Question

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

Answer 1

$L.H.S.={\cos }^{2}x+{\cos }^2(x+\frac{\pi }{3})+{\cos }^2(x-\frac{\pi }{3})=\frac{1+\cos 2x}{2}+\frac{1+\cos (2x+\frac{2\pi }{3})}{2}+\frac{1+\cos (2x-\frac{2\pi }{3})}{2}=\frac{1}{2}(3+\cos 2x+\cos (2x+\frac{2\pi }{3})+\cos (2x-\frac{2\pi }{3}))=\frac{1}{2}(3+\cos 2x+2\cos \left(\frac{2x+\frac{2\pi }{3}+2x-\frac{2\pi }{3}}{2}\right)\cos \left(\frac{2x+\frac{2\pi }{3}-2x+\frac{2\pi }{3}}{2}\right))=\frac{1}{2}(3+\cos 2x+2\cos (2x\left)\cos \left(\frac{2\pi }{3}\right)\right)=\frac{1}{2}(3+\cos 2x-2\cos (2x\left)\left(\frac{1}{2}\right)\right)=\frac{1}{2}(3+\cos 2x-\cos (2x\left)\right)=\frac{3}{2}=R.H.S$

Hence proved.