Question

Prove: $\cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}\cos \frac{16\pi }{15}=\frac{1}{16}$

Answer 1

We know that,

$\sin (a+b)=\sin a\cos b+\sin b\cos a$ (1)

For $a=b$,

$\sin 2a=2\sin a\cos a$ (2)

Let $a=\frac{16\pi }{15}$ (3)

From (2), we have

$\sin 2a=2\sin a\cos a$

$=2\left(\sin a\right)\cos a$

$=2\left(2\sin \frac{a}{2}\cos \frac{a}{2}\right)\cos a$

$=4\sin \frac{a}{2}\cos \frac{a}{2}\cos a$

$=4\left(\sin \frac{a}{2}\right)\cos \frac{a}{2}\cos a$

$=4\left(2\sin \frac{a}{4}\cos \frac{a}{4}\right)\cos \frac{a}{2}\cos a$

$=8\sin \frac{a}{4}\cos \frac{a}{4}\cos \frac{a}{2}\cos a$

$=8\left(\sin \frac{a}{4}\right)\cos \frac{a}{4}\cos \frac{a}{2}\cos a$

$=8\left(2\sin \frac{a}{8}\cos \frac{a}{8}\right)\cos \frac{a}{4}\cos \frac{a}{2}\cos a$

$=16\sin \frac{a}{8}\cos \frac{a}{8}\cos \frac{a}{4}\cos \frac{a}{2}\cos a$ (5)

Since, $a=\frac{16\pi }{15}$

So, $\sin 2a=\sin (2\cdot \frac{16\pi }{15})=\sin \left(\frac{32\pi }{15}\right)=\sin (2\pi +\frac{2\pi }{15})=\sin \frac{2\pi }{15}$ [Since $\sin (2\pi +\theta )=\sin \theta$]

and $\sin \frac{a}{8}=\sin \left(\frac{\frac{16\pi }{15}}{8}\right)=\sin \frac{2\pi }{15}$

On substituting $\sin 2a$ and $\sin \frac{a}{8}$ in (5), we get

$\Rightarrow \sin \frac{2\pi }{15}=16\sin \frac{2\pi }{15}\cos \frac{a}{8}\cos \frac{a}{4}\cos \frac{a}{2}\cos a$

$\Rightarrow 1=16\cos \frac{a}{8}\cos \frac{a}{4}\cos \frac{a}{2}\cos a$

$\Rightarrow \cos \frac{a}{8}\cos \frac{a}{4}\cos \frac{a}{2}\cos a=\frac{1}{16}$

Now substituting $a$ with $\frac{16\pi }{15}$, we get

$\Rightarrow \cos \frac{\frac{16\pi }{15}}{8}\cos \frac{\frac{16\pi }{15}}{4}\cos \frac{\frac{16\pi }{15}}{2}\cos \frac{16\pi }{15}=\frac{1}{16}$

$\Rightarrow \cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}\cos \frac{16\pi }{15}=\frac{1}{16}$ Proved