Question

The value of $\cos {12}^{\circ }+\cos {84}^{\circ }+\cos {156}^{\circ }+\cos {132}^{\circ }$ is :

Answer 1

$\cos {12}^{\circ }+\cos {84}^{\circ }+\cos {156}^{\circ }+\cos {132}^{\circ }$

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

Using the transformation angle formula,$cosC+\cos D=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$

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

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

$=2\cos {72}^{\circ }\cos {60}^{\circ }+2\cos ({180}^{\circ }-{60}^{\circ })\cos {36}^{\circ }$

$=2\cos {72}^{\circ }\cos {60}^{\circ }-2\cos {60}^{\circ }\cos {36}^{\circ }$

$=2\cos {72}^{\circ }\times \frac{1}{2}-2\times \frac{1}{2}\cos {36}^{\circ }$

$=\cos {72}^{\circ }-\cos {36}^{\circ }$

$=\cos ({90}^{\circ }-{18}^{\circ })-\cos {36}^{\circ }$

$=\sin {18}^{\circ }-\cos {36}^{\circ }$

$=\frac{\sqrt{5}-1}{4}-\frac{\sqrt{5}+1}{4}=\frac{-2}{4}=-\frac{1}{2}$