How do you derive this trig identity from the common ones? $\cos^2x=\frac{1+\cos2x}{2}$ $$\cos^2x=\frac{1+\cos2x}{2}$$
Just came across this identity one today.  Where does this come from? Is this an easy derivation from the more popular identities, or is this one you just take it at face value and memorize?
 A: How about using the double angle formula for $\cos 2x:$
$$\cos 2x = \cos^2 x - \sin^2 x = \cos^2 x - \underbrace{(1 - \cos^2 x)}_{ \large =\,\sin^2 x} = 2\cos ^2 x - 1 \iff \cos^2 x=\dfrac{1 + \cos 2x}{2}$$
And the double-angle formula is a special case of the angle sum formula for $\cos(\alpha + \beta)$, $$\cos(\alpha +\beta)=\cos \alpha \cos \beta -\sin \alpha\sin \beta$$ where $\alpha = \beta$.
A: HINT:
Do you know $$\cos(A+B)=\cos A\cos B-\sin A\sin B$$
Put $A=B$ and use $\cos^2x+\sin^2x=1$
A: We have
$$e^{2ix}=e^{ix}e^{ix}=(\cos x+i\sin x)^2=\cos^2 x-\sin^2x+2i\cos x\sin x$$
Now take the real part and combine the result with the identity
$$\cos^2x+\sin^2x=1$$
A: One of teh basic formulas is: $cos(2x)=cos^2x-sin^2x$
Using one more basic formula $sin^2x+cos^2x=1$
We get:
$cos(2x)=cos^2x+cos^2x-1$
$cos(2x)=2cos^2x-1$
$2cos^2x=1+cos(2x)$
$cos^2x=\frac{1+cos(2x)}{2}$
A: Familiar with complex numbers? If so then 'de  Moivre'  is very useful here:
$(\cos x+i \sin x)^{n}=\cos nx +i\sin nx $. 
In your case $n=2$ and $(\cos x+i \sin x)^{2}=\cos^{2} x -\sin^{2} x +2i \sin x \cos x $  
This makes it clear immediately that $\cos^{2} x -\sin^{2} x=\cos 2x$ (and $2 \sin x \cos x=\sin 2x$). Combining it with the well-known equality  $\cos^{2} x + \sin^{2} x =1$ leads to  $\cos^2x=\frac{1+\cos2x}{2}$.
De Moivre is very handsom and can also be used to find expressions for $\cos 3x$, $\sin 3x$, $\cos 4x$ , $\sin 4x$ et cetera.
A: Ok, how's this?
Ok, how's this?

