Prove that $\cos^4(\theta)-\sin^4(\theta)=\cos(2\theta)$ The problem is to prove that $\cos^4(\theta)-\sin^4(\theta)=cos(2\theta)$.
So, here are my steps so far:
$(\cos^2\theta)^2-(\sin^2\theta)^2=\cos(2\theta)$
$(\cos^2\theta+\sin^2\theta)(\cos^2\theta-\sin^2\theta)=\cos(2\theta)$
$(\cos^2\theta+\sin^2\theta)(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)=\cos(2\theta)$
I don't know where to go from here. Please help! I feel like I'm missing something super obvious but I don't know what...
 A: Well, $\cos^2(\theta)+\sin^2(\theta)=1$, and $\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)$ is a standard Double-Angle identity. Can you continue from there?
A: There are a couple of nice ways we can do it. Firstly, note that:
$$\cos^4(\theta)-\sin^4(\theta)=\bigl(\cos^2\theta-\sin^2\theta\bigr)\bigl(\cos^2\theta+\sin^2\theta\bigr)$$
and since:
$$\cos^2\theta+\sin^2\theta=1\tag{1}$$
we can say that:
$$\cos^4\theta-\sin^4\theta=\cos^2\theta-\sin^2\theta=\cos2\theta$$
which is the double angle formula for cosine. This can be proved geometrically a number of ways but also as shown below

We can also take advantage of De Moivre's theorem, which states that:
$$\left(\cos x+i\sin x\right)^n=\cos(nx)+i\sin(nx)\tag{2}$$
Using the fact that:
$$e^{ix}=\cos(x)+i\sin(x)\tag{3}$$
We can do this by saying:
$$\cos(2x)+i\sin(2x)=\left(\cos x+i\sin x\right)^2$$
$$\cos(2x)+i\sin(2x)=\cos^2x+2i\cos x\sin x+i^2\sin^2x$$
$$\cos(2x)+i\sin(2x)=\cos^2x-\sin^2x+2i\cos x\sin x$$
Now if we separate the real and imaginary parts we are left with:
$$\cos(2x)=\cos^2x-\sin^2x,\,\sin(2x)=2\sin(x)\cos(x)\tag{4}$$
A: We have 
$$\cos^4(\theta)-\sin^4(\theta) = (\cos^2\theta+\sin^2\theta)(\cos^2\theta-\sin^2\theta) $$ where $(\cos^2\theta+\sin^2\theta) = 1$ and $(\cos^2\theta-\sin^2\theta) = 2\cos^2(\theta)-1 = \cos(2\theta)$
A: Hint:$$\cos(t) = \frac{e^{it} + e^{-it}}{2}$$
and
$$\sin(t) = \frac{e^{it} - e^{-it}}{2i}$$
