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

• Theofilius
In summary: I don't know. You are supposed to use the real part of a sum to find the difference between two sums.I don't know how.I don't know how.
Theofilius

## Homework Statement

Hi

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

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

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?

Last edited:
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?

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

Any help?

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.

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.

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

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

How will I get rid of "1"?

How will I go next?

Dick, someone?

Theofilius said:
Dick, someone?

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

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

Theofilius said:
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)+...?

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

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

Can you sum a geometric series?

I don't know how.

Theofilius said:
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.

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.

Theofilius said:
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.

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

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

I just TOLD you. Review geometric series.

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?

Theofilius said:
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.

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

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.

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?

Theofilius said:
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?

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.

Theofilius said:
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).

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?

"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.

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

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

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

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

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?

"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?

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

Physicsissuef said:
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.

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

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

Last edited:
Physicsissuef said:
exp(2*i*x)=cos2x+isin2x ?

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

No. What happened to the imaginary part?

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