Question

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

Answer 1

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

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

Now use this formula

$\cos \left(x+y\right)=\cos x\cos y-\sin x\sin y$

$\cos \left(x-y\right)=\cos x\cos y+\sin x\sin y$

$thenLHS={\cos }^{2}x+{\left[\cos x\cos \frac{\pi }{3}-\sin x\sin \frac{\pi }{3}\right]}^2+{\left[\cos x\cos \frac{\pi }{3}+\sin x\sin \frac{\pi }{3}\right]}^2$

= ${\cos }^{2}x+{\left[\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x\right]}^2+{\left[\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\right]}^2$

$(\therefore \cos \frac{\pi }{3}=\frac{1}{2},\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2})$

$={\cos }^{2}x+\frac{1}{4}{\cos }^{2}x+\frac{3}{4}{\sin }^{2}x-\frac{\sqrt{3}}{2}\cos x\sin x+\frac{1}{4}{\cos }^{2}x+\frac{3}{4}{\sin }^{2}x+\frac{\sqrt{3}}{2}\sin x\cos x$

$={\cos }^{2}x+\frac{{\cos }^{2}x}{2}+\frac{3{\sin }^{2}x}{2}$

$=\frac{3}{2}{\cos }^{2}x+\frac{3}{2}{\sin }^{2}x$

$=\frac{3}{2}\left({\cos }^{2}x+{\sin }^{2}x\right)$

$=\frac{3}{2}=RHS$