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I was looking at a solution of a Fourier Transform question and following property was used, if: $$ x(t)\rightarrow X(jw) $$ then:

$$ e^{jw_ot}x(t)\rightarrow X(j(w-w_0)) $$ $$ x(t)\sin(w_0t)\rightarrow \frac{1}{2j}X(j(w-w_0)) - \frac{1}{2j}X(j(w+w_0)) $$ If the above statements are true, can we say that for cos:

$$ x(t)\cos(w_0t)\rightarrow \frac{1}{2}X(j(w-w_0)) - \frac{1}{2}X(j(w-w_0)) $$

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    $\begingroup$ You ask whether $$ x(t)cos(w_0t)\rightarrow \frac{1}{2j}X(j(w-w_0)) - \frac{1}{2j}X(j(w-w_0)) $$ is a true statement. Have you noticed that the right hand side is of the form $\alpha-\alpha$ and so must equal $0$? That is, you are asking whether $$ x(t)cos(w_0t)\rightarrow 0$$ is true. Can you answer your question for yourself? $\endgroup$ – Dilip Sarwate Nov 23 '18 at 18:56
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For $\cos$, assuming $\omega_0$ is real, the identity is: $$ x(t) \cos(\omega_0)t = \frac{1}{2} X(j(\omega - \omega_0)) + \frac{1}{2} X(j(\omega + \omega_0)) $$

This is because $$ \cos(\omega_0 t) = \frac{1}{2}e^{j \omega_0 t} + \frac{1}{2}e^{-j \omega_0 t} $$

Use this expression with your first identity and the superposition property of the Fourier transform to arrive at this result.

As an aside, also note that $$ \sin(\omega_0 t) = \frac{1}{2j}e^{j \omega_0 t} - \frac{1}{2j}e^{-j \omega_0 t} $$ By the same reasoning, this is how you arrive at your second identity.

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