# Fourier Transform of $\sin(3x)$ using a delta relation

I want to find the Fourier Transform of $\sin(3x)$ which I can do normally however I need to find this using this delta relation:

$$\delta(k) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{ikx}dx$$

Basically just need to present in simplified form.

Any ideas?

• hint: $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$ – Veridian Dynamics May 11 '17 at 14:11
• How do you define the Fourier transform of $e^{ix}$ ? The formula you linked to $\delta(k) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{ikx}dx$ is not rigorous and correct. If you suppose that the Fourier inversion theorem works for such things (distributions) then look at the inverse Fourier transform of $\delta(x-a)$ – reuns May 11 '17 at 14:14
• @user1952009: Maybe if you use fourier transform, it will multiply a exp(-ikx) with the function and then you use the relation to identify rhs and change it in 𝛿(k) form? – DSOU May 12 '17 at 0:17

$$\delta(k) = \frac{1}{2\pi} \int_{-\infty}^\infty 1\,.e^{ikx}dx$$ implies

$$\delta(x)\stackrel{\mathcal{F}}\longleftrightarrow 1$$

so using the time shifting property, we have

$$\delta(x-x_0)\stackrel{\mathcal{F}}\longleftrightarrow e^{-ix_0\omega}$$

This can be used in conjunction with the duality property to yield

$$e^{i\omega_0x}\stackrel{\mathcal{F}}\longleftrightarrow 2\pi\delta(\omega-\omega_0)$$

Thus, $$\sin(3x)=\frac{e^{i3x}-e^{-i3x}}{2i}\stackrel{\mathcal{F}}\longleftrightarrow \frac{2\pi(\delta(\omega-3)-\delta(\omega+3))}{2i}$$

or

$$\sin(3x)\stackrel{\mathcal{F}}\longleftrightarrow \pi i(\delta(\omega+3)-\delta(\omega-3))$$