Why $\cos^2 (2x) = \frac{1}{2}(1+\cos (4x))$? Why: $$\cos ^2(2x) = \frac{1}{2}(1+\cos (4x))$$
I don't understand this, how I must to multiply two trigonometric functions?
Thanks a lot.

Recall the formula $$\cos(2 \theta) = 2 \cos^2(\theta) - 1$$ This gives us $$\cos^2(\theta) = \dfrac{1+\cos(2 \theta)}{2}$$ Plug in $\theta = 2x$, to get what you want.
EDIT
The identity
$$\cos(2 \theta) = 2 \cos^2(\theta) - 1$$ can be derived from $$\cos(A+B) = \cos(A) \cos(B) - \sin(A) \sin(B)$$
Setting $A = B = \theta$, we get that
$$\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = \cos^2(\theta) - (1-\cos^2(\theta)) = 2 \cos^2(\theta) - 1$$
