Question

Question on Parametric Equations:

How do we get from 2sin(t)cos(t)
to
+-sqrt(1-y^2)y

(details in attachment)

Answer 1

First attachment=Question
Second=Answer

Answer 2

guess this is a pretty difficult one

Answer 3

it is not...where are you stuck?

Answer 4

Just wondering what we use to get from:
2sin(t)cos(t)
to
+-sqrt(1-y^2)y

Answer 5

let \[z=\sin(t)\]

then \[z^2=\sin^2(t)=1-\cos^2(t)\]

so

\[z=\pm\sqrt{1-\cos^2(t)}=\pm\sqrt{1-y^2}\]

Answer 6

@Zarkon That's how I would've solved the problem too were the given equation x=sin(t).

However, it's x=sin(2t). Can we still use a similar approach in this case?

Answer 7

\[x=\sin(2t)=2\sin(t)\cos(t)=2(\pm\sqrt{1-y^2})y\]


just plug in what I did above

Answer 8

ahh ok got it, thanks