Question

Solve : $\cos {12}^{\circ }\cos {24}^{\circ }\cos {36}^{\circ }\cos {48}^{\circ }\cos {60}^{\circ }\cos {72}^{\circ }\cos {96}^{\circ }$

Answer 1

${\cos 12}^0.\cos {24}^0.\cos {36}^0.\cos {48}^0.\cos {60}^0.{\cos 72}^0.\cos {96}^0=?$

From trigonometry we have a following identity,

$\cos \theta .\cos ({60}^0-\theta ).\cos ({60}^0+\theta )=\frac{1}{4}\cos 3\theta$

So, if $\theta ={12}^0$

$\cos {12}^0.\cos ({60}^0-{12}^0).\cos ({60}^0+{12}^0)=\frac{1}{4}\cos (3\times 12)$

$\Rightarrow \cos {12}^0.\cos {48}^0.\cos {72}^0=\frac{1}{4}\cos {36}^0$

$\Rightarrow \cos {12}^0.\cos {48}^0.\cos {72}^0=\frac{\sqrt{5}+1}{16}⟶\left(1\right)$

if $\theta ={24}^0$

${\cos 24}^0.\cos ({60}^0-{24}^0)\cos ({60}^0+{24}^0)=\frac{1}{4}\cos (3\times {24}^0)$

$\Rightarrow \cos {24}^0.\cos {36}^0.\cos {84}^0=\frac{1}{4}\cos {72}^0$

$\Rightarrow \cos {24}^0.\cos {36}^0.\cos {84}^0=\frac{\sqrt{5}-1}{16}⟶\left(2\right)$

Multiplying $\left(1\right)$ & $\left(2\right)$ we get

$\cos {12}^0.\cos {24}^0.\cos {36}^0.\cos {48}^0.\cos {72}^0.\cos {84}^0=\left(\frac{\sqrt{5}+1}{16}\right)\left(\frac{\sqrt{5}-1}{16}\right)⟶\left(3\right)$

We are asked to find the value of :

${\cos 12}^0.\cos {24}^0.\cos {36}^0.\cos {48}^0.\cos {60}^0.\cos {72}^0.\cos {96}^0$

$={\cos 12}^0.\cos {24}^0.\cos {36}^0.\cos {48}^0.\cos {72}^0.\cos ({180}^0-{84}^0).\cos {60}^0$

$=\cos {12}^0.\cos {24}^0.\cos {36}^0.\cos {48}^0.\cos {72}^0.\left({-\cos 84}^0\right).\cos {60}^0$

$=\frac{(\sqrt{5}+1)(1-\sqrt{5})}{16\times 16}\times \frac{1}{2}$ (from $\left(3\right)$ & $\cos {60}^0=1/2$)

$=\frac{1-5}{32\times 16}$

$=\frac{-4}{32\times 16}$

$=-\frac{1}{128}$