How Do You Convert These Polar Equations to Rectangular Form?

[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.

 

[Pyrrhus]

r = 2 \sin (3 \theta)

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

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

 

[Dooh]

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

 

[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

 

[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?

 

[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)