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My question is whether someone could provide a proof for the following identity:

$$ \frac{1 - e^{int}}{1 - e^{it}} = e^{i(n-1)t/2} \frac{\sin(nt/2)}{\sin(t/2)} $$

Motivation:

The left hand side is the sum of the geometric series of complex exponentials, which is itself the fourier transform of a finite Dirac Comb. This sum appears in the derivation of the Frauenhofer Diffraction pattern in Physics.

$$ \mathcal{F}[\sum_{n=0}^{n=N-1} \delta(t - n)] \sum_{n=0}^{n=N-1}e^{int} = \frac{1 - e^{int}}{1 - e^{it}} $$

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$$\frac{1 - e^{int}}{1 - e^{it}} =\frac{e^{int/2}(e^{-int/2}-e^{int/2})}{e^{it/2}(e^{-it/2}-e^{it/2})}=e^{i(n-1)t/2}\frac{-2i\sin(nt/2)}{-2i\sin(t/2)}= e^{i(n-1)t/2} \frac{\sin(nt/2)}{\sin(t/2)}$$ I think you're missing a factor of $1/2$ in your final exponential?

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Hint

$$1-e^{ix}=e^{i\frac{x}{2}}(e^{-i\frac{x}{2}}-e^{i\frac{x}{2}})$$

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