# Polar to Rectangular Conversion

1. May 1, 2005

### Dooh

Can someone help me with these problems? It's been bugging me i can't seem to solve it.

Lets assume T = theta

I can't seem to find a way to convert these polar equations into rectangular form.

r = 2 sin 3T

r = 6 / 2 - 3 sinT

If possible, can someone help me with this and list it in a step-by-step fashion so i can see how you get the answer. Thanks.

Last edited: May 1, 2005
2. May 1, 2005

### Pyrrhus

$$r = 2 \sin (3 \theta)$$

$$r^2 = 2r \sin (3 \theta)$$

then try the triple angle formulas to reduce $\sin (3 \theta)$

3. May 1, 2005

### Dooh

Sorry but the furthest that our teacher had taught us in the double angle formula.

4. May 1, 2005

### Pyrrhus

It's not hard. Why don't you try it?

$$\sin (3 \theta) = \sin (\theta + 2 \theta)$$

Use the sum of angles for the sin

$$\sin (\alpha + \beta) = \sin \alpha cos \beta + \cos \alpha \sin \beta$$

5. May 1, 2005

### Dooh

Ok, so i got:

$$r^2 = 2r (\sin \theta \cos 2 \theta + \cos \theta \sin 2 \theta )$$

can i take out the $$\sin \theta \cos \theta$$

and get

$$r^2 = 2r \sin \theta \cos \theta (\cos \theta + \sin \theta )$$

or should i further expand the equation?

6. May 1, 2005

### Pyrrhus

No that's wrong it will be

$$\sin (3 \theta) = 3 \sin \theta - 4 \sin^{3} \theta$$

thus

$$r^2 = 2r (3 \sin \theta - 4 \sin^{3} \theta)$$