anonymous
  • anonymous
convert to polar form. x^2=8y
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
x=rcos(theta) y=rsin(theta)
anonymous
  • anonymous
r^2cos^2(theta)=8rsin(theta)
anonymous
  • anonymous
so its r=8sin(theta)/cos^2(theta)?

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anonymous
  • anonymous
right
anonymous
  • anonymous
so the answer is in r= form when there are no set points. But with set points you write it as (r, theta) ?
anonymous
  • anonymous
I don't know what you mean by set points. In polar form there is a magnitude and a direction/angle.
anonymous
  • anonymous
and i'd rather not write up a new question for these, but im also struggling the other way around. converting polar to rectangular. 1) r^2*cos(2theta)=1 2)r=2cos(theta)-4sin(theta) think you can help out?
anonymous
  • anonymous
by set points in meant (1,4) or something to that extent. I was able to answer that though.
anonymous
  • anonymous
when you run into something with a coefficient of theta use your trig identities to get rid of it so you can use x=rcos(theta) y=rsin(theta) to convert back to rectangular.
anonymous
  • anonymous
actually i have 2. and what trig identity would i use here? cos(2A) = cos²A − sin² ?
anonymous
  • anonymous
half angle
anonymous
  • anonymous
cos(A/2)
anonymous
  • anonymous
im not sure how this applies if i have r^2 * cos(2*theta)
anonymous
  • anonymous
cos(theta/2)= sqrt((1+cos(theta))/2)
anonymous
  • anonymous
actually you were right, you don't use the half angle...
anonymous
  • anonymous
you use the one that matches, double angle.
anonymous
  • anonymous
there are 3 for cosine, i cant tell which one would work in this scenario
anonymous
  • anonymous
cos(2a)=cos^2*a)+sin^2(a)
anonymous
  • anonymous
when you put that one in you'll end up with two terms both multiplied by r^2
anonymous
  • anonymous
then you can simply substitute them for x^2 and y^2
anonymous
  • anonymous
so I'm at r^2cos^2(theta)- r^2sin^2(theta)=1. which turns to x^2-y^2=1...and thats the answer on the sheet. Got it, thanks!
anonymous
  • anonymous
You're welcome

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