# Thread: convert to polar form

1. ## convert to polar form

x^2-y^2=5

I see examples of people adding r, but I'm not understanding how. this is my first experience with trig taking in precalc.

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2. ## Re: convert to polar form Originally Posted by necdir x^2-y^2=5

I see examples of people adding r, but I'm not understanding how. this is my first experience with trig taking in precalc.

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The way to convert rectilinear coordinates (x, y) to polar coordinates is $\displaystyle ( r, \theta)$ is
$\displaystyle x = r~cos(\theta)$
$\displaystyle y = r~sin(\theta)$

So what is $\displaystyle x^2 - y^2 = (r~cos( \theta))^2 - (r~sin(\theta))^2$

Can you take it from here?

-Dan

3. ## Re: convert to polar form

This is exactly where I get lost in the problem. Are we able to send pictures so I can show. I basically got as far as you showed but equal to 5. Now I'm not sure how to proceed.

4. ## Re: convert to polar form

You have $\displaystyle r^2cos^2(\theta)- r^2sin^2(\theta)= 5$. You can factor out $\displaystyle r^2$ from both sides to get $\displaystyle r^2(cos^2(\theta)- sin^2(\theta))= 5$. If you do not consider that sufficient as an answer, you can use the trig identity $\displaystyle cos^2(\theta)- sin^2(\theta)= cos(2\theta)$ to write $\displaystyle r^2 cos(2\theta)= 5$.

5. ## Re: convert to polar form Originally Posted by HallsofIvy You have $\displaystyle r^2cos^2(\theta)- r^2sin^2(\theta)= 5$. You can factor out $\displaystyle r^2$ from both sides to get $\displaystyle r^2(cos^2(\theta)- sin^2(\theta))= 5$. If you do not consider that sufficient as an answer, you can use the trig identity $\displaystyle cos^2(\theta)- sin^2(\theta)= cos(2\theta)$ to write $\displaystyle r^2 cos(2\theta)= 5$.
Thank you so much! Exactly what I needed

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