ASN4 -Fourier Transform property

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${ℱ}[cos(2\pi f_{0}t)g(t)]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}cos(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt$

Using Euler's cosine identity

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}cos(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}[\frac{1}{2}{e}^{j2\pi f_{0}t}+\frac{1}{2}{e}^{-j2\pi f_{0}t}]g(t)e^{-j2\pi ft}dt$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}cos(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{2}{e}^{j2\pi f_{0}t}+\frac{1}{2}{e}^{-j2\pi f_{0}t}g(t)e^{-j2\pi ft}dt$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}cos(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{2}{e}^{j2\pi f_{0}t}{e}^{-j2\pi ft}dt+\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{2}{e}^{-j2\pi f_{0}t}g(t)e^{-j2\pi ft}dt$

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

Identifying that the above equation contains Fourier Transforms the solution is

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