1. The problem statement, all variables and given/known data

Hi :smile:

Find the (finite,ultimate,definitive,peremptory,eventua,con clusive) sum:

a) cos^2x+cos^22x+...+cos^2nx ; b) sin^2x+sin^22x+...+sin^2nx

2. Relevant equations

z=A+Bi

3. The attempt at a solution

A=cos^2x+cos^22x+...+cos^2nx

B=sin^2x+sin^22x+...+sin^2nx

z=cos^2x+cos^22x+...+cos^2nx+i(sin^2x+sin^22x+...+ sin^2nx)=(cos^2x+isin^2x)+(cos^22x+isin^22x)+...+( cos^2nx+isin^2nx)=(cos^2x+isin^2x)+
(cos^2x+isin^2x)^2+...+(cos^2x+isin^2x)^n

How will I continue?

You have a geometric progression.
To find the sum of a GP is not difficult. However you will have a complex denominator which is not fun to deal with.
multiply the top and bottom by the complex conjugate of the denominator to geta real denominator.
Equate real and imaginary parts.

What should I multiply with? Can you write it please? Thank you.

Hi Yoda.

Sorry, but this step

(cos^2x+isin^2x)+(cos^22x+isin^22x)+...+(cos^2nx+i sin^2nx)
=(cos^2x+isin^2x)+(cos^2x+isin^2x)^2+...+(cos^2x+i sin^2x)^n


does not look valid to me.

Hi Yoda.

Sorry, but this step




does not look valid to me.

http://en.wikipedia.org/wiki/De_Moivre's_formula

The End.







Anyway OP, from there... consider

S = D + D^2 + D^3 + ... + D^n
= D(1 + D + D^2 + ... + D^{n-1})
= D(1 + S - D^{n})

Solving for S we get:

S = \frac{D - D^{n+1}}{1-D}

I think this will answer your question.

Yes, that is great. Just I have one question.

How did you get the third equation D(1+S-D^n)? Why D^n?

Absane:

De Moivre’s formula states

(\cos{x}+\mathrm{i}\sin{x})^n=\cos{nx}+\mathrm{i}\ sin{x}

NOT

(\cos^2{x}+\mathrm{i}\sin^2{x})^n=\cos^2{nx}+ \mathrm{i}\sin^2{nx}

3. The attempt at a solution

A=cos^2x+cos^22x+...+cos^2nx

B=sin^2x+sin^22x+...+sin^2nx

z=cos^2x+cos^22x+...+cos^2nx+i(sin^2x+sin^22x+...+ sin^2nx)\\
=(cos^2x+isin^2x)+(cos^22x+isin^22x)+...+(cos^2nx+ isin^2nx)
At this point you can use:
\cos( \alpha )^2 = \frac{1}{2} \left( 1 + \cos( 2 \alpha ) \right)

\sin( \alpha )^2 = \frac{1}{2} \left( 1 - \cos( 2 \alpha ) \right)
This gets rather messy if you stick with the elementary trig. functions though. I think substituting:
\cos( \alpha ) = \frac{1}{2} \left( e^{i \alpha} + e^{- i \alpha} \right)
into (a) right from the beginning is a simpler approach.

Absane had great tip, but I don't know where to go out from there, or I don't know how he get the 3rd and the 4th step...

przyk, if I use your tip, I will get mesed up. Anyway thanks for the tip.


algebraic topology please check again wikipedia, or some book, you'll see that you're wrong.

Yoda, please read the Wikipedia carefully. The Wikipedia article clearly states:

(\cos{x}+\mathrm{i}\sin{x})^n=\cos{nx}+\mathrm{i}\ sin{x}

\cos^2{x} in your formula IS NOT THE SAME AS \cos{x} in the Wikipedia formula.

And neither is \sin^2{x} the same as \sin^2{nx}

Check the link up from Absane...
Also, if you know where the De Moivre's formula comes from, you'll know that you're not right...

Check the link up from Absane
I did. Did you? Then I suggest you check the link yourself and see for yourself.

Go and get some book then. Sorry for insulting you, but are you here to make me nervous and iritating?

Okay, for the last time, I’m going to say this.

(1) (\cos{x}+\mathrm{i}\sin{x})^n=\cos{nx}+\mathrm{i}\ sin{x}
is correct. This is De Moivire’s formula. It’s just cos and sin, no ².

(2) (\cos^2{x}+\mathrm{i}\sin^2{x})^n=\cos^2{nx}+ \mathrm{i}\sin^2{nx}
is wrong. This is not De Moivire’s formula. You have cos² and cos² this time, therefore it’s different from the other formula.

You still don’t see the difference, do you?

Well, I am sorry folks, but DeMoivre in my high-school text is given as (\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n \theta. (I have no idea what Wiki says, but it a bad idea to regard it as a deity.)

That is correct, QuarkHead. What you have there is De Moivre’s formula, and it’s the correct one.

Yoda on the other hand has been trying to do something different, and I’ve been trying to show him why what he’s doing is wrong.

Ahhh. I see the difference now. Sorry, you're right. But how will I solve this task?

But how will I solve this task?
Are you comfortable with complex exponentials - ie. have you seen that e^{i x} \equiv \cos( x ) + i \sin( x )? Using this and Absane's formula, it's not too difficult (meaning I did it this way without any need for aspirin). I don't see another straightforward way of solving this.

But de moevier formula is not valid here since we have cos^2x+isin^2x and it is not cosx+isinx.

Basically, if you had a sum like cos(x)+cos(2x)+...+cos(nx), you could apply DeMoivre's theorem and sum it. But your series is more complicated because all of the terms are squared. So what you should do, as suggested above, is use the following conversions:

cos^2(kx)=\frac{cos(2kx)+1}{2},\ sin^2(kx)=\frac{1-cos(2kx)}{2} to convert each of the terms in your series into terms with no squares in them. At that point you have all the tools you need to compute the sum.

(\frac{1+cos2x}{2}+i\frac{1-cos2x}{2})+(\frac{1+cos4x}{2}+i\frac{1-cos4x}{2})+(\frac{1+cosnkx}{2}+i\frac{1-cosnkx}{2})

\frac{1}{2}(1+cos2x+i(1-cos2x))+\frac{1}{2}(1+cos2x+i(1-cos2x))+\frac{1}{2}(1+cos4x+i(1-cos4x))+\frac{1}{2}(1+cosnkx+i(1-cosnkx))

How will I get rid of "1"?

How will I go next?

somebody?

Separate the 1's out and sum them separately.

\frac{3+cos2x+cos4x+cosnx}{2}+

I don't know how to this whole this thing.

Here’s my suggestion. Forget about complex numbers. Just deal with the sums directly:

A=\cos^2{x}+\cos^2{2x}+\ldots+\cos^2{nx}

B=\sin^2{x}+\sin^2{2x}+\ldots+\sin^2{nx}

\therefore\ A+B=(\cos^2{x}+\sin^2{x})+(\cos^2{2x}+\sin^2{2x})+ \ldots+(\cos^2{nx}+\sin^2{nx})

\color{white}.\hspace{57mm}.=\underbrace{1+1+\ldot s+1}_{n\mbox{ of them}}

\color{white}.\hspace{57mm}.=n

\color{white}.\quad. A-B=(\cos^2{x}-\sin^2{x})+(\cos^2{2x}-\sin^2{2x})+\ldots+(\cos^2{nx}-\sin^2{nx})

\color{white}.\hspace{57mm}.=\cos{2x}+\cos{4x}+\ld ots+\cos{2nx}

Now try and find a formula for the last sum. You have have to consider n even and n odd separately. Once you have a formula, you can solve the simultaneous equations for A and B.

\frac{3+cos2x+cos4x+cosnx}{2}+

I don't know how to this whole this thing.
First, forget about A + i B and just calculate A on its own. Now try adding a corresponding i \sin(k x) for every \cos(k x) in your summation. At the end you'll be able to get A back by taking the real part of your result.

Here’s my suggestion. Forget about complex numbers. Just deal with the sums directly:

A=\cos^2{x}+\cos^2{2x}+\ldots+\cos^2{nx}

B=\sin^2{x}+\sin^2{2x}+\ldots+\sin^2{nx}

\therefore\ A+B=(\cos^2{x}+\sin^2{x})+(\cos^2{2x}+\sin^2{2x})+ \ldots+(\cos^2{nx}+\sin^2{nx})

\color{white}.\hspace{57mm}.=\underbrace{1+1+\ldot s+1}_{n\mbox{ of them}}

\color{white}.\hspace{57mm}.=n

\color{white}.\quad. A-B=(\cos^2{x}-\sin^2{x})+(\cos^2{2x}-\sin^2{2x})+\ldots+(\cos^2{nx}-\sin^2{nx})

\color{white}.\hspace{57mm}.=\cos{2x}+\cos{4x}+\ld ots+\cos{2nx}

Now try and find a formula for the last sum. You have have to consider n even and n odd separately. Once you have a formula, you can solve the simultaneous equations for A and B.

Thank you. Here is the formula:

\frac{1-cos^n^+^1x}{1-cosx} Like this??

I don’t think so. That formula is for a geometric progression, whereas the sum \cos{2x}+\cos{4x}+\ldots+\cos{2nx} is not a geometric progression. We must turn it into a geometric progression first.

I think this is where complex numbers come in. Let

C=\cos{2x}+\cos{4x}+\ldots+\cos{2nx}

S=\sin{2x}+\sin{4x}+\ldots+\sin{2nx}

\therefore\ C+ \mathrm{i}S=(\cos{2x}+ \mathrm{i}\sin{2x})+(\cos{4x}+ \mathrm{i}\sin{4x})+\ldots+(\cos{2nx}+ \mathrm{i}\sin{2nx})

\color{white}.\hspace{64mm}. =\mathrm{e}^{2x\mathrm{i}}+ \mathrm{e}^{4x \mathrm{i}}+\ldots+ \mathrm{e}^{2nx \mathrm{i}}

\color{white}.\hspace{64mm}. =\frac{\mathrm{e}^{2x\mathrm{i}}\(1-\mathrm{e}^{2nx\mathrm{i}}\)}{1-\mathrm{e}^{2x\mathrm{i}}}

C is equal to the real part of the RHS.

...

IMHO, the very last step in the quoted post was a bit over the top in terms of giving out answers versus giving gentle nudges to help the student solve the problem on his own. Yoda should be able to calculate a geometric series on his own. Based on postings on this exact problem in another forum, he can't.

IMHO, the very last step in the quoted post was a bit over the top in terms of giving out answers versus giving gentle nudges to help the student solve the problem on his own. Yoda should be able to calculate a geometric series on his own. Based on postings on this exact problem in another forum, he can't.

Yes, I can't, you're right. Can somebody please once forever where the formula comes from. Why I got the wrong formula? How will I take the real part out?

Do a search on Wikipedia for "Geometric Series" and it should have a good explanation. The series you were summing isn't geometric, you have to convert it into a geometric sum by doing what Algebraic Topology did.

IMHO, the very last step in the quoted post was a bit over the top in terms of giving out answers versus giving gentle nudges to help the student solve the problem on his own. Yoda should be able to calculate a geometric series on his own. Based on postings on this exact problem in another forum, he can't.
I’m sorry, I didn’t know. In any case, I haven’t answered the whole question for Yoda. There is still some work to be done to get the answers to the original question.

After 1h and 30min. I found

C=http://pic.mkd.net/images/284673untitled1.JPG

S=http://pic.mkd.net/images/620553untitled12.JPG

But my problem is with cos^2x+isin^2x

After 1h and 30min. I found
C =
\frac {\cos\left(\frac{n+1}2 x\right) \sin\left(\frac{nx}2\right)}
{\sin\left(\frac x 2\right)}
S =
\frac {\sin\left(\frac{n+1}2 x\right) \sin\left(\frac{nx}2\right)}
{\sin\left(\frac x 2\right)}
But my problem is with cos^2x+isin^2x
That is obviously wrong. Try it with n=1.

[url]
Anyway OP, from there... consider

S = D + D^2 + D^3 + ... + D^n
= D(1 + D + D^2 + ... + D^{n-1})
= D(1 + S - D^{n})

Solving for S we get:

S = \frac{D - D^{n+1}}{1-D}

I think this will answer your question.


Yes, that is great. Just I have one question.

How did you get the third equation D(1+S-D^n)? Why D^n?


I think the 3rd step can be obtained from here:

S = D + D^2 + D^3 + ... + D^n
= D(1 + D + D^2 + ... + D^{n-1})
= D(1 + D + D^2 + ... + D^{n-1} + D^{n}-D^{n})
= D(1 + S - D^{n})

Absane had great tip, but I don't know where to go out from there, or I don't know how he get the 3rd and the 4th step...



Whereas for the 4th step, I think you can get it from here:

S= D(1 + S - D^{n})

S= D + DS - D^{n+1}

S - DS = D - D^{n+1}

S(1 - D) = D - D^{n+1}

S = \frac{D - D^{n+1}}{1-D}

I think the 3rd step can be obtained from here:

S = D + D^2 + D^3 + ... + D^n
= D(1 + D + D^2 + ... + D^{n-1})
= D(1 + D + D^2 + ... + D^{n-1} + D^{n}-D^{n})
= D(1 + S - D^{n})

But in the third step S is not equal to D + D^2 + ... + D^{n-1}+D^{n}

Why do you substitute for D + D^2 + ... + D^{n-1}+D^{n}?

That is obviously wrong. Try it with n=1.

Why it is wrong?

Yoda. Consider it another way.
(1)S=d+d^{2}+\dots+d^{n}
(2)Sd=d^{2}+d^{3}+\dots+d^{n+1}
(2)-(1)Sd-S=S(d-1)=d^{n+1}-d
And the result follows.
As tothe rest of the problem. Sorry I fell into the trap of misapplying DeMoivres Thm in my brief reply.

But you seem to be almost there now. Good luck.

PS is this a textbook question? Itseems a bit misleading when it claims all you need is Z=A+iB. If this is an assignment question how many points do you get for it?

It says:

Find the final sum of:

a)cos^2x+cos^22x+...+cos^2nx ; b)sin^2x+sin^22x+...+sin^2nx

I don't know if I need to find the separate sums.

But in the third step S is not equal to D + D^2 + ... + D^{n-1}+D^{n}. Why do you substitute for D + D^2 + ... + D^{n-1}+D^{n}?

Yoda, since you are having difficulties with the geometric sum, here is an annotated version of inzomnia's post.


\begin{eqnarray}
S &= D + D^2 + D^3 + ... + D^n
\qquad \qquad \qquad \qquad \qquad \qquad
\qquad \qquad \qquad \qquad \qquad \qquad \quad
&\text{definition of S} \\
&= D(1 + D + D^2 + ... + D^{n-1})
\qquad \qquad \qquad \qquad \qquad \qquad
\qquad \qquad \qquad \quad
& \text{factor out D} \\
&= D(1 + D + D^2 + ... + D^{n-1} + D^{n}-D^{n})
\quad \;
&\text{add}\ D^{n}-D^{n}=0\\
&= D(1 + \underbrace{(D + D^2 + ... + D^{n-1} + D^{n})}{}-D^{n}) &\text{highlighted section is just}\ S\\
&= D(1 + S - D^{n}) & \text{replace highlighted section with S}
\end{eqnarray}

C=\cos{2x}+\cos{4x}+\ldots+\cos{2nx}
S=\sin{2x}+\sin{4x}+\ldots+\sin{2nx}
After 1h and 30min. I found
C =
\frac {\cos\left(\frac{n+1}2 x\right) \sin\left(\frac{nx}2\right)}
{\sin\left(\frac x 2\right)}
S =
\frac {\sin\left(\frac{n+1}2 x\right) \sin\left(\frac{nx}2\right)}
{\sin\left(\frac x 2\right)}
But my problem is with cos^2x+isin^2x
That is obviously wrong. Try it with n=1.That is obviously wrong. Try it with n=1.
Why it is wrong?

With n=1, your expression evaluates to

\begin{eqnarray}
C &=\frac {\cos\left(\frac{1+1}2 x\right) \sin\left(\frac{x}2\right)}
{\sin\left(\frac x 2\right)} = \cos x \\
S &= \frac {\sin\left(\frac{1+1}2 x\right) \sin\left(\frac{x}2\right)}
{\sin\left(\frac x 2\right)} = \sin x
\end{eqnarray}

while they should evaluate to C=\cos 2x, S=\sin 2x. Show your work and we will help you find find the error.

It says: Find the final sum of:
a)cos^2x+cos^22x+...+cos^2nx ; b)sin^2x+sin^22x+...+sin^2nx
I don't know if I need to find the separate sums.
You have previously labeled these two parts as

\begin{eqnarray}
A &=\cos^2x + \cos^22x + \cdots + \cos^2nx \\
B &=\sin^2x + \sin^22x + \cdots + \sin^2nx
\end{eqnarray}

The whole point of this problem is to calculate A and B, not their sum (A+B[/itex]). Calculating A and B directly is very difficult. Calculating the sum of the two sums, [tex]A+B (Note: *not* A+Bi) is trivial (do it). If you can also calculate the difference A-B, then you can calculate A and B via

\begin{eqnarray}
A &= \frac{(A+B)+(A-B)} 2 \\
B &= \frac{(A+B)-(A-B)} 2\end{eqnarray}

Calculating the sum is trivial. Calculating the difference is a chore. How to accomplish this calculation is what the bulk of this thread has been about.

I would imagine so. Finding the combined sum is pretty trivial.

The reason I asked where this question came from was that I am pretty sure I saw the same or similar question a long time ago. But we were meant to use Fourier series to solve it.( I dont think its possible to use F series for this question, as Itried and went round in circles, so it was probably a different question.) I was just interested what course this was for.

Wait, wait. Is correct that I was searching for cosx+isinx. Why I need to search for cos2x+isin2x? I want to ask you is a) and b) together ? Since in my problem the task is separated on a) and b), just look at the original task. I have one example in my book, and it is solved with A+Bi. I don't what to do. I have never learned this stuff before.

You are searching for something that will help you resolve the sums a) and b). What exactly that something is is up to you. We have given you plenty of hints, far too many hints, in fact. To recap,

The two sums you want to determine:

A \equiv \cos^2x+\cos^22x+\cdots+\cos^2nx
B \equiv \sin^2x+\sin^22x+\cdots+\sin^2nx

Rather than solving these sums independently and directly, solve for the two sums simultaneously via some indirect method. To this end, define

C \equiv A + B
D \equiv A - B

If you can determine these two quantities you can compute A and B via

A = \frac{C+D} 2
B = \frac{C-D} 2

I will leave solving the sum, C=A+B up to you. The difference is

D \equiv A-B
= (\cos^2x+\cos^22x+\cdots+\cos^2nx) -
(\sin^2x+\sin^22x+\cdots+\sin^2nx)
=\cos^2x-\sin^2x + \cos^22x-\sin^22x+\cdots+\cos^2nx-\sin^2nx
= \cos^2x-\sin^2x + \cos^22x-\sin^22x+\cdots+\cos^2nx-\sin^2nx
= \cos(2x) + \cos(2*2x)+\cdots+\cos(2*nx)

The first step is the definition of the difference. The second step arises from substituting A and B with their definitions. The third step is just a rearrangement of the terms. The final step is a repeated application of a simple trigonometric identity. Not quite there yet, because this sum as-is is just as hard to calculate as are A and B. However, this new sum no longer involves squares of trigonometric functions, and it is very closely related to the following very soluble sum:


S = e^{i2x} + e^{i4x} + \cdots + e^{i2nx} = \sum_{k=1}^n e^{i2kx}


This sum is a geometric series. Because \Re(z_1 + z_2) = \Re(z_1)+\Re(z_2) and because \Re(e^{ix}) = \cos x, D=\Re (S). Solve the series, take the real part, and you have D. From that, you can solve for A and B.

Ok I will find A+B:
C=A+B
=(cos^2x+cos^22x+...+cos^2nx)+(sin^2x+sin^22x+...+ sin^2nx)
=cos^2x+sin^2x+cos^22x+sin^22x+...+cos^2nx+sin^2nx
=1+1+...+1

I need to find A and B right?

A=\frac{1+1+...+1+cos2x+cos4x+...+cos2nx}{2}

A=\frac{(1+cos2x)+(1+cos4x)+...+(1+cos2nx)}{2}

A=cos^2x+cos^22x+...+cos^2nx

lol. I got the same using cos^2x=\frac{1+cos2x}{2}

Please be patient with me. Its translation problem.

Ok I will find A+B:
C=A+B
=(cos^2x+cos^22x+...+cos^2nx)+(sin^2x+sin^22x+...+ sin^2nx)
=cos^2x+sin^2x+cos^22x+sin^22x+...+cos^2nx+sin^2nx
=1+1+...+1

So what is that? You can reduce this more.

I need to find A and B right? ...

Not yet. Find A-B first. Sheesh. Read my post I practically gave the answer away.

How will I reduce it more than 1+1+...+1 ?

You found A-B. I need to find A and B using the formulas about right?

How will I reduce it more than 1+1+...+1 ?
Compute the sum.
You found A-B.
I have not told you what this difference evaluates to. You need to determine that.
I need to find A and B using the formulas about right?
Read post #43.

I really cannot give you any more help. I and others have already given far too much. We have practically solved the problem for you.

Calculating A and B directly is very difficult.
Not much more than calculating A - B:

A \, \equiv \, \sum^{n}_{k=1} cos(kx)^2 \, = \, \sum^{n}_{k=1} \frac{1}{2} \left[ 1 + \cos(2kx) \right] \, = \, \frac{n}{2} + \frac{1}{2} \sum^{n}_{k=1} \cos( 2kx )

Good point. It still comes down to evaluating \sum_{k=1}^n \cos(2kx).

I don't even understand what are you telling me. I haven't learn bunch of things, that you posted here. First can we go step by step. So you said that I will find the sums A and B throught A+B and A-B. But I came going into circle. The last part of #43 I don't understand even little, since I have not learned about it. Please give some explanation or tell me what to to exactly. Thanks for your help.

I'll explain the last part of post #43 because you may not know the nomenclature. The operator \Re() simply takes the real part of a complex number. For example, \Re(3+4i)=3. Since e^{ix}=\cos x +i \sin x by Euler's equation, \Re(e^{ix}) = \cos x. It should be obvious that \Re(z_1 + z_2) = \Re(z_1) + \Re(z_2). Thus

\sum_{k=1}^n \cos(2kx) = \sum_{k=1}^n \Re(e^{2ikx})= \Re(\sum_{k=1}^n e^{2ikx})

So. Compute the geometric sum \sum_{k=1}^n e^{2ikx}, take the real part, and viola! you have \sum_{k=1}^n \cos(2kx)

Sorry, but I have never learned about computing geometric sum. Why we are working with sums? I hsbr never learned about sums, and obviously don't know what they mean.

Is it:

cos2x+cos2nx
--------------
1-cos2x

?

Sorry, but I have never learned about computing geometric sum. Why we are working with sums? I hsbr never learned about sums, and obviously don't know what they mean.

Youve never learned about sums? What do you mean? How did you get out of elementary school? Do you mean youve neverworked with sums of trigonometric functions, or youve never workedwith \Sigma notation. Im not trying to be rude but your statement is a little strange for someone heading into complex numbers.

Is it:

cos2x+cos2nx
--------------
1-cos2x

?
Im sorry. Is that what? The final answer?Actually I dont know cus I havent bothered working it out. But the denominator looks wrong. I hope you havent separated realand imaginary pairs from the denominator.
\frac{A}{x+iy}=\frac{A}{x+iy}\frac{x-iy}{x-iy}=\frac{A(x-iy)}{x^{2}+y^{2}}

So the real part is \frac{Ax}{x^{2}+y^{2}

Hope that helps. Good luck.

As far as I remember, I have never deal with it. To be honest, I don't understand what are you talking to me. Please give me the instructions from the start. I am totally confused. I have one example in my book with A+iB. And I don't understand if I can solve it on that way?

Well look at DHs post from　yesterday. The one with the As and Bs.
Look at the part with \cos(2x)+\cos(4x)+\dots+\cos(2nx)

This is alot nicer than the sum we had before, because we got rid of the powers of 2.
Now we use an old trick, that complex numbers let you use.
First note that \cos x+i\sin x=e^{ix}. You know what e is, right?
So lets just pull i\sin(2x)+i\sin(4x)+\dots+i\sin(2nx) out of our hats. We are just gonna add this to our cosine sum at the top.

Then regroup to get\cos(2x)+i\sin(2x)+\cos(4x)+i\sin(4x)\dots+\cos (2nx)+i\sin(2nx)

As I said,we can write this as e^{2ix}+\dots+e^{2nix}.
OK?

Now, call e^{2ix}=K. So we want to find K+K^{2}+\dots+K^{n}. This is the geometric sum you want to find. Youve been shown many times here how to find the sum. I recommend looking up "geometric sum" in a maths textbook. Its pretty standard stuff, and you are gonna need it alot in later maths.

Find the sum in terms of K. Write the sum in terms of e^{2ix}.
I would recommend at this point writing e^{2ix} in terms of sine and cosine, as I showed you above.
Find the real part of this expression (see DHs post about real parts), and be sure to do it correctly. See my previous post.

A-B=\frac{cos(n+1)xsin(nx)}{sinx}+i\frac{sin(n+1)xsi n(nx)}{sinx}

Now what should I do?

How did you get that for C?

C=A+B, right?
Think triangles. Let a=adjacent side of right angled trianle, o=opposite side, h=hypotenuse.\cos(x)=a/h
\sin(x)=o/h
So \cos^{2}x=a^{2}/h^{2} and \sin^{2}x=o^{2}/h^{2}.

So \cos^{2}x+\sin^{2}x=\frac{o^{2}+a^{2}}{h^{2}}

But using pythagoras, we can simplify this to something very, very simple.

Do you see what C might be now?

I modified it. Now it A-B, is the real part only, right?

Finally, I got it. Thanks all for the help.