Finding the Final Sum of Trigonometric Functions: Cosine and Sine Formulas

[Theofilius]

Homework Statement

Hi

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

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

Homework Equations

z=A+Bi

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?

 

[Borek]

Not sure what you are trying to do.

First, you defined two sums - a and b.

Is your task to find sum of those two sums? a+b is what you are looking for?

If so, why do use i?

Do you know what Pythagorean trigonometric identity is?

 

[Theofilius]

I don't know if they are separate or together... I should use "i", since that's my task, to find it through i.

 

[Theofilius]

Any help?

 

[Borek]

You won't get more help without giving more information. At present - at least IMHO - question as posted, and the idea of using i for calculations, doesn't make sense.

 

[Dick]

The real part of exp(i*n*x)^2 is cos(n*x)^2-sin(n*x)^2. So summing the geometric series exp(i*2*n*x) and taking the real part will give you the difference A-B. What's the sum A+B? That's one way to do it.

 

[Theofilius]

(\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?

 

[Theofilius]

Dick, someone?

 

[Dick]

Dick, someone?

Why did you ignore my last message? It had some reasonable suggestions in it.

 

[Theofilius]

I have never learn about it. Is mine correct? How will I continue solving?

 

[Dick]

I have never learn about it. Is mine correct? How will I continue solving?

What you wrote is a version of A+Bi. Do you know how to sum the series
cos(2n)+cos(4n)+cos(6n)+...?

 

[Theofilius]

Actully, I have never learned. So I don't know.

 

[Dick]

Actully, I have never learned. So I don't know.

Can you sum a geometric series?

 

[Theofilius]

I don't know how.

 

[Dick]

I don't know how.

If you haven't been given the sum of the finite series sin(nx) and cos(nx), or been taught how to derive them by summing geometric series like exp(i*n*x), then I don't know how you are supposed to do this problem.

 

[Theofilius]

I know something...
a=cosx+cos2x+...+cosnx

B=sinx+sin2x+...+sinnx

A=Re(z) ; B=Im(z)

z=A+Bi

z=A+iB=(cosx+cos2x+...+cosnx)+i(sinx+sin2x+...+sinnx)= (cosx+isinx)+

+(cos2x+isin2x)+...+(cosnx+isinnx)= (cosx+isinx)+(cosx+isinx)^2+...+(cosx+isinx)^n=

I don't know where to go out of here.

 

[Dick]

I(cosx+isinx)+(cosx+isinx)^2+...+(cosx+isinx)^n=

I don't know where to go out of here.

That's a geometric series with a common ratio of exp(ix). You should know how to write down it's sum. http://en.wikipedia.org/wiki/Geometric_progression
Once you do that the real part of the sum is A and the imaginary part is B.

 

[Theofilius]

For (cosx+isinx) is simpler, but still got no clue how to continue.

 

[Dick]

For (cosx+isinx) is simpler, but still got no clue how to continue.

I just TOLD you. Review geometric series.

 

[Theofilius]

Yes, yes, but I don't understand this step.
Now I have
(cosx+isinx)(1+(cosx+isinx)+...+(cosx+sinx)^n^-^1)
I understand this. But don't understand where the next step comes from?

 

[Dick]

Yes, yes, but I don't understand this step.
Now I have
(cosx+isinx)(1+(cosx+isinx)+...+(cosx+sinx)^n^-^1)
I understand this. But don't understand where the next step comes from?

Why did you do that?? Look, look up the formula for the sum of a geometric series and apply it to this problem. Until you do that, there's not much more I can say.

 

[Theofilius]

But, please. I want to know where the formula comes from.

 

[Dick]

Call the sum S=exp(ix)+exp(ix)^2+...+exp(ix)^n. Calculate S-exp(ix)*S. All but two terms cancel. Now solve for S.

 

[Theofilius]

Ok, in my problem.

\frac{1-(cos^2x+isin^2x)^n^+^1}{1-cos^2x-isin^2x}. But where I will go out of here?

 

[Dick]

Ok, in my problem.

\frac{1-(cos^2x+isin^2x)^n^+^1}{1-cos^2x-isin^2x}. But where I will go out of here?

No. That's wrong. If you want the sum of S=exp(i*2x)+exp(i*4x)+...exp(i*n2x) it's (exp(i*2x)-exp(i*2(n+1)x)/(1-exp(i*2x)). If you manage to take the real part of S, what would it be? What would that have to do with your problem?

 

[Theofilius]

I was talking about the problem in my first post of this thread. I know for cosx+isinx that we can use De Moivre formula.

 

[Dick]

I was talking about the problem in my first post of this thread. I know for cosx+isinx that we can use De Moivre formula.

I was talking about what you just posted. It's wrong. exp(i*2*x) is not equal to cos^2(x)+i*sin^2(x).

 

[Theofilius]

Ok. Using \frac{(cos^2x+isin^2x)(1-(cos^2x+isin^2x)^n)}{1-cos^2x-isin^2x}. But how will I take the real part out?

 

[Dick]

"It's wrong. exp(i*2*x) is not equal to cos^2(x)+i*sin^2(x)." You really aren't paying much attention to me, are you? To get the real part out you generally multiply the denominator by it's complex conjugate. But I wouldn't waste your time doing it on what you just posted. Because it's wrong.

 

[Physicsissuef]

S=d+d^2+...+d^n

Sd=d^2+d^3+...+d^n^+^1

S-Sd=S(1-d)=d-d^n^+^1

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

If we substitute for d=cos^2x+isin^2x

S=\frac{(cos^2x+isin^2x)-(cos^2x+isin^2x)^n^+^1}{1-cos^2x-isin^2x}

Why you say it is not correct?

 

[Dick]

"It's wrong. Because exp(i*2*x) is NOT equal to cos^2(x)+i*sin^2(x)." What part of this don't you understand?

 

[Physicsissuef]

Yes it is not equal. How will we find the sums?

 

[Dick]

Yes it is not equal. How will we find the sums?

By doing what you are doing but using what exp(2*i*x) does equal. Instead of what it doesn't. Duh.

 

[Physicsissuef]

exp(2*i*x)=cos2x+isin2x ?

cos2x-cos2nx
-------------- is the geometric sum?
1-cos2x

 

[Dick]

exp(2*i*x)=cos2x+isin2x ?

cos2x-cos2nx
-------------- is the geometric sum?
1-cos2x

No. What happened to the imaginary part?

 

[Physicsissuef]

But didn't we said that we need to find
the geometric sum cos2x+cos4x+...+cosnx ?

 

[Dick]

But didn't we said that we need to find
the geometric sum cos2x+cos4x+...+cosnx ?

Yes. That's the real part of the sum. But that's NOT (cos(2x)-cos(2nx))/(1-cos(2x)).

 

[Physicsissuef]

I don't understand what you mean. Can you explain how we should find the solution, starting from the beginning, please?

 

[Dick]

I don't understand what you mean. Can you explain how we should find the solution, starting from the beginning, please?

No. I don't have unlimited time to waste on this. I don't notice that you bother to pay any detailed attention to what people tell you anyway. It's all in the previous posts. Go reread them. I'm not going to repeat myself again. I've already done enough of that.

 

[Physicsissuef]

C=\frac{cos(n+1)xsin(nx)}{sinx}

D=\frac{sin(n+1)xsin(nx)}{sinx}

Why I was searching this for?

 

[Dick]

What do those have to do with the problem? Are they supposed to mean something? What?

 

[Physicsissuef]

Where I should go out of here:

\frac{e^2^i^x-e^i^2^n^x}{1-e^2^i^x} ?

 

[Dick]

It should be n+1 power in the numerator. See your post #30. Now do the usual thing. Multiply numerator and denominator by the complex conjugate of the denominator.

 

[Physicsissuef]

\frac{{e}^{2ix}-{e}^{i2(n+1)x}}{1-{e}^{2ix}} \circ \frac{1+{e}^{2ix}}{1+{e}^{2ix}}

Like this?

 

[Dick]

\frac{{e}^{2ix}-{e}^{i2(n+1)x}}{1-{e}^{2ix}} \circ \frac{1+{e}^{2ix}}{1+{e}^{2ix}}

Like this?

No. The complex conjugate of 1-exp(2ix) is 1-exp(-2ix).

 

[Physicsissuef]

<br /> \frac{{e}^{2ix}-{e}^{i2(n+1)x}}{1-{e}^{2ix}} \circ \frac{1-{e}^{-2ix}}{1-{e}^{-2ix}}<br />
How will I multiply all of this ? I have never learn to compute or multiply with Euler's formula.

 

[Dick]

<br /> \frac{{e}^{2ix}-{e}^{i2(n+1)x}}{1-{e}^{2ix}} \circ \frac{1-{e}^{-2ix}}{1-{e}^{-2ix}}<br />
How will I multiply all of this ? I have never learn to compute or multiply with Euler's formula.

Oh, come on. There are just exponentials. Multiply them like you usually multiply exponentials.

 

[Physicsissuef]

<br /> <br /> \frac{{e}^{2ix}-{e}^{i2(n+1)x}}{1-{e}^{2ix}} \circ \frac{1-{e}^{-2ix}}{1-{e}^{-2ix}}=\frac{{e}^{2ix}-{e}^{0}-{e}^{i2(n+1)x}+{e}^{i2nx}}{1-{e}^{-2ix}-{e}^{2ix}+{e}^{0}}

Like this?

 

[Dick]

Yes. Now use deMoivre to get the real part.

 

[Physicsissuef]

\frac{(cos2x+isin2x)-(cos2(n+1)x+isin2(n+1)x)-1}{2(1-cos2x)}

How will I solve this now?