Question

The value of $\cos \left(\frac{\pi }{15}\right)·\cos \left(\frac{2\pi }{15}\right)·\cos \left(\frac{4\pi }{15}\right)·\cos \left(\frac{8\pi }{15}\right)$ is

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

Answer 1

The correct option is B

$-\frac{1}{16}$

Explanation for correct option

Given trigonometric function is $\cos \left(\frac{\pi }{15}\right)·\cos \left(\frac{2\pi }{15}\right)·\cos \left(\frac{4\pi }{15}\right)·\cos \left(\frac{8\pi }{15}\right)$

$$\begin{gathered} \cos \left(\frac{\pi }{15}\right)·\cos \left(\frac{2\pi }{15}\right)·\cos \left(\frac{4\pi }{15}\right)·\cos \left(\frac{8\pi }{15}\right) \\ =\left[\cos \left(\frac{\pi }{15}\right)·\cos \left(\frac{4\pi }{15}\right)\right]·\left[\cos \left(\frac{2\pi }{15}\right)·\cos \left(\frac{8\pi }{15}\right)\right] \\ =\frac{1}{4}\left[2\cos \left(\frac{\pi }{15}\right)·\cos \left(\frac{4\pi }{15}\right)\right]·\left[2\cos \left(\frac{2\pi }{15}\right)·\cos \left(\frac{8\pi }{15}\right)\right] \\ =\frac{1}{4}\left[\cos \left(\frac{4\pi }{15}+\frac{\pi }{15}\right)+\cos \left(\frac{4\pi }{15}-\frac{\pi }{15}\right)\right]·\left[\cos \left(\frac{8\pi }{15}+\frac{2\pi }{15}\right)+\cos \left(\frac{8\pi }{15}-\frac{2\pi }{15}\right)\right]\left[\because 2\cos \left(a\right)\cos \left(b\right)=\cos \left(b+a\right)+\cos \left(b-a\right)\right] \\ =\frac{1}{4}\left[\cos \left(\frac{\pi }{3}\right)+\cos \left(\frac{\pi }{5}\right)\right]·\left[\cos \left(\frac{2\pi }{3}\right)+\cos \left(\frac{2\pi }{5}\right)\right] \\ =\frac{1}{4}\left[\frac{1}{2}+\frac{\sqrt{5}+1}{4}\right]·\left[-\frac{1}{2}+\frac{\sqrt{5}-1}{4}\right]\left[\cos \left(\frac{\pi }{3}\right)=\frac{1}{2},\cos \left(\frac{2\pi }{3}\right)=-\frac{1}{2},\cos \left(\frac{2\pi }{5}\right)=\frac{\sqrt{5}-1}{4}\mathrm{and}\cos \left(\frac{\pi }{5}\right)=\frac{\sqrt{5}+1}{4}\right] \\ =\frac{1}{4}\left[\frac{\sqrt{5}+3}{4}\right]·\left[\frac{\sqrt{5}-3}{4}\right] \\ =\frac{1}{4}\left[\frac{5-9}{16}\right]\left[\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right] \\ =-\frac{1}{16} \end{gathered}$$

Hence, the correct option is $o\mathrm{ption}\left(\mathrm{B}\right)$