trigonometric sum identity

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September 28th 2010, 03:33 AM
magus

trigonometric sum identity

I'm looking for a hint on how to prove that

$\sum \limits _{k=1}^n \cos \left[(2k-1)\theta\right]=\dfrac{\sin \left(2n \theta\right) }{2\sin \theta}$

I know I'm on here a lot but I really have no one to go to for guidance and am trying to learn this material on my own from the text book. Thank you for being patient with me so far and for being so helpful.
September 28th 2010, 03:38 AM
mr fantastic

Quote:

Originally Posted by magus

I'm looking for a hint on how to prove that

$\sum \limits _{k=1}^n \cos \left[(2k-1)\theta\right]=\dfrac{\sin \left(2n \theta\right) }{2\sin \theta}$

I know I'm on here a lot but I really have no one to go to for guidance and am trying to learn this material on my own from the text book. Thank you for being patient with me so far and for being so helpful.

You might consider using the formula for the sum of a geometric series to find $\sum \limits _{k=1}^n e^{i\left[(2k-1)\theta\right]}$ and then getting the real part of the result.

Proof by induction might also work.
September 28th 2010, 03:47 AM
Prove It

I think induction might be the way to go here.

Base step: $n= 1$ ...

$LHS = \cos{\theta}$ .

$RHS = \frac{\sin{2\theta}}{2\sin{\theta}}$

$= \frac{2\sin{\theta}\cos{\theta}}{2\sin{\theta}}$

$= \cos{\theta}$

$= LHS$ .

Inductive step: Assume that the statement is true for $n = r$ , in other words, that $\displaystyle{\sum_{k = 1}^r\cos{[(2r-1)\theta]} = \frac{\sin{(2r\theta)}}{2\sin{\theta}}}$ . We need to show that it's true for $n = r+1$ .

$\displaystyle{LHS = \sum_{k = 1}^{r + 1}\cos{[(2k-1)\theta]}}$

$\displaystyle{= \sum_{k=1}^r\cos{[(2k-1)\theta]} + \cos\{[2(r+1)-1]\theta\}}$

$\displaystyle{ = \frac{\sin{(2r\theta)}}{2\sin{\theta}} + \cos\{[2(r+1)-1]\theta\}}$

See if you can go from here...
September 28th 2010, 10:10 AM
magus

Thanks guys. I was able to do it via induction but I think it was intended that I use some more direct method. It says "obtain the formula" I tried complexification as mr. fantastic recommended and obtained some interesting results but nothing I could use trig identities to simplify into the required form.

Also I was wondering if there is a closed form of $\sum \limits _{k=1}^{n}cos(k)$ or $\sum \limits _{k=1}^{n}sin(k)$ . I figured if there is such a formula it might come in handy here.

I've also tried using the infinite series definition of cosine and sine and again came up with interesting results but nothing that would help me unless I had a formula for the series I just mentioned.

Thank you two for your help thus far. I really just wish my book had more examples that I could build on for these problems. Does anyone know of some catalogue of proofs that I could look at to get some more inspiration?
September 28th 2010, 06:42 PM
mr fantastic

Quote:

Originally Posted by magus

Thanks guys. I was able to do it via induction but I think it was intended that I use some more direct method. It says "obtain the formula" I tried complexification as mr. fantastic recommended and obtained some interesting results but nothing I could use trig identities to simplify into the required form.

[snip]

You should show your working.

I get $\displaystyle \frac{1 - e^{i(2n) \theta}}{e^{-i \theta} - e^{i \theta}}$ and the real part of this is $\displaystyle \frac{\sin (2n \theta) }{2 \sin (\theta)}$ . Your job is to get these two expressions.

Quote:

Originally Posted by magus

[snip]

Also I was wondering if there is a closed form of $\sum \limits _{k=1}^{n}cos(k)$ or $\sum \limits _{k=1}^{n}sin(k)$ .

[snip]

Yes there are. But they are not relevant to the proof you are working on.
September 28th 2010, 08:24 PM
magus

Oh yeah the formula for a geometric sum. But what I get is

$\sum \limits _{k=1} ^n e^{(2k-1)\theta} = \sum \limits _{k=1} ^n e^{2k\theta}e^{-\theta} = \dfrac{\sum \limits _{k=1} ^ne^{(2k)\theta}}{e^{\theta}} = \dfrac{\dfrac{1-e^{2n\theta+1}}{1-e^{2n\theta}}}{e^{\theta}} = \dfrac{1-e^{2n\theta+1}}{e^{\theta}-e^{2n\theta-1+\theta}}=\dfrac{1-\cos(2n\theta+1)}{\cos(\theta)-\cos(2n\theta-1+\theta)}$

I've looked over the identities and I can't see how I get get to the form you got. Did I take a wrong step?
September 29th 2010, 04:28 AM
mr fantastic

Quote:

Originally Posted by magus

Oh yeah the formula for a geometric sum. But what I get is

$\sum \limits _{k=1} ^n e^{(2k-1)\theta} = \sum \limits _{k=1} ^n e^{2k\theta}e^{-\theta} = \dfrac{\sum \limits _{k=1} ^ne^{(2k)\theta}}{e^{\theta}} = \dfrac{\dfrac{1-e^{2n\theta+1}}{1-e^{2n\theta}}}{e^{\theta}} = \dfrac{1-e^{2n\theta+1}}{e^{\theta}-e^{2n\theta-1+\theta}}=\dfrac{1-\cos(2n\theta+1)}{\cos(\theta)-\cos(2n\theta-1+\theta)}$

I've looked over the identities and I can't see how I get get to the form you got. Did I take a wrong step?

You're missing the i's (I was tempted to say you have no i-dea, open your i's etc. ....)

Anyway, it's meant to be $\sum \limits _{k=1} ^n e^{i(2k-1)\theta}$ . Do you see the i? It is important (each word starts with i ....)

Now use the formula for the sum of a geometric series where the first term is $e^{i \theta}$ , $\displaystyle r = e^{i 2 \theta}$ and there are n terms in the series. Then get the real part of the result. Refer to my previous post.
September 29th 2010, 07:00 AM
magus

Darn it! I always make mistakes like those. Thank you for that

so *I* have

$\sum \limits _{k=1} ^n e^{i(2k-1)\theta} = \sum \limits _{k=1} ^n e^{i2k\theta}e^{-i\theta} = \dfrac{\sum \limits _{k=1} ^ne^{i(2k)\theta}}{e^{i\theta}} =\dfrac{\dfrac{1-e^{i 2n\theta+1}}{1-e^{i2n\theta}}}{e^{i\theta}}=\dfrac{1-e^{i 2n\theta+1}}{e^{i\theta}-e^{i2n\theta-i\theta}}$

Is this right so far? Because what comes after when I apply Euler's identity again is a god forsaken mess when I try to get the real part.

Just to show I'm doing the work though

$= \dfrac{1-\cos(2n\theta+1)+i\sin(2n\theta+1)}{\cos\theta+i\s in\theta+(\cos\theta-i\sin\theta)(\cos(2n\theta)+i\sin(2n\theta))}=$

$\dfrac{1-\cos(2n\theta+1)+i\sin(2n\theta+1)}{\cos\theta+i\s in\theta+\cos\theta\cos(2n\theta-1)+i\cos\theta\sin(2n\theta-1)-i\sin\theta\cos(2n\theta-1)-\sin\theta\sin(2n\theta-1)}$

$=\dfrac{1-\cos(2n\theta+1)+i\sin(2n\theta+1)}{\cos\theta+\co s\theta\cos(2n\theta-1)-\sin\theta\sin(2n\theta-1)+i(\sin\theta+\cos\theta\sin(2n\theta-1)-\sin\theta\cos(2n\theta-1))}$

Then comes the process of making the denominator real by multiplying by a conjugate over a conjugate. I'd show you the work I've done for that but unfortunately the size limit for the LaTeX images created prohibits me from doing so.

How does this reduce?
September 29th 2010, 01:35 PM
mr fantastic

Quote:

Originally Posted by magus

Darn it! I always make mistakes like those. Thank you for that

so *I* have

$\sum \limits _{k=1} ^n e^{i(2k-1)\theta} = \sum \limits _{k=1} ^n e^{i2k\theta}e^{-i\theta} = \dfrac{\sum \limits _{k=1} ^ne^{i(2k)\theta}}{e^{i\theta}} =\dfrac{\dfrac{1-e^{i 2n\theta+1}}{1-e^{i2n\theta}}}{e^{i\theta}}=\dfrac{1-e^{i 2n\theta+1}}{e^{i\theta}-e^{i2n\theta-i\theta}}$

Is this right so far? Because what comes after when I apply Euler's identity again is a god forsaken mess when I try to get the real part.

Just to show I'm doing the work though

$= \dfrac{1-\cos(2n\theta+1)+i\sin(2n\theta+1)}{\cos\theta+i\s in\theta+(\cos\theta-i\sin\theta)(\cos(2n\theta)+i\sin(2n\theta))}=$

$\dfrac{1-\cos(2n\theta+1)+i\sin(2n\theta+1)}{\cos\theta+i\s in\theta+\cos\theta\cos(2n\theta-1)+i\cos\theta\sin(2n\theta-1)-i\sin\theta\cos(2n\theta-1)-\sin\theta\sin(2n\theta-1)}$

$=\dfrac{1-\cos(2n\theta+1)+i\sin(2n\theta+1)}{\cos\theta+\co s\theta\cos(2n\theta-1)-\sin\theta\sin(2n\theta-1)+i(\sin\theta+\cos\theta\sin(2n\theta-1)-\sin\theta\cos(2n\theta-1))}$

Then comes the process of making the denominator real by multiplying by a conjugate over a conjugate. I'd show you the work I've done for that but unfortunately the size limit for the LaTeX images created prohibits me from doing so.

How does this reduce?

$\displaystyle \sum_{k = 1}^{n}a r^{k-1} = \frac{a(1 - r^n)}{1 - r}$ (think carefully because this is what you essentially have) and I have told you what a and r are.

After substitution you get $\displaystyle \frac{e^{i \theta} (1 - (e^{i2\theta})^n)}{1 - e^{i\theta}} = \frac{1 - e^{i2n\theta}}{e^{-i\theta} - e^{i\theta}}$ .

Getting the real part of this expression should be trivial at this level (but making careless mistakes will make it seem a lot harder I suppose).
September 29th 2010, 04:21 PM
magus

Quote:

Originally Posted by mr fantastic

$\displaystyle \sum_{k = 1}^{n}a r^{k-1} = \frac{a(1 - r^n)}{1 - r}$ (think carefully because this is what you essentially have) and I have told you what a and r are.

After substitution you get $\displaystyle \frac{e^{i \theta} (1 - (e^{i2\theta})^n)}{1 - e^{i\theta}} = \frac{1 - e^{i2n\theta}}{e^{-i\theta} - e^{i\theta}}$ .

Getting the real part of this expression should be trivial at this level (but making careless mistakes will make it seem a lot harder I suppose).

That's my problem I was using $\sum_{k = 1}^{n} r^{k} = \frac{1 - r^{n+1}}{1 - r}$ and never thought of futtzing with that.

So for $a=e^{i\theta}$ , $r=e^{i2\theta}$ , and we obviously get $\displaystyle \frac{e^{i \theta} (1 - (e^{i2\theta})^n)}{1 - e^{i\theta}}$

Now what I don't see is how you get $\displaystyle \frac{e^{i \theta} (1 - (e^{i2\theta})^n)}{1 - e^{i\theta}} = \frac{1 - e^{i2n\theta}}{e^{-i\theta} - e^{i\theta}}$ because for me if I bring the $e^{-i\theta}$ into the denominator I get

$\displaystyle \frac{ (1 - (e^{i2\theta})^n)}{e^{-i \theta}(1 - e^{i\theta)}}=\frac{1 - e^{i2n\theta}}{e^{-i\theta} - e^{-i\theta}e^{i\theta}}=\frac{1 - e^{i2n\theta}}{e^{-i\theta} - 1}$
September 29th 2010, 06:32 PM
mr fantastic

Quote:

Originally Posted by magus

That's my problem I was using $\sum_{k = 1}^{n} r^{k} = \frac{1 - r^{n+1}}{1 - r}$ and never thought of futtzing with that.

So for $a=e^{i\theta}$ , $r=e^{i2\theta}$ , and we obviously get $\displaystyle \frac{e^{i \theta} (1 - (e^{i2\theta})^n)}{1 - e^{i\theta}}$

Now what I don't see is how you get $\displaystyle \frac{e^{i \theta} (1 - (e^{i2\theta})^n)}{1 - e^{i\theta}} = \frac{1 - e^{i2n\theta}}{e^{-i\theta} - e^{i\theta}}$ because for me if I bring the $e^{-i\theta}$ into the denominator I get

$\displaystyle \frac{ (1 - (e^{i2\theta})^n)}{e^{-i \theta}(1 - e^{i\theta)}}=\frac{1 - e^{i2n\theta}}{e^{-i\theta} - e^{-i\theta}e^{i\theta}}=\frac{1 - e^{i2n\theta}}{e^{-i\theta} - 1}$

If you are following closely you will see that I made a typo in the denominator of $\displaystyle \frac{e^{i \theta} (1 - (e^{i2\theta})^n)}{1 - e^{i\theta}}$ . Recall what r is ....
September 29th 2010, 07:06 PM
magus

Ahhh

So what we really have is $\displaystyle \frac{e^{i \theta} (1 - (e^{i2\theta})^n)}{1 - e^{i2\theta}}$

Which then leaves an $e^{i\theta}$ left in that term when we bring down the $-e^{i\theta}$ so we have

$\displaystyle \frac{ (1 - (e^{i2\theta})^n)}{e^{-i \theta}(1 - e^{2i\theta)}}=\frac{ (1 - (e^{i2\theta})^n)}{(e^{-i \theta} - e^{-i \theta}e^{2i\theta})}=\frac{ 1 - e^{i2n\theta}}{e^{-i \theta} - e^{i\theta}}$

Using Euler again

$\displaystyle \frac{1-\cos(2n\theta)-i\sin(2n\theta)}{\cos\theta-\sin\theta-\cos\theta-i\sin\theta)}=\frac{1-\cos(2n\theta)-i\sin(2n\theta)}{-2i\sin\theta}}=\frac{1-\cos(2n\theta)}{-2i\sin\theta}-\dfrac{i\sin \left(2n \theta\right) }{-2i\sin \theta}=i\frac{1-\cos(2n\theta)}{2\sin\theta}+\dfrac{\sin \left(2n \theta\right) }{2\sin \theta}$

Taking the real part of which leaves me $\dfrac{\sin \left(2n \theta\right) }{2\sin \theta}$

Wow it's finally done. Thank you so so very much for your help. Thank goodness it's finally done!