Question

HElp with 4 part question

Answer 1

any ideas how you would start graphing cos(x)? since they want two periods start from -2pi and go to 2pi

Answer 2

it's really important to know how to do this, I would recommend reviewing sin and cos graphs on your own time

Answer 3

part II: what's the domain and range of cos?

Answer 4

still there? if you don't know the domain and range of cos from memory you must look them up

Answer 5

So all real numbers

Answer 6

sorry i was afk

Answer 7

good so the domain is all real #'s and the range is [-1,1]

Answer 8

okay so the graph you showed me for part I that is the one I used, so for part III can we do the markings on that?

Answer 9

for part III we will be restricting the domain of cos from [0,pi], mark this region on your graph

Answer 10

place an appropriate pair of tick marks at 0 and pi to mark the region from 0 to pi

Answer 11

not -pi and pi

0 and pi

Answer 12

well done, so you will be inverting this region

Answer 13

now, we will draw the line y = x

Answer 14

now we will take the region of the graph we have selected and imagine that the line y = x is a mirror and flip the graph over

Answer 15

(you will need to extend your y-axis up to pi so the whole graph shows up)

the blue region is your new arccos(x) function

Answer 16

and that's it, let me know if you need help with the graph because they give you a graph that's been marked on increments of 1 instead of pi which can be hard to graph

Answer 17

um yeah

Answer 18

i need help graphing it on this type of graph

Answer 19

I think I basically butchered it :/

Answer 20

the key is to identify key points on the graph first

Answer 21

well zero is one

Answer 22

for example, cos(0) = 1 so start with (0,1)

Answer 23

okay got that

Answer 24

awesome, then the next critical point is (pi/2,0), try to find where that would be on the graph, pi/2 is like 1.7 ish

Answer 25

1.57 actually

Answer 26

wait (0,1.57) or (1.57,0)

Answer 27

(pi/2,0)

Answer 28

pi/2 is the x-coordinate

Answer 29

(pi/2,0) means pi/2 is the x-coordinate, so you would estimate where pi/2 would be on the x-axis, then where 0 would be on the y-axis, then draw the appropriate point

Answer 30

anyway the next point after that is (pi,-1)

Answer 31

then connect the points with a curve appropriate for a cos graph and then draw the inverse

Answer 32

yeah that's fine actually (I forgot the tick marks go by 2 not 1)

Answer 33

then you would try and connect them with a curved line

Answer 34

remember that cos only goes up to 1 and down to -1

Answer 35

(0,1), (pi/2,0) and (pi, -1)

Answer 36

yeah that's better

i know the draw tool is a little hard to use but it would be good to try and smooth out the curve a bit more

Answer 37

anyway when you're done with that just try reflecting it across y = x

Answer 38

check the first answer choice again

where on the unit circle would sin(theta)/cos(theta) = -sqrt(3)?

Answer 39

-pi/3

Answer 40

good

all the other ones are right

Answer 41

first one needs to be negative but otherwise good

Answer 42

-1.07

Answer 43

alright for Part I you just need to solve 4sin^2(x) - 3 = 0 for sin^2(x), lmk what you get

Answer 44

pi/3

Answer 45

part I is just asking you to solve for sin^2(x), not x

Answer 46

like, when you take 4sin^2(x) - 3 = 0 and isolate sin^2 (x) you get

sin^2 (x) = 3/4, that's all they want for part A

Answer 47

anyway for part II how would you solve sin^2 (x) = 3/4 for sin(x)? remember we are not solving for x yet, just sin(x)

Answer 48

just as a side note we have a lot of new users today and they're kind of being butts in the chat so if I seem tense it's not you, it's just them getting on my nerves

Answer 49

okay so 3/4

Answer 50

and no worries

Answer 51

sin^2(x) = 3/4 so to solve for sin(x) we would square root each side so sin(x) = sqrt(3)/2 = the sol'n for part II

Answer 52

actually it's +/- sqrt(3)/2

Answer 53

anyway, for part III we just need to find the angles on the UC where sin(theta) = sqrt(3)/2 or -sqrt(3)/2

Answer 54

okay so 7pi/6

Answer 55

be careful, 7pi/6 is where cos(theta) = -sqrt(3)/2 not sin

Answer 56

5pi/6

Answer 57

remember that sin is the y-coordinate

Answer 58

pi/3

Answer 59

good, pi/3 is one

Answer 60

5pi/3

Answer 61

oh okay so theres going to be 4 values

Answer 62

yes

Answer 63

pi/3 , 4pi/3 , 5pi/3 , 2pi/3

Answer 64

good

Answer 65

anyway that's it for parts I-III