Question

For Jim

Answer 1

@jim_thompson5910

Answer 2

\[\sin^-1(\frac{ \sqrt{3} }{ 2 })\]

Answer 3

I can do it on the calculator obviously, but he wants us to know how to do with with a reference triangle or something?

Answer 4

are you given a unit circle? or this reference triangle you refer to?

or is this something he wants you to memorize?

Answer 5

Yes I think he wants us to memorize? He will not be giving us a unit circle which I think is stupid

Answer 6

Ill show you how he does this kind of problem

Answer 7

ok here's the unit circle

https://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Unit_circle_angles_color.svg/2000px-Unit_circle_angles_color.svg.png

which looks like it has a lot of values on it. But honestly, it's not that bad. Notice how we have symmetry going on in a lot of places. For instance the coordinates at 60 degrees and 120 degrees are nearly identical except for x is negative

Answer 8

So all you need to memorize is the stuff in Q1 (quadrant 1)

Everything else will follow from that. If you need to find the cosine of say 150 degrees, then you subtract that from 180 to get

180 - 150 = 30

then you compute the cosine of 30 degrees

cos(30) = -cos(150) = -sqrt(3)/2

Answer 9

This problem is sin though, that picture is just an example of how he does it

Answer 10

Recall that any point on the unit circle is (x,y)

x = cos(theta)
y = sin(theta)

so basically the y coordinate is equal to the sine value

Answer 11

Cant I just plug the answers into my calculator until i get the square root 3 over 2 lol

Answer 12

you could but you might take more time than you want

Answer 13

I just dont get his stupid reference triange he always uses radians and i hate it

Answer 14

if you want to go the calc route, then I would compute

sin(pi/6)
sin(pi/4)
sin(pi/3)

and see which one spits out sqrt(3)/2, which is roughly 0.866

Answer 15

See thats the issue, theres pi/3 and 2pi/3 which give you the same answers so thats why i need to learn the reference triangle thing

Answer 16

let's set up the reference triangle

Answer 17

Okay yay

Answer 18

does it say how theta is restricted?

Answer 19

No

Answer 20

Not in this one

Answer 21

it's a 1, 2, root 3 right angled triangle.?

Answer 22

ok, so the range of `arcsine` is from -pi/2 to pi/2. This spans from Q4 to Q1



as shaded in the drawing

Answer 23

the input is positive, so the result will be positive

meaning whatever theta is, it's in Q1

Answer 24

How would I have known that though?

Answer 25

about the range of arcsine?

Answer 26

yes

Answer 27

I know the quadrants and where each trig function is postive but thats it

Answer 28

Q1 all are positive
Q2 sin and csc
Q3 tan and cot
Q4 cos and sec

Answer 29

this is something that has a bit of explanation to it, which I'm not going to re-invent the wheel on. Instead I'll refer you to a page where I think it could help. Let me know if it does or not

http://www.math.tamu.edu/~austin/section4_6.pdf

see page 3. notice how the domain restriction makes the function one-to-one to allow for an inverse to happen

Answer 30

sine goes -1 to +1 "endlessly". arcsine goes from -infinitity radians to + inifinity radians unless you stay within the "principal values", which are designed to stop this "numerical/algebraic car crash".

Answer 31

I guess you could say that Q4 handles the negative outputs for sine, Q1 does the positive outputs.

Q2 is a complete identical repeat of what Q1 does (positive results)
Q3 is a repeat of Q4

Answer 32

Yeah I just have an issue knowing which quadrant to draw the triangle in

Answer 33

arcsine will end up in either Q1 or Q4

but the input is positive (sqrt(3)/2), so we're going to end up somewhere in Q1

Answer 34

okay got it