Prove that $\cos ^{2} x+\cos ^{2}\left(x+\frac{\pi}{3}\right)+\cos ^{2}\left(x-\frac{\pi}{3}\right)=\frac{3}{2}$
Prove that $\cos ^{2} x+\cos ^{2}\left(x+\frac{\pi}{3}\right)+\cos ^{2}\left(x-\frac{\pi}{3}\right)=\frac{3}{2}$
We have
${\text{L}}{\text{.H}}{\text{.S}}{\text{.}}$
$ = \frac{{1 + \cos 2x}}{2} + \frac{{1 + \cos \left( {2x + \frac{{2\pi }}{3}} \right)}}{2} + \frac{{1 + \cos \left( {2x - \frac{{2\pi }}{3}} \right)}}{2}$
$ = \frac{1}{2}\left[ {3 + \cos 2x + \cos \left( {2x + \frac{{2\pi }}{3}} \right) + \cos \left( {2x - \frac{{2\pi }}{3}} \right)} \right]$
$ = \frac{1}{2}\left[ {3 + \cos 2x + 2\cos 2x\cos \frac{{2\pi }}{3}} \right]$
$=\frac{1}{2}\left[3+\cos 2 x+2 \cos 2 x \cos \left(\pi-\frac{\pi}{3}\right)\right]$
$=\frac{1}{2}\left[3+\cos 2 x-2 \cos 2 x \cos \frac{\pi}{3}\right]$
$=\frac{1}{2}[3+\cos 2 x-\cos 2 x]=\frac{3}{2}= R.H.S.$
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