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?