Suppose we have this signal:
$$x(t)=5 \sin(2\pi 1000 t) \cos(2\pi 10000 t)$$
I calculated the Fourier transform as below:
$$X(f)=\left(j\frac{5}{4}\right)\left[\delta(f-9000)+\delta(f+11000)-\delta(f-11000)-\delta(f+9000)\right]$$
and then I would like to convert it to $\omega$ form. I know that the result is
$$X(\omega )=\left(j\frac{5}{2}\right)\left[\delta(\omega -9000)+\delta(\omega +11000)-\delta(\omega -11000)-\delta(\omega +9000)\right]$$
I know that $\omega=2\pi f T$ where $T$ is sampling rate.
What I cannot understand is how we convert the Fourier transform of our signal to $\omega$ form using above formula.