Why is $\frac{6\sin(2\pi/3)}2 = \frac{3\sqrt3}{2}$ The source of my problem is that I have an integral:
$$\int_{\pi/3}^\pi{(6\cos 2x)dx}$$
For anyone having the same problem as me, to see the boundaries, They are $\pi/3$ and $\pi$
The primitive function of $\,6\cos 2x\,$ is, if I'm not mistaken, $\dfrac{6\sin2x}2$
I haven't figured out how to do large square brackets so I've skipped the enclosed formal primitive function below (Maybe someone could help me with this?)
$$\int_{\pi/3}^\pi{(6\cos 2x)dx} = \frac{6\sin(2\pi)}2 - \frac{6\sin(2\pi/3)}2$$
Well... So far, so good I think. $2\pi$ should be possible to replace with 0, right?, and $2\pi/3$ could be replaced with 120? In that case I would end up with:
$\dfrac{3\sin (120)}2$
Wolfram alpha says: 
$$\frac{6\sin(2\pi/3)}2 = \frac{3\sqrt3}{2}$$
So... the question(s) in short:


*

*How is that last transformation based on?

*Is the calculation correct overall? I tried to figure out how to input it all to wolfram alpha for an answer but weren't able to format it correctly. The reason I'm asking is that I have a couple of sample exam questions, but no answers, and I don't want to learn this wrong from the beginning.  

 A: $$\dfrac {6\sin\left(\frac{2\pi}{3}\right)}{2} = 3 \sin \left(\frac {2\pi}{3}\right) = 3\sin \left(\frac{\pi}{3}\right) = 3\left(\frac{\sqrt 3}{2}\right)$$
Re: your question in the comment below your question: $\sin\left(\frac{\pi}{3}\right) = \sin\left(\pi - \frac {\pi}{3}\right) = \sin \left(\frac{2\pi}{3} \right)$, which can be confirmed using the unit circle.  If it is more intuitive for you to visualize angles in degrees, then the equivalent identity can be stated as follows:
$$\sin\left(60^\circ\right) = \sin\left(180^\circ - 60^\circ\right) = \sin \left(120^\circ \right) = \frac{\sqrt 3}{2}$$
$(2)$ Your evaluation of the integral is correct. Note that $$\begin{align}\int_\frac\pi3^\pi{(6\cos 2x)dx} & = \frac{6\sin(2\pi)}2 - \frac{6\sin(2\pi/3)}2\\ \\ & = {3\sin(2\pi)} - {3\sin(2\pi/3)} \\ \\ & = 0 - 3\left(\frac{\sqrt 3}{2}\right)\\ \\ & = - 3\left(\frac{\sqrt 3}{2}\right)\end{align}$$
A: As $$\sin(\pi-x)=\sin x$$
$$\sin\frac{2\pi}3=\sin\left(\pi-\frac\pi3\right)=\sin\frac\pi3=\frac{\sqrt3}2$$
