ASN4 -Fourier Transform property

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

Using Euler's cosine identity

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

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

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

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

Identifying that the above equation contains Fourier Transforms the solution is

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