# Why is $\sin(t)\cos(t)$ equal to $\frac{1}{2}\sin(2t)$?

I know that $\sin(t)\cos(t)$ is equal to $\frac{1}{2}\sin(2t)$ but I do not understand why, please explain it to me!

-
This question has some explanations "from the first principles". –  Dan Shved Nov 8 '13 at 10:29

$\sin 2t =\sin (t+t)= \sin t \cos t+ \cos t\sin t =\ldots$.
@Dan Shved: Yes, I hope that if OP don't know $\sin(x+y)$, he will ask this. –  Boris Novikov Nov 8 '13 at 10:27
We now that: $$\sin(2t)=\frac{1}{2}\sin t\cos t/:\frac{1}{2}$$ Now we have: $$\frac{1}{2}\sin(2t)=\sin t\cos t$$