complex analysis - How to finish proof of ${1 over 2}+ sum_{k=1}^n cos(kvarphi ) = {sin({n+1 over 2}varphi)over 2 sin {varphi over 2}}$

I'm trying to prove the identity $${1 \over 2}+ \sum_{k=1}^n \cos(k\varphi ) = {\sin({n+1 \over 2}\varphi)\over 2 \sin {\varphi \over 2}}$$

What I've done so far:

From geometric series $\sum_{k=0}^{n-1}q = {1-q^n \over 1 - q}$ for $q = e^{i \varphi}$ and taking the real part on both sides I got

$$\sum_{k=0}^{n-1}\cos (k\varphi ) = {\sin {\varphi \over 2} - \cos (n\varphi) + \cos (n-1)\varphi \over 2 \sin {\varphi \over 2}}$$

I checked all my calculations twice and found no mistake. Then I used trigonometric identities to get

$${1 \over 2} + \sum_{k=1}^{n-1}\cos (k\varphi ) + {\cos n\varphi \over 2 } = {\sin \varphi \sin (n \varphi ) \over 2 \sin {\varphi \over 2}}$$

How to finish this proof? Is there a way to rewrite

$\sin \varphi \sin (n \varphi )$ as

$\sin({n+1 \over 2}\varphi) - \sin {\varphi \over 2} \cos (n \varphi)$?

Answer

There is a mistake in the real part.

$$\frac{q^n - 1}{q - 1} = \frac{e^{in\phi} - 1}{e^{i\phi} - 1} = \frac{e^{i(n-1/2)\phi} - e^{-i\phi/2}}{e^{i\phi/2} - e^{-i\phi/2}} = \frac{- i e^{i(n-1/2)\phi} + i e^{-i\phi/2}} {2\sin{\phi/2}}$$
the real part is

$$\frac{\sin ((n-1/2)\phi) + \sin(\phi/2)} {2\sin{\phi/2}}$$
yielding the right result.

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However, there is a simpler solution:
$$1 + 2\sum_{k=1}^n \cos k\phi = 1+ \sum_{k=1}^n \left(e^{i k\phi} + e^{-i k\phi}\right) = \sum_{k=-n}^n e^{i k\phi}$$
which simplifies to
$$e^{-i n\phi} \frac{1 - e^{i (2n+1)\phi}}{1 - e^{i\phi}} = \frac{e^{-i (n + 1/2)\phi} - e^{i (n + 1/2)\phi}} { e^{-i\phi/2} - e^{i\phi/2}} = \frac{\sin((n + 1/2)\phi)}{\sin(\phi/2)}$$