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

Question

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?

Answer 1

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

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