# Final sum

1. May 20, 2008

### Theofilius

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

Hi

Find the (finite,ultimate,definitive,peremptory,eventua,conclusive) 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?

Last edited: May 20, 2008
2. May 20, 2008

### Staff: Mentor

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?

3. May 20, 2008

### 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.

4. May 21, 2008

### Theofilius

Any help?

5. May 21, 2008

### Staff: Mentor

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.

6. May 21, 2008

### 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.

7. May 23, 2008

### 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?

8. May 23, 2008

### Theofilius

Dick, someone?

9. May 23, 2008

### Dick

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

10. May 23, 2008

### Theofilius

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

11. May 23, 2008

### Dick

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

12. May 23, 2008

### Theofilius

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

13. May 23, 2008

### Dick

Can you sum a geometric series?

14. May 23, 2008

### Theofilius

I don't know how.

15. May 23, 2008

### Dick

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.

16. May 23, 2008

### 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.

17. May 23, 2008

### Dick

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.

18. May 23, 2008

### Theofilius

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

19. May 23, 2008

### Dick

I just TOLD you. Review geometric series.

20. May 23, 2008

### 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?