# Is the following property true?

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:

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

• 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? – Dilip Sarwate Nov 23 '18 at 18:56

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