# Thread: Polar to Cartesian Equations

1. ## Polar to Cartesian Equations

Instructions:
Replace the polar equations in Exercises 23-48 by equivalent Cartesian equations. Then describe or identify the graph.

40. $\displaystyle \cos^2(\theta) = \sin^2(\theta)$

I think I have it solved, but I'm unsure whether I am correct or not.. All problems in this section have an $\displaystyle r$ to convert to Cartesian (ie $\displaystyle r\cos(\theta) = x$). However, this problem does not.

So I multiplied the entire problem by $\displaystyle \frac {r^2}{r^2}$ which gave me the following:

$\displaystyle \frac {r^2\cos^2(\theta)}{r^2} = \frac {r^2\sin^2(theta)}{r^2}$
$\displaystyle \sqrt{\frac {r^2\cos^2(\theta)}{r^2}} = \sqrt{\frac {r^2\sin^2(theta)}{r^2}}$
$\displaystyle \pm \frac {r\cos(\theta)}{r} = \pm \frac {r\sin(\theta)}{r}$
$\displaystyle \pm \frac {x}{r} = \pm \frac {y}{r}$
$\displaystyle \pm x = \pm y$

Is this correct? Can I drop the $\displaystyle \pm$ sign? And if so, why?

Thanks!

2. Originally Posted by Calix
Instructions:
Replace the polar equations in Exercises 23-48 by equivalent Cartesian equations. Then describe or identify the graph.

40. $\displaystyle \cos^2(\theta) = \sin^2(\theta)$

I think I have it solved, but I'm unsure whether I am correct or not.. All problems in this section have an $\displaystyle r$ to convert to Cartesian (ie $\displaystyle r\cos(\theta) = x$). However, this problem does not.

So I multiplied the entire problem by $\displaystyle \frac {r^2}{r^2}$ which gave me the following:

$\displaystyle \frac {r^2\cos^2(\theta)}{r^2} = \frac {r^2\sin^2(theta)}{r^2}$
$\displaystyle \sqrt{\frac {r^2\cos^2(\theta)}{r^2}} = \sqrt{\frac {r^2\sin^2(theta)}{r^2}}$
$\displaystyle \pm \frac {r\cos(\theta)}{r} = \pm \frac {r\sin(\theta)}{r}$
$\displaystyle \pm \frac {x}{r} = \pm \frac {y}{r}$
$\displaystyle \pm x = \pm y$

Is this correct? Can I drop the $\displaystyle \pm$ sign? And if so, why?

Thanks!
$\displaystyle x = r \cos \theta \Rightarrow \cos \theta = \frac{x}{r}$.

$\displaystyle y = r \sin \theta \Rightarrow \sin \theta = \frac{y}{r}$.

Therefore $\displaystyle \cos^2 \theta = \sin^2 \theta \Rightarrow \frac{x^2}{r^2} = \frac{y^2}{r^2} \Rightarrow y^2 = x^2 \Rightarrow y = \pm x, ~ r \neq 0$.

This answer makes perfect sense since the gradient of these two lines is equal to $\displaystyle \pm 1$ and so the angle they make with the x-axis is $\displaystyle \frac{\pi}{4}, ~ \frac{3 \pi}{4}, ~ \frac{5\pi}{4}$ and $\displaystyle \frac{7\pi}{4}$.

3. Hah, I knew I missed something pretty obvious. Going on only a few hours of sleep..

Thanks a lot!