Question

The value of $\left(1+cos \frac{\pi}{6}\right)\left(1+cos \frac{\pi}{3}\right)\left(1+cos\frac{2\pi}{3}\right)\left(1+cos\frac{7\pi}{6}\right)$ is

• A. $\frac{3}{16}$
• B. $\frac{3}{8}$
• C. $\frac{3}{4}$
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

$\left(1+cos\frac{\pi}{6}\right)\left(1+cos\frac{\pi}{3}\right)\left(1+cos\frac{2\pi}{3}\right)\left(1+cos\frac{7\pi}{6}\right)$

$=\left(1+\frac{\sqrt{3}}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{2}\right)\left(1-\frac{\sqrt{3}}{2}\right)$

$=\left(1-\frac{3}{4}\right)\left(1-\frac{1}{4}\right)$

$=\frac{1}{4}\times\frac{3}{4}=\frac{3}{16}$