Proving $\sum\limits_{k=0}^{n}\cos(kx)=\frac{1}{2}+\frac{\sin(\frac{2n+1}{2}x)}{2\sin(x/2)}$ I am being asked to prove that $$\sum\limits_{k=0}^{n}\cos(kx)=\frac{1}{2}+\frac{\sin(\frac{2n+1}{2}x)}{2\sin(x/2)}$$
I have some progress made, but I am stuck and could use some help.
What I did:
It holds that
$$\sum\limits_{k=0}^{n}\cos(kx)=\sum\limits_{k=0}^{n}Re(\cos(kx))=\sum\limits_{k=0}^{n}Re(\cos(x)^{k})=Re(\sum\limits_{k=0}^{n}\cos(x)^{k})=Re\left(\cos(0)\cdot\frac{\cos(x)^{n}-1}{\cos(x)-1}\right)=Re\left(\frac{\cos(x)^{n}-1}{\cos(x)-1}\right)
$$
For any $z_{1},z_{2}\in\mathbb{C}$ we have it that if $z_{1}=a+bi,z_{2}=c+di$
then $$\frac{z_{1}}{z_{2}}=\frac{z_{1}\overline{z2}}{|z_{2}|^{2}}=\frac{(a+bi)(c-di)}{|z_{2}|^{2}}=\frac{ac-bd+i(bc-ad)}{|z_{2}|^{2}}$$
hence $$Re\left(\frac{z_{1}}{z_{2}}\right)=\frac{Re(z_{1})Re(z_{2})-Im(z_{1})Im(z_{2})}{|z_{2}|^{2}}$$
Thus, $$Re\left(\frac{\cos(x)^{n}-1}{\cos(x)-1}\right)=\frac{(\cos(nx)-1)(\cos(x)-1)-\sin(nx)\sin(x)}{(\cos(x)-1)^{2}+\sin^{2}(x)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{\cos^{2}(x)-2\cos(x)+1+\sin^{2}(x)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2\cos(x)+2}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2(\cos(x)-1)}=
\frac{=\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2(-2\cdot\sin^{2}(x/2))}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{4\sin^{2}(x/2)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{4\sin^{2}(x/2)}=\frac{\cos(x(n+1))-\cos(nx)-\cos(x)+1}{4\sin^{2}(x/2)}
$$
This is the part where I am stuck, I would appriciate any help or hint on how to continue.
Edit: Given the corrections by André I get:
$$(\cos(nx+x)-1)(\cos(x)-1)+\sin(nx+x)\sin(x)=\cos(nx+x)\cos(x)-\cos(nx)-\cos(x)+1+\sin(nx+x)\sin(x)$$
so $$\cos(nx+x)\cos(x)+\sin(nx+x)\sin(x)=\cos(xn+x-x)-\cos(nx)=0$$
Edit 2: I found anoter mistake in the above, I will try to correct
Edit 3: When multiplying correctly the above it works out :-)
 A: Here is the simplest and fastest way I've found.
\begin{align}
1+2\sum_{k=1}^n\cos(\theta) & = \sum_{k=-n}^n e^{ik\theta} \\
& = \frac{e^{i(n+1/2)\theta}-e^{-i(n+1/2)\theta}}{e^{i\theta/2}-e^{-i\theta/2}} \\
& = \frac{\sin((n+1/2)\theta)}{\sin(\theta/2)} \\
\end{align}
which can easily be rearranged to get the desired identity.
See Lagrange's trigonometric identities.
A: There are faster ways to go, but if you want to continue along the path you have taken, you are quite close to the end. Please see the remark for a couple of small things that need to be corrected in the calculation.  Essentially the same trigonometric tricks continue to work.
By a double angle formula for the cosine, we have 
$$\cos x=1-2\sin^2(x/2),$$ so 
$$\frac{1-\cos x}{4\sin^2(x/2)}=\frac{1}{2},$$ part of what you are aiming for. However, this could have been obtained in a simpler way in the third displayed formula after the "Thus," in the OP.
And the front part will "simplify" by a difference of $\cos$ formula, obtained from $$\cos(a+b)=\cos a\cos b-\sin a\sin b,\qquad \cos(a-b)=\cos a\cos b+\sin a\sin b.$$ Subtract. We get 
$$\cos(a+b)-\cos(a-b)=-2\sin a\sin b.$$
Let $a+b=x(n+1)$, and $a-b=nx$. So $a=\dfrac{x(2n+1)}{2}$ and $b=\dfrac{x}{2}$.
Remark: There is a little sign glitch in the calculation of $(a+bi)(c-di)$. Note that the real part should be $ac+bd$. Also, when you are summing the geometric progression, we need $\text{cis}^{n+1}$.
A: $\require{cancel}$
Start by being inspired by the product-to-sum identity:
$$
2 \cos a \sin b = \sin(a+b) - \sin(a-b) ,
$$
then naturally let $a = kx$, next, in order for the right-hand side terms to be eliminated when multiple equalities are superimposedly added, which eventually making the remaining terms as few as possible, the value of $((a+b)-(a-b))$ should be the smallest integer multiple of $x$, hence let $b=\dfrac{x}{2} ,$ so the equality changes as
$$
2 \cos kx \sin\dfrac{x}{2}
=
\sin\left(k+\dfrac{1}{2}\right)x - \sin\left(k-\dfrac{1}{2}\right)x .
$$
And then, use a table to visualize the cumulative destructive process of the right-hand side terms:
$$\begin{array}{c|cc}
{\small(+)~}\ k+\dfrac{1}{2} & \cancel{\dfrac{3}{2}} & \cancel{\dfrac{5}{2}} & \cancel{\dfrac{7}{2}} & \cdots & \cancel{n-\dfrac{1}{2}} & n+\dfrac{1}{2}\\
\hline
{\small(-)~}\ k-\dfrac{1}{2} & \dfrac{1}{2} & \cancel{\dfrac{3}{2}} & \cancel{\dfrac{5}{2}} & \cdots & \cancel{n-\dfrac{3}{2}} & \cancel{n-\dfrac{1}{2}}
\end{array}
\\
⇓
\\
\sum_{k=1}^n \left[\sin\left(k+\dfrac{1}{2}\right)x - \sin\left(k-\dfrac{1}{2}\right)x\right]
 = \sin\left(n+\dfrac{1}{2}\right)x - \sin\dfrac{x}{2} .
$$
Thus
$$
\sum_{k=1}^n \cos kx
 = \left.\left[\sin\left(n+\dfrac{1}{2}\right)x - \sin\dfrac{x}{2}\right] \middle/
   \left(2 \sin\dfrac{x}{2}\right)\right. .
$$

P.S. Similarly,
$$\sum_{k=1}^n \sin kx
 = \left.\left[\cos\dfrac{x}{2} - \cos\left(n+\dfrac{1}{2}\right)x\right] \middle/
   \left(2 \sin\dfrac{x}{2}\right)\right.$$
can be derived from $2 \sin a \sin b = \cos(a-b) - \cos(a+b)$ through the same steps.
A: $$\sum_{0\le r\le n}e^{ikx}=\frac{e^{i(n+1)x}-1}{e^{ix}-1}$$
$$=\frac{e^{\frac{i(n+1)x}2}}{e^{\frac{ix}2}}\frac{\left(e^{\frac{i(n+1)x}2}-e^{-\frac{i(n+1)x}2}\right)}{\left( e^{\frac{ix}2}-e^{-\frac{ix}2}\right)}$$
$$=e^{\frac{inx}2}\frac{2i\sin\frac{(n+1)x}2}{2i\sin{\frac{x}2}}$$ as $e^{iy}-e^{-iy}=2i\sin y,$
$$=(\cos\frac{nx}2+i\sin\frac{nx}2)\frac{\sin\frac{(n+1)x}2}{\sin{\frac{x}2}}$$ using Euler's identity. 
Its real part is $$\cos\frac{nx}2 \frac{\sin\frac{(n+1)x}2}{\sin{\frac{x}2}}=\frac{2\cos\frac{nx}2\sin\frac{(n+1)x}2}{2\sin{\frac{x}2}}=\frac{\sin\frac{(2n+1)x}2+\sin{\frac{x}2}}{2\sin{\frac{x}2}}$$ using $2\sin A\cos B=\sin(A+B)+\sin(A-B)$
A: Just multiply both sides by $2\sin(x/2)$ and use Briggs' formula:
$$ 2 \sin(x/2)\cos(kx) = \sin((k+1/2)x)-\sin((k-1/2)x)$$
to get a telescoping sum.
A: $$\begin{align}
\sum_{k=0}^n2\cos k\theta&=\sum_{k=0}^n(e^{i\theta k}+e^{-i\theta k})\\
&=\frac{e^{i\theta(n+1)}-1}{e^{i\theta}-1}+\frac{e^{-i\theta(n+1)}-1}{e^{-i\theta}-1}\\
&=\frac{e^{in\theta}+e^{-in\theta}-e^{i(n+1)\theta}-e^{-i(n+1)\theta}-e^{i\theta}-e^{-i\theta}+2}{2-e^{i\theta}-e^{-i\theta}}\\
&=\frac{2\cos n\theta-2\cos(n+1)\theta+2-2\cos\theta}{2-2\cos\theta}\\
&=\frac{\cos n\theta-\cos(n+1)\theta+1-\cos\theta}{1-\cos\theta}=\frac{\sin(n+\frac 1 2)\theta}{\sin\frac\theta2}+1
\end{align}$$
A: We wish the compute the sum
$$S=\sum_{k=0}^n \cos(k\theta)$$
We can use the properties of the Chebyshev polynomials of the first kind. Recall that the $n$th Chebyshev polynomial $T_n$ satisfies the relation
$$T_n(\cos\theta)=\cos(n\theta).$$
With this definition your sum can be restated as
$$S=\sum_{k=0}^n T_k(\cos\theta)$$
We can let $\cos(\theta):=x$ and deduce that
$$T_0=1 \text{ and }T_1(x)=x$$
And because of the following identity,
$$\cos((n+1)\theta) + \cos((n-1)\theta) = 2\cos (\theta) \cos (n\theta)$$
Which can be proved via sum-difference formulas for cosine,  we can again let $\cos(\theta):=x$ and shift our index to obtain
$$T_{n+2}=2xT_{n+1}-T_n$$
So let's return to the sum
$$g_n(x)=\sum_{k=0}^n T_k(x)$$
We can start by summing both sides of the recurrence relation:
$$\sum_{k=0}^n T_{k+2}=2x\sum_{k=0}^nT_{k+1}-\sum_{k=0}^nT_k$$
We can mix and match different terms in the sums:
$$g_n-T_0-T_1+T_{n+1}+T_{n+2}=2x(g_n-T_0+T_{n+1})-g_n$$
Some algebra,
$$2(1-x)g_n=T_0-2xT_0+T_1-T_{n+1}+2xT_{n+1}-T_{n+2}$$
However, using the recurrence, one can notice $2xT_{n+1}-T_{n+2}=T_n.$ Furthermore, $T_0=1$ and $T_1=x$, as stated before. Thus,
$$g_n(x)=\frac{1}{2}+\frac{T_n(x)-T_{n+1}(x)}{2(1-x)}$$
Letting $x=\cos(\theta)$,
$$g_n(\cos\theta)=S=\frac{1}{2}+\frac{\cos(n\theta)-\cos((n+1)\theta)}{2(1-\cos\theta)}$$
It remains to be shown that
$$\frac{\cos(n\theta)-\cos((n+1)\theta)}{1-\cos\theta}=\frac{\sin\left(\frac{2n+1}{2}x\right)}{\sin(x/2)}$$
Which I'm actually not sure how to verify, though I have checked it in Desmos and on Wolfram.
