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How do I get x^2-y^2 into polar coordinates?
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x^2+y^2=r^2 so x^2-y^2=-r^2
Ooh wow.. that was so simple it was hard. Okay thank you! haha
you are welcome. just multiply both sides by negative one
\(\bf \color{blue}{x= rcos(\theta)\qquad y = rsin(\theta)}\\ \quad \\ x^2-y^2\implies [rcos(\theta)]^2-[rsin(\theta)]^2\implies [r^2cos^2(\theta)]^2-[r^2sin^2(\theta)]\\ \quad \\ r^2[cos^2(\theta)-sin^2(\theta)]\implies r^2[cos(2\theta)]\)
how does it equal \[\cos (2\theta)\]
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it is r^2(cos(2theta)) cos^2(x) - sin^2(x) = cos(2x)
trig id: jag was not right btw
Yay i got it right! Thanks :)
woops i thought it was -x^2-y^2. sorry
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