# Homework Help: Simplifying cosine and sine expressions

1. Dec 8, 2011

### Evo8

1. The problem statement, all variables and given/known data
So im a little rusty on my basic math and I am trying to simplify this expression. In case you were curious the expression came from a signal block diagram that I calculated the transfer function H(z) to then found the frequency response H(f) and I think i have done that correctly I just need to reduce this further. I have attached a picture of what I have left. The theta stands for exp(j2pifT) if that maters to you.

2. Relevant equations

3. The attempt at a solution

So this is where I get confused and cant remember the proper rules. These are super basic I know. Unfortunately I never learned them that well to begin with which is why im having trouble now..

the terms that are j*sin(-2*theta) can I bring the -2 out of the sin function? Same with the cosine? I want to say no but part of me feels like you can..... If so im thinking this might help me cross out a few terms?

Or if i break it up into sections i get something like this for the first few terms. -2/4 * cos(-theta)/cos(-theta). I then feel like i can cross out the cosine terms and be left with -1/2? Same with the -2jsin(theta)/4jsin(theta) etc.

I really appreciate any help here guys/girls.

Thanks

p.s. Sorry if my handwriting is messy. I tried my best to write it down as neat as possible. I figured it would be easier to see if i took a picture rather then type it out here and have you try to understand what im looking at.

Last edited: Dec 8, 2011
2. Dec 8, 2011

### Fredrik

Staff Emeritus
I don't see an attachment. There are formulas that tell you how to evaluate sin(2x) and cos(2x). If you remember the formulas for sin(x+y) and cos(x+y), you can use them with y=x. Also, keep in mind that sin is an odd function, and cos is an even function.

3. Dec 8, 2011

### Evo8

Sorry about that i had a link to the picture but it must be just me that can see it because it was local or something like that. Ive uploaded it to a flikr type account and changed the link. You should be able to see it now.

I saw some of these forumulas but im not 100% Im still looking at them as we speak.

4. Dec 8, 2011

### Fredrik

Staff Emeritus
I can see it now. You can certainly use the formulas I mentioned, but my first thought is that it doesn't look like the result will be any simpler.

5. Dec 8, 2011

### Evo8

Hmm maybe that's it then. The reason I said I want 100% was because I did try those formulas and it did not make the eq more simple so I suspected that I was doing something wrong. If it looks in simplest form to you then maybe that's the case.

Thanks for taking a look at this for me.

6. Dec 8, 2011

### Fredrik

Staff Emeritus
You should at least use the fact that sin is odd and cos even to get rid of the minus signs in front of the θ and 2θ.

I had another look at this, and I think the formulas I mentioned can simplify this a little. The expression cos θ-j sin θ shows up in a couple of places. Is j2=-1? In that case, cos θ-j sin θ=e-jθ.

The final result I got has 4(cos θ+1) as the denominator (after cancelling a factor of) e-jθ), and the numerator is at least a little bit simpler than what you started with.

7. Dec 8, 2011

### I like Serena

Hi Evo8!

It seems a bit weird to me that you have a cosine with exp(j2pifT) as argument.
And that for a transfer function H(z)?

If you expand part of what you have, you get:
$$\cos(-\theta)+j\sin(-\theta)=e^{-j\theta}=e^{-j e^{j2\pi f T}}$$

I don't know... it just doesn't look like part of a transfer function...

8. Dec 8, 2011

### Joffan

I pretty sure that Evo8 meant to say that $θ=2πfT$, given the pattern of cos+j.sin combinations.

$\frac{-2cos(-θ)-2j.sin(-θ)+cos(-2θ)+j.sin(-2θ)}{2+4cos(-θ)+4j.sin(-θ)+2cos(-2θ)+2j.sin(-2θ)}$

$\ldots = \frac{-2e^{-jθ}+e^{-j2θ}}{2+4e^{-jθ}+2e^{-j2θ}}$

$\ldots = \frac{e^{-jθ}(e^{-jθ}-2)}{2(e^{-jθ}+1)^{2}}$

Hmm, things behave badly when $θ=π$, or using my assumed identity above when $2fT=1$

9. Dec 9, 2011

### Evo8

Well for starters my transfer function was derived wrong...:( I derived it from a block diagram and now realize the mistake. So this is irrelivent but at least i get some more experience with this.

Yes in this case i had theta=(2pifT). Its a method my book used so i tried mimicking it.

I appreciate all the help guys. You have been super helpful.