Question
Prove that: $\cos ^{2} x+\cos ^{2}\left(x+\frac{\pi}{3}\right)+\cos ^{2}\left(x-\frac{\pi}{3}\right)=\frac{3}{2}$

Answer

Taking L.H.S, we have
$\text { L.H.S. }=\frac{1+\cos 2 x}{2}+\frac{1+\cos \left(2 x+\frac{2 \pi}{3}\right)}{2}+\frac{1+\cos \left(2 x-\frac{2 \pi}{3}\right)}{2}$
$=\frac{1}{2}\left[3+\cos 2 x+\cos \left(2 x+\frac{2 \pi}{3}\right)+\cos \left(2 x-\frac{2 \pi}{3}\right)\right]$
$=\frac{1}{2}\left[3+\cos 2 x+2 \cos 2 x \cos \frac{2 \pi}{3}\right]$
$=\frac{1}{2}\left[3+\cos 2 x+2 \cos 2 x \cos \left(\pi-\frac{\pi}{3}\right)\right]$
$=\frac{1}{2}\left[3+\cos 2 x-2 \cos 2 x \cos \frac{\pi}{3}\right]$
$=\frac{1}{2}[3+\cos 2 x-\cos 2 x]=\frac{3}{2}=\mathrm{R.H.S.}$

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