Exponential function goes to sin function - but how?

Hi op.

There is an identity that says:

[latex]\frac{e^{iz}-e^{-iz}}{2i}=\sin(z)[/latex]

With this in mind,

[latex]e^{iz}-e^{-iz}=2i\sin(z)[/latex]

If we let [latex]z=\frac{\Delta \omega}{2}t[/latex] we have:

[latex]e^{i\frac{\Delta \omega}{2}t}-e^{-i\frac{\Delta \omega}{2}t}}=2i\sin(\frac{\Delta \omega}{2}t)[/latex]

No if we consider your problem we have:

[latex]\frac{1}{it}\left(e^{i\left(\frac{\Delta \omega}{2}t\right)}-e^{-i\left(\frac{\Delta \omega}{2}t\right)}\right)[/latex]

We can therefore replace all that stuff in the brackets to get:

[latex]\frac{1}{it}\left(2i\sin\left(\frac{\Delta \omega}{2}t\right)[/latex]

Cancelling the [latex]i[/latex] gives us what we are looking for.

[latex]\frac{2\sin\left(\frac{\Delta \omega t}{2}\right)}{t}[/latex]

No probs. Glad I could help.