Fourier Transform - solving $\cos(60 \pi t) \cdot \sin(20 \pi t)$

Question

How can i calculate the Fourier transform of $\cos(60 \pi t) \cdot \sin(20 \pi t)$?

By applying the formula for $\cos$ and $\sin$ I get $$ \frac{1}{4i}[\delta(f-60) + \delta(f + 60)] \cdot [\delta(f - 20) - \delta(f + 20)]. $$ Is it possible to simplify this even more?

Answer 1

Using trigonometric identities, note that $\cos(60 \pi t) \cdot \sin(20 \pi t) = 0.5 \sin(80 \pi t) - 0.5\sin(40 \pi t)$. From this point, you can obtain the Fourier transform.