# Thread: Equations in Polar Coordinates

1. ## Equations in Polar Coordinates

Write the following equation in polar coordinates (using t for ):

(x2 + y2)2 = x2 - y2 becomes

____?____ - cos(2t) = 0

I know that r2=x2+y2 but i'm not sure where everything else leads me.

2. Originally Posted by mickles
Write the following equation in polar coordinates (using t for ):

(x2 + y2)2 = x2 - y2 becomes

____?____ - cos(2t) = 0

I know that r2=x2+y2 but i'm not sure where everything else leads me.
$x^2+y^2=r^2$

$x=r\cos(t)$

$y=r\sin(t)$

Try this subs and see what happens

3. Originally Posted by TheEmptySet
$x^2+y^2=r^2$

$x=r\cos(t)$

$y=r\sin(t)$

Try this subs and see what happens
Okay i did this:

r^2 = r^2cos^2(x)+r^2sin^2(x)
1 = cos^2(x) +sin^2(x)
So 1 = cosx + sinx

4. I also inserted x and y into the equation in my first post and i got this:

cos^2(x)-sin^2(x) = (r^2cos^4(x)+2r^2cos^2(x)sin^2(x)+r^2sin^4(x))

5. Originally Posted by mickles
Write the following equation in polar coordinates (using t for ):

(x2 + y2)2 = x2 - y2 becomes

____?____ - cos(2t) = 0

I know that r2=x2+y2 but i'm not sure where everything else leads me.
Remember that $\cos^2(t)-\sin^2(t)=\cos(2t)$

so we get

$(r^2)^2=r^2\cos^2(t)-r^2\sin(t) \iff r^4=r^2(\cos^2(t)-\sin^2(t))$

$r^4=r^2(\cos(2t)) \iff r^2=\cos(2t) \iff r^2-\cos(2t)=0$

6. Originally Posted by TheEmptySet
Remember that $\cos^2(t)-\sin^2(t)=\cos(2t)$

so we get

$(r^2)^2=r^2\cos^2(t)-r^2\sin(t) \iff r^4=r^2(\cos^2(t)-\sin^2(t))$

$r^4=r^2(\cos(2t)) \iff r^2=\cos(2t) \iff r^2-\cos(2t)=0$
Oh okay i understand now. I forgot the identity where cos^2(2t)-sin^2(2t) = cos(2t)

Thank you very much i understand now!