# Homework Help: More complex numbers, more trigs

1. Sep 18, 2008

### rock.freak667

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

The angles $\theta$ and $\phi$ lie on the interval (-pi/2,pi,2) and

$z=(cos\theta +cos\phi)+i(sin\theta +sin\phi)$

Show that |z|=2cos($\frac{\theta - \phi}{2}$) and find arg(z)

2. Relevant equations
If z=x+iy

$$|z|=\sqrt{x^2 +y^2}$$

3. The attempt at a solution

$$|z|= \sqrt{(cos\theta +cos\phi)^2+(sin\theta +sin\phi)^2}$$

$$=2+2cos(\theta - \phi)$$

I don't know how to get rid of the additional two and how to make the angle halved.

2. Sep 18, 2008

### rock.freak667

Solved the first part

2nd part giving trouble

arg(z)=tan-1{Im(z)/Re(z)}

$$tan^{-1}(\frac{sin\theta + sin\phi}{cos\theta + cos\phi})$$

I think I'll need to simplify this further.

I tried multiplying by the conjugate of the denominator but that didn't lead anywhere feasible.

Last edited: Sep 18, 2008
3. Sep 19, 2008

### Dick

(sin(a)+sin(b))/(cos(a)+cos(b))=tan((a+b)/2). I'm going to the wikipedia page of trig identities and fishing around. You might notice this is related to the third problem of your other post if you put phi=0.

4. Sep 19, 2008

### rock.freak667

oh my god! I completely forgot factor formula....seems that 1 month away from math has completely worked my brain in the wrong way.... thank you.

5. Sep 19, 2008

### Dick

I don't remember stuff like that either. I just figured that if the problem called for it then something like that must exist.

6. Sep 19, 2008

### rock.freak667

Well I knew it had to be arctan of the tan of something, but I thought the something would have been just with theta and phi and not half angled. Was looking to make a tan(A+B) thing.

7. Sep 19, 2008

### Dick

See, they tricked you. Seriously, some trig identities are just tricks. It's hard to see them coming. Like that one. If the problem were even slightly different it wouldn't work. Very specific solution to a very specific problem. I think the other problems in your other set are more instructive.