Cfrgtkky

I am being asked to prove that $$sumlimits_{k=0}^{n}cos(kx)=frac{1}{2}+frac{sin(frac{2n+1}{2}x)}{2sin(x/2)}$$

I have some progress made, but I am stuck and could use some help.

What I did:

It holds that
$$sumlimits_{k=0}^{n}cos(kx)=sumlimits_{k=0}^{n}Re(cos(kx))=sumlimits_{k=0}^{n}Re(cos(x)^{k})=Re(sumlimits_{k=0}^{n}cos(x)^{k})=Releft(cos(0)cdotfrac{cos(x)^{n}-1}{cos(x)-1}right)=Releft(frac{cos(x)^{n}-1}{cos(x)-1}right)
$$

For any $z_{1},z_{2}inmathbb{C}$ 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 $$Releft(frac{z_{1}}{z_{2}}right)=frac{Re(z_{1})Re(z_{2})-Im(z_{1})Im(z_{2})}{|z_{2}|^{2}}$$

Thus, $$Releft(frac{cos(x)^{n}-1}{cos(x)-1}right)=frac{(cos(nx)-1)(cos(x)-1)-sin(nx)sin(x)}{(cos(x)-1)^{2}+sin^{2}(x)}=frac{cos(nx)cos(x)-cos(nx)-cos(x)+1-sin(nx)sin(x)}{cos^{2}(x)-2cos(x)+1+sin^{2}(x)}=frac{cos(nx)cos(x)-cos(nx)-cos(x)+1-sin(nx)sin(x)}{-2cos(x)+2}=frac{cos(nx)cos(x)-cos(nx)-cos(x)+1-sin(nx)sin(x)}{-2(cos(x)-1)}=
frac{=cos(nx)cos(x)-cos(nx)-cos(x)+1-sin(nx)sin(x)}{-2(-2cdotsin^{2}(x/2))}=frac{cos(nx)cos(x)-cos(nx)-cos(x)+1-sin(nx)sin(x)}{4sin^{2}(x/2)}=frac{cos(nx)cos(x)-cos(nx)-cos(x)+1-sin(nx)sin(x)}{4sin^{2}(x/2)}=frac{cos(x(n+1))-cos(nx)-cos(x)+1}{4sin^{2}(x/2)}
$$

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 :-)

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
$$cos x=1-2sin^2(x/2),$$ so
$$frac{1-cos x}{4sin^2(x/2)}=frac{1}{2},$$ 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 $$cos(a+b)=cos acos b-sin asin b,qquad cos(a-b)=cos acos b+sin asin b.$$ Subtract. We get
$$cos(a+b)-cos(a-b)=-2sin asin b.$$
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 $text{cis}^{n+1}$.

Just multiply both sides by $2sin(x/2)$ 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.

$$sum_{0le rle n}e^{ikx}=frac{e^{i(n+1)x}-1}{e^{ix}-1}$$

$$=frac{e^{frac{i(n+1)x}2}}{e^{frac{ix}2}}frac{left(e^{frac{i(n+1)x}2}-e^{-frac{i(n+1)x}2}right)}{left( e^{frac{ix}2}-e^{-frac{ix}2}right)}$$

$$=e^{frac{inx}2}frac{2isinfrac{(n+1)x}2}{2isin{frac{x}2}}$$ as $e^{iy}-e^{-iy}=2isin y,$

$$=(cosfrac{nx}2+isinfrac{nx}2)frac{sinfrac{(n+1)x}2}{sin{frac{x}2}}$$ using Euler's identity.

Its real part is $$cosfrac{nx}2 frac{sinfrac{(n+1)x}2}{sin{frac{x}2}}=frac{2cosfrac{nx}2sinfrac{(n+1)x}2}{2sin{frac{x}2}}=frac{sinfrac{(2n+1)x}2+sin{frac{x}2}}{2sin{frac{x}2}}$$ using $2sin Acos B=sin(A+B)+sin(A-B)$

Here is the simplest way I've found.
begin{align}
1+2sum_{k=1}^ncos(theta) & = sum_{k=-n}^n e^{iktheta} \
& = frac{e^{i(n+1/2)theta}-e^{-i(n-1/2)theta}}{e^{itheta /2}-e^{-itheta / 2}} \
& =frac{sin(n+1/2)theta}{sin(theta/2)} \
end{align}
which can easily be rearranged to get the desired identity.
See https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Lagrange's_trigonometric_identities