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F [ c o s ( 2 π f 0 t ) g ( t ) ] = ∫ − ∞ ∞ c o s ( 2 π f 0 t ) g ( t ) e − j 2 π f t d t {\displaystyle {\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
∫ − ∞ ∞ c o s ( 2 π f 0 t ) g ( t ) e − j 2 π f t d t = ∫ − ∞ ∞ 1 2 e j 2 π f 0 t + 1 2 e − j 2 π f 0 t g ( t ) e − j 2 π f t d t {\displaystyle \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\!}
∫ − ∞ ∞ c o s ( 2 π f 0 t ) g ( t ) e − j 2 π f t d t = ∫ − ∞ ∞ 1 2 e j 2 π f 0 t e − j 2 π f t d t + ∫ − ∞ ∞ 1 2 e − j 2 π f 0 t g ( t ) e − j 2 π f t d t {\displaystyle \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\!}
∫ − ∞ ∞ c o s ( 2 π f 0 t ) g ( t ) e − j 2 π f t d t = ∫ − ∞ ∞ 1 2 e − j 2 π ( f − f 0 ) t g ( t ) d t + 1 2 ∫ − ∞ ∞ 1 2 e − j 2 π ( f + f 0 ) t g ( t ) d t {\displaystyle \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
F [ c o s ( 2 π f 0 t ) g ( t ) ] = 1 2 G ( f − f 0 ) + 1 2 [ G ( f + f 0 ) {\displaystyle {\mathcal {F}}[cos(2\pi f_{0}t)g(t)]={\frac {1}{2}}G(f-f_{0})+{\frac {1}{2}}[G(f+f_{0})\!}
![{\displaystyle {\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\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2141863fea59d36f0fb6be6452b53a62cc58accd)
![{\displaystyle {\mathcal {F}}[cos(2\pi f_{0}t)g(t)]={\frac {1}{2}}G(f-f_{0})+{\frac {1}{2}}[G(f+f_{0})\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3860455af7a9378c44f2e45201bbca36c3ffb589)