# Points of intersection to find area inside a region

1. Apr 12, 2010

### edough

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

Use a dounble integral to find the area of the region inside r = 1+ cos (theta) and outside r = 2sin (theta). sketch region and indecate the points of intersection.

I'm confused how to find the points of intersection of these two equations

2. Relevant equations
I've tried several trig identities and i just can't seem to get it. Can anyone help me get started on this problem??

2. Apr 12, 2010

### lanedance

have you tried some double/half angle formula?

3. Apr 13, 2010

### lanedance

also are there any limits put on theta in the question. Note $r = 2 sin(\theta)$ is only positive for $\theta \in (0,\pi)$

4. Apr 13, 2010

### edough

Theres no limits to theta. I have tried half and double angle formulas but I keep getting stuck.

5. Apr 13, 2010

### lanedance

ok, well a negative radius doesn't make much sense to me, so assume we're only working in
$\theta \in [0,\pi]$

so how about equating $r = 1- cos(\theta)$and $r = 2sin(\theta)$, then substituting:
$$cos(\theta) = 1-2sin^2(\theta/2)$$
$$sin(\theta) = 2sin(\theta/2)cos(\theta/2)$$

6. Apr 13, 2010

### edough

So after substituting that, I used sin(Θ/2) = ( 1-cosΘ / 2)^(1/2) and cos(Θ/2) = (1+cosΘ / 2)
at the end i got 2 = 2 (1-cos^2(Θ) )^(1/2) + (1-cosΘ) / 2
I'm not sure what to do after this. Did I take it a whole wrong direction?

7. Apr 13, 2010

### lanedance

i think so, just try the first substitution & some simplification and see how you go, i got to an equation something like
$$tan(\theta/2) = \frac{1}{2}$$
and thought that was enough to solve for