Why is Fourier transform of sin(2t) not 2? I have been told that the Fourier transform outputs the frequencies of a given function. So I would have imagined to get $2$ from the function $\sin(2t)$, for it is its frequency. But from calculation (online) i get a whole different answer.
My professor told me that I will get 2 only for calculating $C_n$, that are the Fourier coefficients for the Fourier series.
So what DOES the Fourier transform output for a given function?
 A: 
I have been told that the Fourier transform outputs the frequencies of a given function.

What you have been told is not quite correct. 
The Fourier transform does tell you about the frequencies of a function, but it doesn't do so by outputting the frequencies of that function. Instead, the Fourier transform $\hat{f}$ of a function $f$ takes any possible frequency as an input, and outputs what is essentially the magnitude of that frequency in $f$.
In the case of the Fourier transform of $f = \sin (2t)$, the Fourier transform $\hat{f}(\nu)$ is $0$ almost everywhere, because the magnitude of almost every frequency is $0$. However, at $\nu = -2$ and $\nu = 2$, $\hat{f}(\nu)$ looks like the Dirac delta function, which is thought of as being an infinitely large value. This is because the magnitude of the frequency $2$ is infinite (which is because the function $f$ maintains a frequency of $2$ for an infinite amount of time).
(The reason that the frequency I give is $2$ instead of $\frac{2}{2 \pi}$ is that I'm using the angular frequency definition of the Fourier transform. The reason that both $2$ and $-2$ appear in the output function is that the Fourier transform is based not on sine waves, but on "helixes" that look like $\cos (a x) + i \sin (a x)$; we need to add two of these "helixes" together to make the cosines cancel out and just leave us the sines.)
In order for a function to have a Fourier transform of $2$, it would have to contain every frequency, with a magnitude of $2$. The distribution $2 \delta(x)$ does exactly that, so its Fourier transform is $2$. (Here, $\delta$ means the Dirac delta function—which is actually a distribution, not a function.) 
