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Mathematics 19 Online
OpenStudy (anonymous):

How do I get x^2-y^2 into polar coordinates?

OpenStudy (anonymous):

x^2+y^2=r^2 so x^2-y^2=-r^2

OpenStudy (anonymous):

Ooh wow.. that was so simple it was hard. Okay thank you! haha

OpenStudy (anonymous):

you are welcome. just multiply both sides by negative one

OpenStudy (jdoe0001):

\(\bf \color{blue}{x= rcos(\theta)\qquad y = rsin(\theta)}\\ \quad \\ x^2-y^2\implies [rcos(\theta)]^2-[rsin(\theta)]^2\implies [r^2cos^2(\theta)]^2-[r^2sin^2(\theta)]\\ \quad \\ r^2[cos^2(\theta)-sin^2(\theta)]\implies r^2[cos(2\theta)]\)

OpenStudy (anonymous):

how does it equal \[\cos (2\theta)\]

OpenStudy (anonymous):

it is r^2(cos(2theta)) cos^2(x) - sin^2(x) = cos(2x)

OpenStudy (anonymous):

trig id: jag was not right btw

OpenStudy (anonymous):

Yay i got it right! Thanks :)

OpenStudy (anonymous):

woops i thought it was -x^2-y^2. sorry

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