How to finish proof of $ {1 \over 2}+ \sum_{k=1}^n \cos(k\varphi ) = {\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)
 $?

 A: 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.

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
$$
e^{-i n\phi} \frac{1 - e^{i (2n+1)\phi}}{1 - e^{i\phi}}
= \frac{e^{-i (n + 1/2)\phi} - e^{i (n + 1/2)\phi}}
{ e^{-i\phi/2} - e^{i\phi/2}} = \frac{\sin((n + 1/2)\phi)}{\sin(\phi/2)}
$$
A: Although this isn't answering your question exactly, you might like to know that there is an easier way of proving this. Just multiply both sides by the denominator of the right-hand side and then use $\sin(A+B)-\sin(A-B)=2$cosAsinB. You now have a telescoping series which leads quickly to the result.
A: What you are trying to prove is not correct. Let $n=1$, $\varphi = \frac{\pi}{3} $.
