Question

Another long question.

Answer 1

well, first step would be to sketch the graph of sin(x) from -2pi to 2pi

Answer 2

for "critical points" I couldn't find a precise definition but a good idea would be to label the maximum, minimum, and x-intercepts

Answer 3

that's a good start

when does sin(x) reach its maximum value 1?

Answer 4

2

Answer 5

wouldn't the lines connect?

Answer 6

hm, i'm not sure what you mean

the lines you've plotted are the values when sin(x) = 0 so you would plot (-2pi,0) and (2pi,0) as your starting point

then, you would find the value(s) when sin(x) = 1 and connect the points with a curved line

Answer 7

for example, sin(pi/2) = 1 so we could plot:

Answer 8

and then keep going to find other values when sin(x) = 0, 1, or -1 as our critical values

Answer 9

good, now what's another value when sin(x) would be 0?

Answer 10

well 0

Answer 11

good, what else? it's between 0 and 2pi

Answer 12

um -pi

Answer 13

i mean pi

Answer 14

wrong side sorry

Answer 15

good, both pi and -pi have sin values of 0, so you could plot (pi,0) and connect it

Answer 16

good, then connect (pi/2,1) to (pi/0) using a curved line

Answer 17

then, find a value of sin(x) where it would equal -1 and keep connecting

Answer 18

Okay I got this, a value of sin of -1 would b um 0?

Answer 19

sin(0) = 0

check the unit circle for a value of x where sin(x) = -1

Answer 20

oh -1??

Answer 21

oh okay so 3pi/2

Answer 22

so you can plot (3pi/2, -1) and then connect it to (2pi, 0) with a curved line

Answer 23

Sorry I was getting confused

Answer 24

the curved line needs to be in the other direction like :

Answer 25

then these two points need to be connected like this:

Answer 26

so the overall sin graph is a nice curve like so:

Answer 27

like that kind of

Answer 28

yeah just make it a little neater so the line crosses through the points its supposed to

Answer 29

like the two blue dots

Answer 30

good now connect (pi,0) and (3pi/2, -1) with a smooth curved line

Answer 31

other way around

Answer 32

I would recommend keeping a sin(x) graph next to you as a reference

Answer 33

good

now for the left side, what x-value would sin(x) = -1?

Answer 34

pi

Answer 35

or wait 3pi/2

Answer 36

good but we want to extend the graph on the left side, and if you remember we can subtract/add 2pi to an angle to get an "equivalent" one

so 3pi/2 - 2pi = ?

gives you another value when sin(x) = -1

Answer 37

-pi/2

Answer 38

okay so graphed that

Answer 39

(sorry got distracted by mod stuff)

sin(pi/2) = -1 not 0

Answer 40

something like that

Answer 41

then connect that to (-pi,0) with a curved line

Answer 42

needs to be a curved line

Answer 43

better

so, what x-value would sin(x) = 1 on the left side of the graph?

Answer 44

um 3

Answer 45

as a hint we determined that sin(pi/2) = 1

so pi/2 - 2pi =

gives you another place where sin(x) = 1

Answer 46

pi/2 - 2pi = -3pi/2, plot (-3pi/2, 1) then connect w/ a smooth curved line as usual

Answer 47

and that's it

Answer 48

that kind of took a while but the rest of the worksheet is a lot easier thankfully

like for part IIA) amplitude of 1.5sin(4x) = ?

Answer 49

1.5

Answer 50

awesome

for part IIB) period = 2pi/b where b is the coefficient of x

since the coefficient of x is 4, period = ?

[leave in terms of pi, dont' round to a decimal]

Answer 51

pi/2

Answer 52

awesome

for C) just compare the amplitudes of 1.5sin(4x) and sin(x) by stating "the amplitude of 1.5sin(4x) is 1.5 while the amplitude of sin(x) is 1", similar process for part D

Answer 53

The period of 1.5sin(4x) is 1.5 while the period of sin(x) is 1



woul it be this

Answer 54

yes, and then repeat for the periods (recall that the period of sin(x) is 2pi and the period of 1.5sin(4x) is pi/2 like we calculated before

Answer 55

The period of 1.5sin(4x) is pi/2 while the period of sin(x) is 2pi

Answer 56

good

for part E) do you know how you would approach graphing it?

Answer 57

part III I mean

Answer 58

graph 1.5sin(4x) right

Answer 59

yes, do you need any help with that?

Answer 60

Let me try and and then yu can check it if that is alright

Answer 61

sure

Answer 62

Dont want to fish for answers, ya know

Answer 63

Would it be something like this

Answer 64

two things

the amplitude is 1.5 so the graph needs to go up to 1.5 and down to -1.5

the period is pi/2 and it only wants 2 periods so the graph should start at -pi/2 and end at pi/2

Answer 65

that's a good attempt, but 2 periods only means 2 cycles of the graph, so:

Answer 66

that's the first cycle (sorry if it's hard to see, I can re-draw it neater)

Answer 67

and then the second cycle

Answer 68

something like that

Answer 69

almost there

just make sure the maximum/minimum is 1.5 and -1.5 respectively

also the graph needs to start and end on (-pi/2,0), so: (pi/2,0)

Answer 70

the meat is there it just needs to be tidied up a bit, you can put lines at -1.5 and 1.5 to help guide you as long as you delete them later

Answer 71

Phew, I think this is it

Answer 72

this is the graph

Answer 73

almost there ^_^ just need to trim off those ends and make the graph start and end at (-pi/2,0) and (pi/2,0)

Answer 74

perfect

Answer 75

Ugh thank you sooo miuch, really is so dreading