Question

Prove that ${\cos }^{2}A+co{s}^2(A+120)+{\cos }^2(A-120)=\frac{3}{2}$

Answer 1

${\cos }^{2}A+{\cos }^2(A+120)+{\cos }^2(A-120)={\cos }^{2}A+{\cos }^{2}A{\cos }^{2}120+{\sin }^{2}A{\sin }^{2}120-2\cos A\cos 120\sin A\sin 120+{\cos }^{2}A{\cos }^{2}120+{\sin }^{2}A{\sin }^{2}120+2\cos A\cos 120\sin A\sin 120={\cos }^{2}A+\frac{{\cos }^{2}A}{4}+\frac{3{\sin }^{2}A}{4}+\frac{{\cos }^{2}A}{4}+\frac{3{\sin }^{2}A}{4}=\frac{3{\cos }^{2}A}{2}+\frac{3{\sin }^{2}A}{2}=\frac{3}{2}({\cos }^{2}A+{\sin }^{2}A)=\frac{3}{2}$