Question

How do I convert r^2= sin2theta to rectangular form?

Answer 1

sin 2t = sin (t+t) = sin(t)cos(t) + sin(t)cos(t)
= 2sin(t)cos(t)

might help

Answer 2

Im actually really confused on how you get to 2sin(t)cos(t).
x^2 + y^2 = 2sin(t)cos(t) ^2? Is that how you continue on?

Answer 3

its a trig identity, it was my first thought in order to get some sort of sin(t) and cos(t) construct

Answer 4

It's been a while since I took trig. I don't remember much of the identities haha.

Answer 5

r^2 = 2 sin cos, so dont square it

Answer 6

Oh. I read the equation wrong .
x^2+y^2 = 2sincos.
does that mean
x= 2sincos (cos)
and y = 2sincos ( sin)?

Answer 7

if anything:

r^2 = xy , such that x = sqrt(2) cos(t), y = sqrt(2) sin(t)

Answer 8

Im actually confused on how you go from r^2 = xy.

Answer 9

... just remembering stuff is all

we should know that x = r cos(t), and y = r sin(t) are fundamental properties between polar and cartesian

Answer 10

Ah. This boggles my brain. I understand x= r cos(t) and y = r sin(t) but when I was taught this earlier today, I couldn't get my mind around anything that was happening. Even if it is
x=sqrt(2)cos(t) and y = sqrt(2) sin(t) I don't understand how it got there.

Answer 11

\[\sqrt{2}*\sqrt{2}=2\]

Answer 12

since sin(2t) = 2 sin cos, that can break up into \[\sqrt{2}*\sqrt{2}*sin(t)*cos(t)=\sqrt{2}sin(t)*\sqrt{2}cos(t)=xy\]

Answer 13

Uh alright. I'll try to finish the rest on my own. Thanks!

Answer 14

good luck, if you see anything wrong with what i remembered, or if i find a different setup, ill let you know