Express $\sin 4\theta$ by formulae involving $\sin$ and $\cos$ and its powers. I have an assignment question that says "Express $\sin 4\theta$ by formulae involving $\sin$ and $\cos$ and its powers."
I'm told that $\sin 2\theta = 2 \sin\theta \cos\theta$ but I don't know how this was found. 
I used Wolfram Alpha to get the answer but this is what I could get :
$$
4\cos^3\theta\sin\theta- 4\cos\theta \sin^3\theta
$$
How can I solve this problem?

(Note that this method is pretty explanatory and slow, you can do it faster).
Let $u = 2\theta$, then we have:
$$ \sin 4\theta = \sin 2u $$
We know that:
$$ \sin 2u = 2\sin u\cos u$$
Now put $u = 2\theta$ back in:
$$ \sin (2 \cdot 2\theta) = 2\sin 2\theta \cos 2\theta $$
$$ \sin (4\theta) = 2\sin 2\theta \cos 2\theta $$
We know that $\sin 2\theta = 2\sin\theta\cos\theta$, so:
$$ \sin (4\theta) = 4\sin \theta \cos \theta \cos 2\theta $$
Still, we must get rid of that pesky $\cos 2\theta$. You should know the other double angle sum formula for $\cos$:
$$\cos 2\theta = \cos^2 \theta - \sin^2 \theta$$
So:
$$ \sin (4\theta) = 4\sin \theta \cos \theta \left( \cos^2 \theta - \sin^2 \theta \right)$$
$$ \sin (4\theta) = 4\sin \theta \cos^3 \theta - 4\sin^3 \theta \cos \theta$$
