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. 
A: I shall prove the generalization of $$\cos x+\cos(x+120^\circ)+\cos(x+240^\circ)=0$$  (here $x=40^\circ$)
$$S=\sum_{r=0}^{n-1}\cos\left(x+\dfrac{360^\circ r}n\right)=0$$
Method $\#1:$
$S=$ real part of $\sum_{r=0}^{n-1} \exp i\left(x+\dfrac{360^\circ r}n\right)$
Now $\displaystyle \sum_{r=0}^{n-1} \exp i\left(x+\dfrac{360^\circ r} n \right) = \frac{e^{i360^\circ}-1}{\exp i\left(x+\dfrac{360^\circ(-n)}n\right)-1}=0$
Now equate the real & the imaginary parts.
Method $\#2:$
Using multiple angle formula cosine:
$$\cos(nx)=2^{n-1}\cos^nx-n2^{n-3}\cos^{n-2}x+\cdots$$
Now if $\cos nx=\cos ny, nx=360^\circ m\pm ny$ where $m$ is any integer
$x= y+\dfrac{360^\circ m}n$ where $m\equiv0,1,2,\ldots, n-1\pmod n$
So, the roots of  $$2^{n-1}\cos^nx-n2^{n-3}\cos^{n-2}x+\cdots-\cos ny=0$$ are
$\cos\left(y+\dfrac{360^\circ m}n\right)$ where $m\equiv0,1,2,\ldots n-1\pmod n$
Using Vieta's formula,
$$\sum_{r=0}^{n-1}\cos\left(y+\dfrac{360^\circ m}n\right)=\dfrac0{2^{n-1}}$$
