How to prove this trigonometry I need to prove that $$\cos^2(\beta -\gamma)+ \cos^2( \gamma - \alpha) +\cos^2(\alpha -\beta) = 1+2 \cos(\beta- \gamma) \cos( \gamma - \alpha)\cos(\alpha -\beta)  $$
To do this I have used the formula $2 cos^2 \theta  = 1+ \cos2\theta$
But after using the formula  it seems that, the whole equation is getting complex. Can you please help me with this? 
 A: I'll rewrite it as
$\cos^2(b -c)+ \cos^2( c - a) +\cos^2(a -b) = 1+2 \cos(b- c) \cos( c - a)\cos(a -b)
$
because I am lazy.
Since
$\cos(2u)
=2\cos^2(u)-1
$
and
$\cos(u)\cos(v)
=\frac12(\cos(u-v)+\cos(u+v))
$,
applying these whenever possible
(without much thinking),
$\begin{array}\\
\cos(b- c) \cos( c - a)\cos(a -b)
&=\frac12(\cos(b-2c+a)+\cos(b-a))\cos(a-b)\\
&=\frac12(\cos(b-2c+a)\cos(a-b)+\cos(b-a)\cos(a-b))\\
&=\frac14((\cos(2b-2c)+\cos(2a-2c))+2\cos^2(b-a))\\
&=\frac14((2\cos^2(b-c)-1)+(2\cos^2(a-c)-1))+2\cos^2(b-a))\\
&=\frac12(\cos^2(b-c)+\cos^2(a-c)-1+\cos^2(b-a))\\
\end{array}
$
Q.E.D.
(to my surprise).
A: $$F=\cos^2A+\cos^2B+\cos^2C=\cos^2A-\sin^2B+1+\cos^2C$$
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$, $$F=\cos(A+B)\cos(A-B)+1+\cos^2C$$
If $\cos(A+B)=\cos C\iff A+B=2m\pi\pm C,$ (where $m$ is any integer)
$\displaystyle F=\cos C\cos(A-B)+\cos(A+B)\cos C+1$
$\displaystyle=1+2\cos C[\cos(A+B)+\cos(A-B)]$
$\displaystyle=1+2\cos A\cos B\cos C$
Set $\beta-\gamma=A,\gamma-\alpha=B,\alpha-\beta=C$
