Question

The value of $\cos \frac{\pi }{15}\cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}$ is

• A. $\frac{1}{16}$
• B. $-\frac{1}{16}$
• C. $1$
• D. $0$

Answer 1

Answer: (B)

$-\frac{1}{16}$

$\cos \frac{\pi }{15}.\cos \frac{2\pi }{15}.\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}$
$=\frac{1}{4}\left(2\cos \frac{4\pi }{15}\cos \frac{\pi }{15}\right)\left(2\cos \frac{8\pi }{15}\cos \frac{2\pi }{15}\right)$
$=\frac{1}{4}(\cos {60}^{\circ }+\cos {36}^{\circ })(\cos {120}^{\circ }+\cos {72}^{\circ })$
$=\frac{1}{4}(\frac{1}{2}+\frac{\sqrt{5}+1}{4})(-\frac{1}{2}+\frac{\sqrt{5}-1}{4})$
$(\because \cos {36}^{\circ }=\frac{\sqrt{5}+1}{4}and\cos {72}^{\circ }=\frac{\sqrt{5}-1}{4})$
$=\frac{1}{4}[-\frac{1}{4}+\frac{1}{2}(\frac{\sqrt{5}-1}{4}-\frac{\sqrt{5}+1}{4})+\frac{5-1}{16}]$
$=\frac{1}{4}\left[\frac{1}{2}(-\frac{1}{2})\right]=-\frac{1}{16}$