Question

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

Solution

## $$\cos{{12}^{\circ}}+\cos{{84}^{\circ}}+\cos{{156}^{\circ}}+\cos{{132}^{\circ}}$$$$=\left(\cos{{12}^{\circ}}+\cos{{132}^{\circ}}\right)+\left(\cos{{84}^{\circ}}+\cos{{156}^{\circ}}\right)$$Using the transformation angle formula,$$cos{C}+\cos{D}=2\cos{\left(\dfrac{C+D}{2}\right)}\cos{\left(\dfrac{C-D}{2}\right)}$$$$=2\cos{\left(\dfrac{{12}^{\circ}+{132}^{\circ}}{2}\right)}\cos{\left(\dfrac{{12}^{\circ}-{132}^{\circ}}{2}\right)}+2\cos{\left(\dfrac{{84}^{\circ}+{156}^{\circ}}{2}\right)}\cos{\left(\dfrac{{84}^{\circ}-{156}^{\circ}}{2}\right)}$$$$=2\cos{{72}^{\circ}}\cos{{60}^{\circ}}+2\cos{{120}^{\circ}}\cos{{36}^{\circ}}$$$$=2\cos{{72}^{\circ}}\cos{{60}^{\circ}}+2\cos{\left({180}^{\circ}-{60}^{\circ}\right)}\cos{{36}^{\circ}}$$$$=2\cos{{72}^{\circ}}\cos{{60}^{\circ}}-2\cos{{60}^{\circ}}\cos{{36}^{\circ}}$$$$=2\cos{{72}^{\circ}}\times\dfrac{1}{2}-2\times\dfrac{1}{2}\cos{{36}^{\circ}}$$$$=\cos{{72}^{\circ}}-\cos{{36}^{\circ}}$$$$=\cos{\left({90}^{\circ}-{18}^{\circ}\right)}-\cos{{36}^{\circ}}$$$$=\sin{{18}^{\circ}}-\cos{{36}^{\circ}}$$$$=\dfrac{\sqrt{5}-1}{4}-\dfrac{\sqrt{5}+1}{4}=\dfrac{-2}{4}=-\dfrac{1}{2}$$Mathematics

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