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

[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?

 

[Dick]

You are missing the exp(2inx) part. Now drop the imaginary parts. You just want the real part.

 

[DavidWhitbeck]

Dick is about to earn the 2008 award for homework helper just for this thread! What patience! :-)

 

[Dick]

Dick is about to earn the 2008 award for homework helper just for this thread! What patience! :-)

It's not over yet. If this is still going on in 2009 I can use it to claim the 2009 award as well.

 

[Physicsissuef]

\frac{cos2x+cos2nx-cos2(n+1)x-1}{2(1-cos2x)}+\frac{isin2x+isin2nx-isin2(n+1)x}{2(1-cos2x)}
I have no idea how will I solve this.

 

[Dick]

Now you dropped the (-1). You used to have it. Now just take the real part. You don't have to solve it. It IS the solution. It's the formula for the sum cos(2x)+cos(4x)+...+cos(2nx). How does this help you with the original problem?

 

[Physicsissuef]

In my textbook results it is:

\frac{cos(n+1)xsin(nx)}{sinx}+i\frac{sin(n+1)xsin(nx)}{sinx}

I am wondering how did they found this results...

 

[Dick]

In my textbook results it is:

\frac{cos(n+1)xsin(nx)}{sinx}+i\frac{sin(n+1)xsin(nx)}{sinx}

I am wondering how did they found this results...

They used trig identities. It's the same thing you have written in a different form. Let's try and solve the problem first and then worry about how to simplify it.

 

[Physicsissuef]

I think I know how to solve it.
A=cos^2x+cos^22x+...+cos^2nx

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

C=A+B

D=A-B

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

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

I found C

C=n

D is the real part of what we were doing now.

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

With substitution:

A=\frac{n+\frac{cos(n+1)xsin(nx)}{sinx}}{2}

B is same just with opposite sign (instead of + it is -)

Am I right?

 

[D H]

Dick is about to earn the 2008 award for homework helper just for this thread! What patience! :-)

Does helping the exact same person on the exact same question with the exact same frustration level but in another forum count?

 

[Dick]

Finally. Yes, that's how to finish it.