Prove that, cos306° +cos234° +cos162° +cos18°=0

Prove that, cos306° +cos234° +cos162° +cos18°=0

Solution:

L.H.S = (cos306° +cos234°)+(cos162° +cos18°)

= 2 cos $\frac{306° + 234°}{2}$ cos $\frac{306° – 234°}{2}$ + 2 cos $\frac{162° + 18°}{2}$ cos $\frac{162° – 18°}{2}$

= 2 cos $\frac{540°}{2}$ cos $\frac{72°}{2}$ + 2 cos $\frac{180°}{2}$ cos $\frac{144°}{2}$

= 2 cos 270° cos36° + 2 cos90° cos72°

= 2 sin0° cos36°

= 0 = R.H.S [Proved]