Reduce formula using Euler's?

Question

I am performing a self-study, and I am lost as to a derivation that has taken place.

I basically started with this equation:

$$ \Upsilon(\phi) = e^{-j\frac{N-1}{2}\phi} \ \Big[ \frac{1 - e^{j N \phi}}{1 - e^{j \phi}} \Big] $$

Somehow, the result is that this above equation can be reduced to this:

$$ \Upsilon(\phi) = \frac{\sin(\frac{N \phi}{2})}{\sin(\frac{\phi}{2})} $$

I am at a loss as to how this was accomplished.

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What I have tried: I understand and have tried to use euler's formulas, $\sin(x) = \frac{e^{j \ x} - e^{-j x}}{2 \ j}$, etc, but I was only able to recover the numerator of the first equation. I am not sure really if there is a simpler way that I am simply not getting. Would appreciate any insight. I also tried expanding by multiplying by the conjugate of the denominator, but still no luck... Thanks.

Answer 1

Hint Use this equality $$1-e^{jx}=e^{jx/2}(e^{-jx/2}-e^{jx/2})=-2j\sin(x/2)e^{jx/2}$$

Answer 2

$$e^{-j\frac{N-1}{2}\phi}=e^{-j\frac{N}{2}\phi}e^{j\frac{1}{2}\phi}$$

Now take this and apply it to the full equation:

$$\Upsilon(\phi) = e^{-j\frac{N-1}{2}\phi} \left[ \frac{1 - e^{j N \phi}}{1 - e^{j \phi}} \right]$$

$$\Upsilon(\phi)=e^{-j\frac{N}{2}\phi}\left[ \frac{(1 - e^{j N \phi})}{e^{-j\frac{1}{2}\phi} - e^{j \frac{1}{2}\phi}} \right]$$

Can you take it from here?