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Find F [ c o s ( 2 π f 0 t ) g ( t ) ] {\displaystyle {\mathcal {F}}[cos(2\pi f_{0}t)g(t)]\!}
Recall w 0 = 2 π f 0 {\displaystyle w_{0}=2\pi f_{0}\!} , so
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 + 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 f_{0}t}]g(t)e^{-j2\pi ft}dt\!}
Now F [ c o s ( 2 π f 0 t ) g ( t ) ] = 1 2 ∫ − ∞ ∞ e − j 2 π ( f − f 0 ) t g ( t ) d t + 1 2 ∫ − ∞ ∞ e − j 2 π ( f + f 0 ) t g ( t ) d t {\displaystyle {\mathcal {F}}[cos(2\pi f_{0}t)g(t)]={\frac {1}{2}}\int _{-\infty }^{\infty }e^{-j2\pi (f-f_{0})t}g(t)dt\ +\ {\frac {1}{2}}\int _{-\infty }^{\infty }e^{-j2\pi (f+f_{0})t}g(t)dt\!}
So F [ c o s ( 2 π f 0 t ) g ( t ) ] = 1 2 [ G ( f − f 0 ) + G ( f + f 0 ) ] {\displaystyle {\mathcal {F}}[cos(2\pi f_{0}t)g(t)]={\frac {1}{2}}[G(f-f_{0})+G(f+f_{0})]\!}
![{\displaystyle {\mathcal {F}}[cos(2\pi f_{0}t)g(t)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e2df42cc264afce1eae7d5bdcdcdda53516f72c)
, so
![{\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 f_{0}t}]g(t)e^{-j2\pi ft}dt\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aa67da17dc1d9f04fa8b72c6d0cf3a712f40c70f)
![{\displaystyle {\mathcal {F}}[cos(2\pi f_{0}t)g(t)]={\frac {1}{2}}\int _{-\infty }^{\infty }e^{-j2\pi (f-f_{0})t}g(t)dt\ +\ {\frac {1}{2}}\int _{-\infty }^{\infty }e^{-j2\pi (f+f_{0})t}g(t)dt\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/553ec68dad91d376d485594e651e85d3840e58d9)
![{\displaystyle {\mathcal {F}}[cos(2\pi f_{0}t)g(t)]={\frac {1}{2}}[G(f-f_{0})+G(f+f_{0})]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/752447d704f375e795524f3d8f8b5524e50b535c)