Question

The value of $\cos {12}^0\cos {24}^0\cos {36}^0\cos {48}^0\cos {72}^0\cos {96}^0$ equals

• A. $-\frac{1}{2^6}$
• B. $\frac{1}{2^8}$
• C. $\frac{1}{2^7}$
• D. $-\frac{1}{2^7}$

Answer 1

The correct option is A $-\frac{1}{2^6}$

We need to find value of $\cos {12}^o\cos {24}^o\cos {36}^o\cos {48}^o\cos {72}^o\cos {96}^o$

Multiplying and dividing by $\sin {12}^0$ gives us

$\frac{1}{\sin {12}^0}[\sin {12}^0.\cos {12}^0.\cos {24}^0.\cos {48}^0.\cos {96}^0].\cos {36}^0.\cos {72}^0$

$=\frac{1}{\sin {12}^0}\left[\frac{\sin {24}^0\cos {24}^0\cos {48}^0.\cos {96}^0}{2}\right]\cos {36}^0\cos {72}^0$

$=\frac{1}{\sin {12}^0}[\frac{\sin {48}^0.\cos {48}^0}{4}.\cos {96}^0]\cos {36}^0.\cos {72}^0$

$=\frac{1}{\sin {12}^0}\left[\frac{\sin {96}^0.\cos {96}^0}{8}\right]\cos {36}^0.\cos {72}^0$

$=\frac{\sin {192}^0}{16\sin {12}^0}.\cos {36}^0.\cos {72}^0$

$=\frac{-1}{16}.\cos {36}^0.\cos {72}^0$

$=\frac{-1}{16\sin {36}^0}.\frac{\sin {72}^0.\cos {72}^0}{2}$

$=\frac{-1}{16\sin {36}^0}.\frac{\sin {114}^0}{4}$

$=\frac{-\sin {144}^0}{64\sin {36}^0}$

$=\frac{-1}{64}$