Fourier Transform of $\sin(2 \pi f_0 t)$ using only the Fourier transform of $\cos(2 \pi f_0 t)$

Question

So I have to get the Fourier Transform of $\sin(2 \pi f_0 t)$ using differentiation and $$\mathcal{F} \{ \cos(2 \pi f_0 t\} = \frac{1}{2} (\delta (f+f_0) + \delta(f-f_0)) $$

Differentiation gives me $$\frac{d}{dt} \cos(2 \pi f_0 t\ ) = (-2\pi f_0) \sin(2 \pi f_0 t) $$

My "Solution" gives me this:

$$(j2\pi f) \mathcal{F}\{ \cos(2 \pi f_0 t\ ) \} = (-2\pi f_0) \mathcal{F} \{\sin(2 \pi f_0 t)\}$$

I do not understand what is happening here. I know that the Diff g'(x) of a Function g(x) when it is Fourier transformed gives me this: $\mathcal{F}\{g^{(n)}(t) \} = (j2\pi f)^n \cdot \hat g(f)$

But here both sides have have their Fourier transform taken. How does this work ?

Answer 1

You have an identity: $$\frac{d}{dt} \cos(2 \pi f_0 t\ ) = (-2\pi f_0) \sin(2 \pi f_0 t)$$

Then you take the Fourier transform of both sides to get another identity: $$\mathcal{F}\{\frac{d}{dt} \cos(2 \pi f_0 t\ )\} = \mathcal{F}\{(-2\pi f_0) \sin(2 \pi f_0 t)\}$$

Then you use a Fourier transform identity to rewrite the derivative in the left hand side: $$j 2\pi f \, \mathcal{F}\{ \cos(2 \pi f_0 t\ )\} = \mathcal{F}\{(-2\pi f_0) \sin(2 \pi f_0 t)\}$$

Answer 2

Nevermind I think I figured it out. $$sin(2 \pi f t) \cdot (-j2 \pi t) $$ is being used as the time function. So the left side of this is the sin:

$$\mathcal{F}\{g^{(n)}(t) \} = (j2\pi f)^n \cdot \hat g(f)$$

Is that correct ?