Question

Help

Answer 1

well, in order to "undo" the cos which inverse trig function would you use?

Answer 2

sin

Answer 3

well that's a good guess but if we take the sin of both sides we just get

sin(cos(3x)) = sin(-1) which doesn't really help us much

but we can take the arccos of both sides to undo the cos, what do you get when you take the arccos of both sidse?

Answer 4

-3

Answer 5

hm

Answer 6

Ugh :/

Answer 7

so we start with cos(3x) = -1

and take the arccos of both sides we get

arccos[cos(3x)] = arccos(-1)

make sense so far?

Answer 8

yes , i think so

Answer 9

the arccos and the cos cancel out to get

3x = arccos(-1)

so that "defines" the value of 3x and is a sufficent sol'n for part I

Answer 10

anyway for part II: they want the angle sol'ns for 3x

so what angle(s) correspond to arccos(-1)?

Answer 11

angle as in on the unit circle?

Answer 12

yes, what angle gives a cos value of -1?

Answer 13

3pi/2

Answer 14

that's sin not cos

Answer 15

cos is the x-coordinate so check to see where the x-coordinate is -1

Answer 16

oh pi

Answer 17

sorry

Answer 18

awesome so part II is just 3x = pi

then for part III is just algebra, solve this equation for x

Answer 19

x=pi/3

Answer 20

good ^^

I kind of forgot that cos(theta) is an even function so part II has pi AND -pi as its sol'ns and part III has pi/3 and -pi/3 as its' sol'ns ;;

Answer 21

good

for part II, "3x = pi" and "3x = -pi" is a better way to express the information

Answer 22

alrighty then for part I how would you solve tan(theta) = -1 for theta?

Answer 23

inverse? :/

Answer 24

good, you'd take the arctan of each side, lmk what you get for theta

Answer 25

arctan[tan(theta) = arctan (-1)

Answer 26

good, so the arctan and theta cancel out on the left side to get theta = arctan(-1)

Answer 27

that is part I right

Answer 28

yes

then for part II simply solve arctan(-1) by finding the angle where tan = -1

Answer 29

3pi/2

Answer 30

keep in mind tan(theta) = sin(theta)/cos(theta)

so you want the angles where sin and theta have the same magnitude but opposite sides

Answer 31

it would be a y value right

Answer 32

tan = y/x

Answer 33

Pi :/

Answer 34

at pi, sin = 0 and cos = -1 so sin/cos = 0 not -1

Answer 35

notice how these two locations have sin and cos equal in magnitude but opposite in sign? your sol'ns are therefore 7pi/4 and 3pi/4

Answer 36

Alright and an expression would be one that would invlude 7pi/4 and 3pi/4 right

Answer 37

yeah, for part III since tan repeats itself every pi units it's

7pi/4 +/- pi and 3pi/4 +/- pi

Answer 38

for number same process so i would take the inverse

Answer 39

yes, the arcsin of both sides

Answer 40

arcsin[sin(theta/2) = arctan (1/2)

Answer 41

it's arcsin on the right side not arctan

Answer 42

Oops

Answer 43

anyway, that gives us theta/2 = arcsin(1/2) as your sol'n for part I

Answer 44

fixed that

Answer 45

anyway for part II just need to solve for theta/2 by finding angles on the UC where sin(theta) = 1/2

Answer 46

2pi/3

Answer 47

sin is the y-coordinate and it needs to be 1/2 not -1/2

Answer 48

so pi/6 and 5pi/6 are your sol'ns for part II

Answer 49

then for part III if theta/2 = pi/6 and theta/2 = 5pi/6 solve for your two theta values

Answer 50

1/2?

Answer 51

wait i would use the unit circle

Answer 52

sqrt3/2

Answer 53

you don't need to use the UC here, this is just simple algebra

if theta/2 = pi/6 what does theta = ?

Answer 54

pi/3

Answer 55

good, and if theta/2 = 5pi/6 then theta = 5pi/3 so your sol'ns for part III are pi/3 and 5pi/3

Answer 56

then for part IV you just need to consider all possible sin values, since sin repeats itself every 2pi units it's

5pi/3 +/- 2pi * n and pi/3 +/- 2pi*n

Answer 57

also you need to go back to the tan problem (#6) and change part III to 7pi/4 +/- n*pi and 3pi/4 +/- n*pi ( I forgot the n's)

Answer 58

Perfect, you must get going now !! =) Thank you !!