How to obtain the CTFT of a sine in terms of f in hertz (from the formula in terms of $ \omega $)
Recall:
$ x(t)= \sin(\omega_0 t) $
$ \mathcal{X}(\omega)=\frac{\pi}{i} \left[\delta (\omega - \omega_0) - \delta (\omega + \omega_0)\right] $
To obtain X(f), use the substitution
$ \omega= 2 \pi f $.
More specifically
$ \begin{align} X(f) &=\mathcal{X}(2\pi f) \\ &=\frac{\pi}{i} \left[\delta (2\pi f- \omega_0) - \delta (2\pi f+ \omega_0)\right] \\ &=\frac{1}{2i} \left[\delta (f - \frac{\omega_0}{2\pi}) - \delta (f + \frac{\omega_0}{2\pi})\right] \end{align} $
$ Since\ k\delta (kt)=\delta (t),\forall k\ne 0 $
