Polar Coordinates into Rectangular Form

[chelser13]

Convert r^2 = Cos 2 theta into rectangular form

I know that x^2 + y^2 = r^2 and x = r Cos theta
First, I'm not exactly sure how to convert and the 2 theta is messing me up on where it goes.
This is what I have so far:
x^2 + y^2 = Cos 2 theta (but i'm not sure what to do with the 2 theta)

 

[DrMike]

You should use the rule

\(\displaystyle \cos2\theta=\cos^2\theta-\sin^2\theta\)

Let us know if you're still stuck..

 

[chelser13]

That does help but what do I do with the squared on Cos and Sin? and I feel like I need r so I can get x = r Cos theta and y = r Sin theta?

 

[Deleted member 4993]

Convert r^2 = Cos 2 theta into rectangular form

I know that x^2 + y^2 = r^2 and x = r Cos theta
First, I'm not exactly sure how to convert and the 2 theta is messing me up on where it goes.
This is what I have so far:
x^2 + y^2 = Cos 2 theta (but i'm not sure what to do with the 2 theta)

\(\displaystyle \cos(2\theta) \, = \, \frac{1 \, - \tan^2(\theta)}{1 \, + \tan^2(\theta)}\)

 

[soroban]

Hello, chelser13!

\(\displaystyle \text{Convert }\:r^2 \:=\:\cos2\theta\:\text{ into rectangular form.}\)

\(\displaystyle \text{Using a Double-angle identity: }\;r^2 \;=\;\cos^2\!\theta - \sin^2\!\theta\)

\(\displaystyle \text{Multiply by }r^2\!:\quad r^2\cdot\,r^2 \;=\;r^2\cos^2\!\ttheta - r^2\sin^2\!\theta \quad\Rightarrow\quad (r^2)^2 \;=\;(r\cos\theta)^2 - (r\sin\theta)^2\)

. . \(\displaystyle \text{Convert:} \;(x^2+y^2)^2 \;=\;x^2 - y^2\)