Q:

Prove that: $\cos^2 20° + \cos^2 40° +\cos^2 80° = \sin^2 20° + \sin^2 40° + \sin^2 80°$

Prove that: $\cos^2 20° + \cos^2 40° +\cos^2 80° = \sin^2 20° + \sin^2 40° + \sin^2 80°$
My Attempt:
$$L.H.S=\cos^2 20° + \cos^2 40° + \cos^2 80°$$
$$=\dfrac {1+\cos 40}{2}+\dfrac {1+\cos 80}{2} + \dfrac {1+\cos 160°}{2}$$
$$=\dfrac {3+\cos 40°+\cos 80°+\cos 160°}{2}$$
I.could not solve further from here..

A:

We know the identity $$\cos^2 x - \sin^2 x = \cos 2x \tag {1}$$ and $$\cos A - \cos B = 2\cos \frac {A+B}{2} \cos \frac {A-B}{2} \tag {2} $$
We have $$(\cos^2 20^\circ - \sin^2 20^\circ) + (\cos^2 40^\circ - \sin^2 40^\circ) + (\cos^2 80^\circ - \sin^2 80^\circ) $$ $$= \cos 40^\circ + \cos 80^ \circ  + \cos 160^ \circ $$ (using $(1)$) $$ =2\cos 60^\circ \cos 20^\circ  +\cos (90^\circ + 70^\circ) $$ (using $(2)$) $$ = \cos 20^\circ  - \sin 70^\circ =0$$
where we used the identity  $\cos (90^\circ + x) = -\sin x $. 
Hope it helps. 

