Question

Find the value of cos$\frac{\pi }{6}$.cos$\frac{\pi }{3}$.cos$\frac{4\pi }{6}$.cos$\frac{8\pi }{6}$.cos$\frac{16\pi }{6}$.cos$\frac{32\pi }{6}$

• A. $\frac{\sqrt{3}}{32}$
• B. $\frac{\sqrt{3}}{16}$
• C. $\frac{\sqrt{3}}{8}$
• D. $\frac{\sqrt{3}}{64}$

Answer 1

The correct option is D

$\frac{\sqrt{3}}{64}$

We know cos$\theta$. cos$2\theta$.cos$2^2\theta$.cos$2^3\theta$................cos$2^{n-1}\theta$ = $\frac{sin{2}^n\theta }{2^{n}sin\theta }$

we can easily derive it by multiplying and dividing by 2 sin$\theta$ and then using the formula sin$2\theta$ = 2 sin$\theta$.cos$\theta$

cos$\frac{\pi }{6}$.cos$\frac{2\pi }{6}$.cos$\frac{4\pi }{6}$.cos$\frac{8\pi }{6}$.cos$\frac{16\pi }{6}$.cos$\frac{32\pi }{6}$

= $\frac{sin{2}^6\frac{\pi }{6}}{2^{6}sin\frac{\pi }{6}}$
$=\frac{sin\frac{32\pi }{3}}{2^{6}sin{30}^{\circ }}=\frac{sin(15\times 2\pi \times \frac{2\pi }{3})}{2^6\times \frac{1}{2}}$

= $\frac{sin\frac{2\pi }{3}}{2^5}$ = $\frac{sin(\pi -\frac{\pi }{3})}{32}$ = $\frac{sin\frac{\pi }{3}}{32}$

= $\frac{\sqrt{3}}{2\times 32}$ = $\frac{\sqrt{3}}{64}$