CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image