# 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 $$\sum\limits_{k=0}^{n}\cos(kx)=\sum\limits_{k=0}^{n}Re(\cos(kx))=\sum\limits_{k=0}^{n}Re(\cos(x)^{k})=Re(\sum\limits_{k=0}^{n}\cos(x)^{k})=Re(\cos(0)\cdot\frac{\cos(x)^{n}-1}{\cos(x)-1})=Re(\frac{\cos(x)^{n}-1}{\cos(x)-1})$$

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 $$Re(\frac{z_{1}}{z_{2}})=\frac{Re(z_{1})Re(z_{2})-Im(z_{1})Im(z_{2})}{|z_{2}|^{2}}$$

Thus, $$Re(\frac{\cos(x)^{n}-1}{\cos(x)-1})=\frac{(\cos(nx)-1)(\cos(x)-1)-\sin(nx)\sin(x)}{(\cos(x)-1)^{2}+\sin^{2}(x)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{\cos^{2}(x)-2\cos(x)+1+\sin^{2}(x)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2\cos(x)+2}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2(\cos(x)-1)}= \frac{=\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2(-2\cdot\sin^{2}(x/2))}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{4\sin^{2}(x/2)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{4\sin^{2}(x/2)}=\frac{\cos(x(n+1))-\cos(nx)-\cos(x)+1}{4\sin^{2}(x/2)}$$

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 :-)

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– Hans Lundmark Oct 31 '12 at 9:07
Hmm... I'm unsure about this being a duplicate. (a) the formulas are not exactly the same, and (b) a large part of this question is helping the OP with their attempt. – Douglas S. Stones Oct 31 '12 at 10:36
Replaced \Sigma by \sum. – Did Oct 31 '12 at 11:36

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 $\text{cis}^{n+1}$.

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I am having a little problem with this, I get the answer with a minus sign. can you please help me with telling me where I did wrong ? $\cos(x(n+1))-\cos(nx)=-2\sin(xn+\frac{x}{2})\sin(\frac{x}{2})$ thus $\frac{\cos(x(n+1))-\cos(nx)}{4\sin^{2}(x/2)}=\frac{-2\sin(xn+\frac{x}{2})\sin(\‌​frac{x}{2})}{4\sin^{2}(x/2)}=-\frac{\sin(xn+\frac{x}{2})}{2\sin(\frac{x}{2})}$ – Belgi Oct 31 '12 at 10:09
Am I doing the last step wrong or do you think I have a previous minus sign error somewhere ? – Belgi Oct 31 '12 at 10:17
I corrected that error after you last comment, but its still not working right: $(\cos(nx)-1)(\cos(x)-1)+\sin(nx)\sin(x)=\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1+\sin‌​(nx)\sin(x)$ so $\cos(nx)\cos(x)+\sin(nx)\sin(x)=\cos(xn-x)-\cos(nx)=\cos(x(n-1))-\cos(nx)$. If $a+b=xn-x,a-b=xn$ then $a=xn-\frac{x}{2},b=-\frac{x}{2}$ so $-2\sin(a)\sin(b)=-2\sin(x(n-1))\sin(-\frac{x}{2})=+2\sin(x(n-1))\sin(\frac{x}{2‌​})$. So when $\sin(\frac{x}{2})$ cancels out we get $\frac{\sin(x(n-1))}{2\sin(\frac{x}{2})}$ – Belgi Oct 31 '12 at 10:38
I'll try to correct the second mistake you found, thanks! – Belgi Oct 31 '12 at 10:39
I still have a problem after your last correction: $(\cos(nx+x)-1)(\cos(x)-1)+\sin(nx+x)\sin(x)=\cos(nx+x)\cos(x)-\cos(nx)-\cos(x)+‌​1+\sin(nx+x)\sin(x)$ so $\cos(nx+x)\cos(x)+\sin(nx+x)\sin(x)=\cos(xn+x-x)-\cos(nx)=0$ – Belgi Oct 31 '12 at 10:44

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

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

$$=e^{\frac{inx}2}\frac{2i\sin\frac{(n+1)x}2}{2i\sin{\frac{x}2}}$$ 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 $2\sin A\cos B=\sin(A+B)+\sin(A-B)$

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Can you please provide some insight on how did you know to factor the denomenator numerator in that way in the second line ? – Belgi Oct 31 '12 at 10:56
@Belgi, we know $e^{2y}-1=e^y(e^y-e^{-y})$ and my target was to utilize $e^{iy}-e^{-iy}=2i\sin y$. You may have a look into the last approach of math.stackexchange.com/questions/183859/… – lab bhattacharjee Oct 31 '12 at 11:10

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.

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Isn't there a nice way to continue my work ? – Belgi Oct 31 '12 at 9:18