I'm no sure how to prove this equality true.
(Oh and I used an image instead of latex, because I don't know how, someone also tell me how to make to the power and cos?)
It doesn't. You need to use the identity
$\displaystyle \displaystyle \cos{(\alpha \pm \beta)} = \cos{\alpha}\cos{\beta} \mp \sin{\alpha}\sin{\beta}$.
So that means
$\displaystyle \displaystyle \cos{(x + y)}\cdot \cos{(x - y)} = (\cos{x}\cos{y} - \sin{x}\sin{y})(\cos{x}\cos{y} + \sin{x}\sin{y})$
$\displaystyle \displaystyle = \cos^2{x}\cos^2{y} - \sin^2{x}\sin^2{y}$
$\displaystyle \displaystyle = \cos^2{x}\cos^2{y} - (1 - \cos^2{x})(1 - \cos^2{y})$
$\displaystyle \displaystyle = \cos^2{x}\cos^2{y} - (1 - \cos^2{y} - \cos^2{x} + \cos^2{x}\cos^2{y})$
$\displaystyle \displaystyle = \cos^2{y} + \cos^2{x} - 1$.
When you get to
$\displaystyle cos^2x\;cos^2y-sin^2x\;sin^2y=cos^2x-cos^2y$
you can combine some of the squares to get 1.
$\displaystyle cos^2x\;cos^2y-sin^2x\left(1-cos^2y\right)=cos^2x-cos^2y$
$\displaystyle cos^2x\;cos^2y-sin^2x+sin^2x\;cos^2y=cos^2x-cos^2y$
$\displaystyle cos^2y\left(cos^2x+sin^2x\right)-sin^2x=cos^2x-cos^2y$
$\displaystyle cos^2y-sin^2x=cos^2x-cos^2y$
$\displaystyle 2cos^2y=cos^2x+sin^2x=1$
Note that there is a difference between an equation and an identity.
An identity is true for all x.
An equation is true for certain x for which we solve.
Start by noting that $\displaystyle \displaystyle \cos{(5\theta)} = \cos{(3\theta + 2\theta)}$ and use the angle sum identity to simplify.
Then note that $\displaystyle \displaystyle \cos{(3\theta)} = \cos{(2\theta + \theta)}$ and again use the angle sum identity to simplify.
Alternatively, get everything in terms of "Sine", using the following identities
$\displaystyle cosA-cosB=-2sin\left[\frac{A+B}{2}\right]sin\left[\frac{A-B}{2}\right]$
$\displaystyle sinAcosB=\frac{1}{2}\left[sin(A+B)+sin(A-B)\right]$
Then...
$\displaystyle sin\theta\;sin2\theta\;cos2\theta=sin\theta\left[\frac{1}{2}\left(sin4\theta+sin0\right)\right]=\frac{1}{2}sin\theta\;sin4\theta$
$\displaystyle \frac{1}{4}\left[cos3\theta-cos5\theta\right]=\frac{1}{4}\left[-2sin\left(\frac{8\theta}{2}\right)sin\left(-\frac{2\theta}{2}\right)\right]=-\frac{1}{2}\left[sin4\theta\left(-sin\theta\right)\right]$