ASN4 -Fourier Transform property

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Find ${ℱ}[cos(2\pi f_{0}t)g(t)]$

Recall $w_0=2\pi f_0$, so

Using Euler's cosine identity $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}cos(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{2}[e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}]g(t)e^{-j2\pi ft}dt$

Now $$\begin{gathered} {ℱ}[cos(2\pi f_{0}t)g(t)]=\frac{1}{2}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-j2\pi (f-f_0)t}g(t)dt \\ + \\ \frac{1}{2}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-j2\pi (f+f_0)t}g(t)dt \end{gathered}$$

So ${ℱ}[cos(2\pi f_{0}t)g(t)]=\frac{1}{2}[G(f-f_0)+G(f+f_0)]$