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

[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.

 

[D H]

Am I right?

Yes!

 

[Dick]

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

It counts double.

 

[Physicsissuef]

But can you tell me please how did they simplify it?

 

[Dick]

I think you should try it first. I don't see any '2' factors in the final result. So you might start with a double angle formula.

 

[Physicsissuef]

\frac{cos^2x-sin^2x+cos^2nx-sin^2nx-cos^2(n+1)x+sin^2(n+1)x-sin^2x-cos^2x}{4sin^2x}

\frac{-2sin^2x+cos^2nx-sin^2nx-cos^2(n+1)x+sin^2(n+1)x}{4sin^2x}

Like this?