Question

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

Answer 1

${\cos }^{2}x+{\cos }^2(x+\frac{\pi }{3})+{\cos }^2(x-\frac{\pi }{3})=\frac{3}{2}$

We know,

$\cos 2x=2{\cos }^{2}x-1$

$\cos 2x+1=2{\cos }^{2}x$

$\frac{\cos 2x+1}{2}={\cos }^{2}x$

${\cos }^{2}x=\frac{1}{2}(\cos 2x+1)$

Replace $x$ with $x+\frac{\pi }{3}.$

${cos}^2(x+\frac{\pi }{3})=\frac{1}{2}cos2(x+\frac{\pi }{3})+\frac{1}{2}=\frac{cos(2x+\frac{2\pi }{3})+1}{2}$

Similarly replace $x$ with $x-\frac{\pi }{3}.$

${cos}^2(x-\frac{\pi }{3})=\frac{1}{2}cos(2x-\frac{2\pi }{3})+\frac{1}{2}$

Now, $L.H.S$

$⟹\frac{1}{2}[1+\cos 2x+1+\cos (2x+\frac{2\pi }{3})+1+\cos (2x-\frac{2\pi }{3})]$

$⟹\frac{1}{2}[3+\cos 2x+\cos (2x+\frac{2\pi }{3})+\cos (2x-\frac{2\pi }{3})]$

$⟹\frac{1}{2}[3+\cos 2x+2\cos 2x\cos \frac{2\pi }{3}]$

$⟹\frac{1}{2}[3+\cos 2x+2\cos 2x\frac{1}{2}]$

$⟹\frac{1}{2}[3+\cos 2x-\cos 2x]$

$⟹\frac{3}{2}=R.H.S$ $\left(proved\right)$