Question

Please help. I am studying for an exam for tomorrow. find the sinx of y=2x^2-2

Answer 1

What do you think? Isolate x?

Answer 2

well actually they are asking for sin theta. I had questions where we just drew the triangle within for the linear equation and used soacahtoa, but this is a parabola, that is what is confusing me

Answer 3

if you isolate x then it would be \[\sqrt{(y+2)/2}\]

Answer 4

You will not do well on this exam if this is the sort of question you will be required to answer. It makes very little or no sense. Can you provide the ENTIRE question?

Answer 5

ok one sec

Answer 6

find sin theta and cosine theta if r(v,u) is the point where the terminal side of theta in standard position intersects the unit circle, and u and v satisfy the given conditions. the equation is v=2u^2-2, and u<0, and v<0

Answer 7

SO much more sense.

"unit circle" \(u^{2} + v^{2} = 1\)
"parabola" \(v = 2u^{2}-2\)

Please find where those intersect. with \(u < 0\;and\;v<0\)

Answer 8

ok just a second

Answer 9

sorry it is taking me some time. my teacher never explained this to us, but included it on the review

Answer 10

Did you study quadratic Equations in a previous class? That's all we're doing, so far.

Answer 11

yes so we solve by elimination?

Answer 12

Sort of. You solve by any means available to you. These are not linear equations. A judicious substitution might be simple enough.

Answer 13

u=-1, and v=0

Answer 14

right?

Answer 15

How did you get that?

You have these:

"unit circle" \(u^{2}+v^{2}=1\)
"parabola" \(v=2u^{2}−2\)

From the circle: \(u^{2} = 1-v^{2}\)
Substituting into the parabola: \(v = 2(2-v^{2}) - 2\) or \(2v^{2} + v - 2 = 0\)
Then: \(v = \dfrac{-1\pm\sqrt{1 + 4(2)(2)}}{4} = \dfrac{-1\pm\sqrt{17}}{4}\)
We need a negative v, so \(v = \dfrac{-1-\sqrt{17}}{4}\)
Find a corresponding negative u and we're almost there.

Answer 16

oh ok now I get it, I will find the other one then one moment

Answer 17

please be patient with me if I did this wrong but does u then - \[\sqrt{5}/2\]

Answer 18

How did you do that? Is it right? Where is the work that I can see? You MUST show your work.

Answer 19

ok sorry, I am just not a quick typeset when it comes to math, and this is the first time I am using open study. I will type it

Answer 20

It takes a little practice, but it's well worth the effort.

Answer 21

At some point, you should notice that you are ON the Unit Circle. When you find u and v, you HAVE the cosine and sine. Done!

Answer 22

Now that you said that it makes a lot of sense

Answer 23

\(u^{2} - v^{2} = 1\) That's not a circle.

\(u^{2} + v^{2} = 1\) That's a circle.

Answer 24

oh no. That was a really dumb mistake on my part. ok I will fix that then

Answer 25

Yikes! Are you SURE those four actually are solutions to BOTH equations? You had better check that.

Hint: Your solution promoted the problem from quadratic to quartic! That's almost never a good idea.

Hint: u = u^2 This is not a good substitution. VERY confusing. Pick a new variable, don't reuse an old one.