Question

The value of $\cos {12}^0+\cos {84}^0+\cos {156}^0+\cos {132}^0$ $=-\frac{1}{a}$. Find $a.$

Answer 1

We have, $\cos {12}^0+\cos {84}^0+\cos {156}^0+\cos {132}^0$

$=(\cos {12}^0+\cos {132}^0)+(\cos {84}^0+\cos {156}^0)$

$=2\cos \left(\frac{{12}^0+{132}^0}{2}\right)\cos \left(\frac{{132}^0-{12}^0}{2}\right)+2\cos \left(\frac{{84}^0+{156}^0}{2}\right)\cos \left(\frac{{156}^0-{84}^0}{2}\right)$

$=2\cos {72}^0\cos {60}^0+2\cos {120}^0\cos {36}^0$

$=2\sin {18}^0\cos {60}^0+2\cos {120}^0\cos {36}^0$

$=2\left(\frac{\sqrt{5}-1}{4}\right)\frac{1}{2}+2(-\frac{1}{2})\left(\frac{\sqrt{5}+1}{4}\right)=-\frac{1}{2}$
Ans: a=2