At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
so its r=8sin(theta)/cos^2(theta)?
so the answer is in r= form when there are no set points. But with set points you write it as (r, theta) ?
I don't know what you mean by set points. In polar form there is a magnitude and a direction/angle.
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?
by set points in meant (1,4) or something to that extent. I was able to answer that though.
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.
actually i have 2. and what trig identity would i use here? cos(2A) = cos²A − sin² ?
im not sure how this applies if i have r^2 * cos(2*theta)
actually you were right, you don't use the half angle...
you use the one that matches, double angle.
there are 3 for cosine, i cant tell which one would work in this scenario
when you put that one in you'll end up with two terms both multiplied by r^2
then you can simply substitute them for x^2 and y^2
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!