Question

$\cos \frac{\pi }{6}\cos \frac{\pi }{3}\cos \frac{4\pi }{6}\cos \frac{8\pi }{6}\cos \frac{16\pi }{6}\cos \frac{32\pi }{6}=$

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

Answer 1

The correct option is B $\frac{\sqrt{3}}{64}$

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 $\sin 2\theta =2\sin \theta \cos \theta$

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

$=\frac{\sin 2^6\frac{\pi }{6}}{2^6\sin \frac{\pi }{6}}$

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

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

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