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**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!