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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:


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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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
up vote 4 down vote accepted

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

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

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

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

$$=e^{\frac{inx}2}\frac{2i\sin\frac{(n+1)x}2}{2i\sin{\frac{x}2}}$$ 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… – lab bhattacharjee Oct 31 '12 at 11:10

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