Question

Question on Trig. Identities on this problem that involves converting rectangular equation to polar form for xy=16.

Answer 1

\[xy=16 (1st step), (rcos \theta)(rsin \theta)=16 (2nd step), (\cos \theta)(\sin \theta)r^2=16 (3rd step), \frac{ 1 }{ 2 }(\sin2 \theta)r^2=16 (4th step).\] How do I get from the 3rd step to the 4th step?

Answer 2

x = r cos Θ
y = r sin Θ

xy = 16
(r sin Θ)(r cos Θ) = 16
r² sin Θ cos Θ = 16

Answer 3

From the third step to the 4th?
You apply your Sine Double Angle Identity :)

Answer 4

\[\Large\rm \color{orangered}{2\sin \theta \cos \theta=\sin(2\theta)}\]We want to make that show up in our problem.
When we're only given this:\[\Large\rm \sin \theta \cos \theta\]We'll make a 2 show up by double two things at once,
multiplying by 2, and dividing by 2.\[\Large\rm \frac{1}{2}\cdot2\cdot\sin \theta \cos \theta\]And we can apply our identity from there :)\[\Large\rm \frac{1}{2}\cdot\color{orangered}{2\sin \theta \cos \theta}\]

Answer 5

by doing* two things at once,
blah typo

Answer 6

\[(\cos \theta)(\sin \theta)r^2=16\] Could u do the next step for that?

Answer 7

Just a tiny bit confused on the double angle identity part

Answer 8

\[(\cos \theta)(\sin \theta)r^2=16\]\[\sin \theta \cos \theta r^2=16\]\[\frac{1}{2}\cdot2\cdot \sin \theta \cos \theta r^2=16\]\[\frac{1}{2}\cdot\color{orangered}{2\cdot \sin \theta \cos \theta} r^2=16\]\[\frac{1}{2}\cdot\color{orangered}{\sin (2 \theta)} r^2=16\]

Answer 9

Why do we multiply by 2 and divide by 2 for the 2? That is the only part that I don't get, other than that, I understand on how to do rest of the problem.

Answer 10

Well you're probably used to this simple trick in math,
do something to one side, do it to the other as well.

We're using a different trick,
do something to one side, undo it at the same time.

Example:
\(\Large\rm x= x-2+2\)
Maybe I really need an x-2 to show up so I can use it in some way.

Like in this example:\[\Large\rm \frac{x}{x-2}\]Yes, we could go through the process of long division, but instead we could add and subtract 2 in the numerator,\[\Large\rm =\frac{x-2+2}{x-2}=\frac{x-2}{x-2}+\frac{2}{x-2}=1+\frac{2}{x-2}\]And that's how we would divide the example I gave.

Answer 11

We're multiplying by 2,
we're also dividing by 2 because we have to keep things balanced.

Then we `use` the 2 which is multiplying to fix our identity.

Answer 12

Ok, thanks, I got it now after looking at your examples

Answer 13

You can think of it like this if it helps maybe,\[\Large\rm \sin \theta \cos \theta=\frac{2}{2}\sin \theta \cos \theta=\frac{1}{2}\cdot 2\sin \theta \cos \theta\]Ok cool :)

Answer 14

yea, that definitely helps by thinking it as "keeping the balance." :D