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Find F [ c o s ( w 0 t ) g ( t ) ] {\displaystyle {\mathcal {F}}[cos(w_{0}t)g(t)]\!}
Recall w 0 = 2 π f 0 {\displaystyle w_{0}=2\pi f_{0}\!} , so
F [ c o s ( w 0 t ) g ( t ) ] = 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(w_{0}t)g(t)]={\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 + 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 ∫ − ∞ ∞ 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 = 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 = 1 2 G ( f − f 0 ) + 1 2 G ( f + f 0 ) {\displaystyle \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={\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={\frac {1}{2}}G(f-f_{0})\ +\ {\frac {1}{2}}G(f+f_{0})\!} So F [ c o s ( w 0 t ) g ( t ) ] = 1 2 [ G ( f − f 0 ) + G ( f + f 0 ) ] {\displaystyle {\mathcal {F}}[cos(w_{0}t)g(t)]={\frac {1}{2}}[G(f-f_{0})+G(f+f_{0})]\!}