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)

$(1+\cos \frac{\pi }{6})\left(1+\cos \frac{\pi }{3}\right)\left(1+\cos \frac{2\pi }{3}\right)\left(1+\cos \frac{7\pi }{6}\right)$
$=\left(1+\frac{\sqrt{3}}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{2}\right)\left(1-\frac{\sqrt{3}}{2}\right)$
$=\left(1-\frac{3}{4}\right)\left(1-\frac{1}{4}\right)$
$=\frac{1}{4}\times \frac{3}{4}=\frac{3}{16}$