anonymous
  • anonymous
Algebra 2 Help! The point (1, −1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of θ? Make sure to show all work.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
zepdrix
  • zepdrix
|dw:1450292176273:dw|Hey Lidayuh :) So here is where we are located.
zepdrix
  • zepdrix
Let's find this radial distance r which measures from the origin to our point.
zepdrix
  • zepdrix
If we think of this as a triangle,|dw:1450292281431:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

zepdrix
  • zepdrix
Our Pythagorean Theorem will give us the length of the missing side, ya?
anonymous
  • anonymous
Yeah, ok.
zepdrix
  • zepdrix
So what do you get for that length? c: Do it!! That information will `complete` our triangle, and from there we'll be able to directly use our SOH CAH TOA or however you've learned it.
anonymous
  • anonymous
What is the Pythagorean Theorem formula?
zepdrix
  • zepdrix
Come on Lidia -_- you gotta tattoo that one into your brain. It's going to show up over and over and over in every single math class you ever take. \(\large\rm a^2+b^2=c^2\) Or for us we'll think of it like this: \(\large\rm x^2+y^2=r^2\)
zepdrix
  • zepdrix
And you're plugging in your coordinate values: \(\large\rm (x,y)=(1,-1)\)
anonymous
  • anonymous
Ok, but \[1^{2}\] + \[-1^{2}\] is equal to 0.... I think I'm confused.
zepdrix
  • zepdrix
No, you're squaring away the negative.
zepdrix
  • zepdrix
\(\large\rm (-1)^2\ne-1\)
anonymous
  • anonymous
So then it would be 1, right?
zepdrix
  • zepdrix
\(\large\rm (-1)^2=1\) Mm k good, so how bout the rest of the calculations? :)\[\large\rm 1^2+(-1)^2=r^2\]\[\large\rm 1+1=r^2\]Remember, we're trying to solve for r, not r squared.
zepdrix
  • zepdrix
\[\large\rm 2=r^2\]How do we solve for r from here? What operation will undo "squaring" ?
anonymous
  • anonymous
2^2?
zepdrix
  • zepdrix
`Square` to undo the `square` on the r? Hmm no, that's not what we're looking for.
zepdrix
  • zepdrix
`Addition` will undo `Subtraction`. `Multiplication` will undo `Division`. ` ` will undo `Square`. Don't remember? :)
anonymous
  • anonymous
No, I don't remember lol
zepdrix
  • zepdrix
`Square root` is what we're looking for.
anonymous
  • anonymous
Ohhhhh that makes sense
zepdrix
  • zepdrix
\[\large\rm \sqrt{2}=\sqrt{r^2}\]So we'll square root each side. We're doing this because `square root` and `square` are opposite operations, they undo one another, you can sort of think of it like, they "cancel out",\[\large\rm \sqrt{2}=r\]
zepdrix
  • zepdrix
|dw:1450293029118:dw|
zepdrix
  • zepdrix
So that gives us all of the sides to our triangle that we traced out.
anonymous
  • anonymous
So now we just label the sine, cosine and tangent?
zepdrix
  • zepdrix
In reference to where our angle theta is, we can think of the sides like this,|dw:1450293148164:dw|
zepdrix
  • zepdrix
Yes, use that information to figure out sine, cosine and tangent. Remember how to set those up?\[\large\rm \sin \theta=\frac{opposite}{hypotenuse}\qquad\to\qquad \sin \theta=\frac{?}{?}\]
anonymous
  • anonymous
\[\frac{ -1 }{ \sqrt{2} }\]
zepdrix
  • zepdrix
Good, as a final step, you should `rationalize` that fraction.
zepdrix
  • zepdrix
We don't like to leave `irrational numbers`, like sqrt(2), in the denominator if we can avoid it. So we find a way to move it up into the numerator.
zepdrix
  • zepdrix
\[\large\rm \frac{-1}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}\quad=\quad?\]
anonymous
  • anonymous
\[-\frac{ \sqrt{2} }{ 2 }\]
anonymous
  • anonymous
Right?
zepdrix
  • zepdrix
yup! :)
anonymous
  • anonymous
So, that is the answer to the whole question, right?
anonymous
  • anonymous
@zepdrix
zepdrix
  • zepdrix
No, that is the answer to one-third of your question. sine theta = -sqrt(2)/2. Now you also have to find cos theta and tan theta

Looking for something else?

Not the answer you are looking for? Search for more explanations.