\cos \left(\frac{\pi }{18}\right)\cos \left(\frac{5\pi }{18}\right)}{8\cos \left(\frac{\pi }{18}\right)\cos \left(\frac{5\pi }{18}\right)\cos \left(\frac{7\pi }{18}\right)}=\frac{1}{8}$
Hence, option 'C' is correct.