Question

Prove ${\cos }^3\theta +{\cos }^3({120}^{\circ }+\theta )+{\cos }^3({220}^{\circ }+\theta )=\frac{3}{4}\cos 3\theta$

Answer 1

We know that $\cos 3A=4{\cos }^{3}A-3\cos A$

$\therefore {\cos }^{3}A=\frac{1}{4}(\cos 3A+3\cos A).$

Also $\cos (2n\pi +\theta )=\cos \theta .$

Applying the above we have

L.H.S. $=\frac{1}{4}[3\cos \theta +\cos 3\theta )+3\cos ({120}^{\circ }+\theta )+\cos ({360}^{\circ }+3\theta )+3\cos ({240}^{\circ }+\theta )+\cos ({720}^{\circ }+3\theta )]$

$=\frac{1}{4}\left[3\cos 3\theta \right]+\frac{3}{4}[\cos \theta +\cos ({120}^{\circ }+\theta )+\cos ({240}^{\circ }+\theta \left)\right]$

$=\frac{3}{4}\cos 3\theta +\frac{3}{4}[\cos \theta +2\cos ({180}^{\circ }+\theta \left)\cos {60}^{\circ }\right]$

$=\frac{3}{4}\cos 3\theta +\frac{3}{4}[\cos \theta +2.\frac{1}{2}(-\cos \theta \left)\right]$

$=\frac{3}{4}\cos 3\theta .$