Question

D??

Answer 1

if sin(x) = 5/13 for this angle what is cos(x)?

Answer 2

well the denom is 13 because it is the hypotenuse

Answer 3

so that would make the answer B

Answer 4

@Vocaloid

Answer 5

good, B is correct

since cos(x) = 12/13 you can plug that into the half angle sin identity

Answer 6

B and C

Answer 7

hm

when I solved for tan(x) I end up with tan(x) = 1/sqrt(3) and tan(x) = -sqrt(3), what x angle values correspond to these values?

Answer 8

so x = pi/6 gives us sin(x) = 1/2 and cos(x) = sqrt(3)/2 giving us tan = sin/cos = 1/2 * 2 / sqrt(3) = 1/ sqrt(3) [answer choice C]

x = 2pi/3 gives us cos(x) = -1/2 and sin(x) = sqrt(3)/2 giving us sin/cos = sqrt(3)/2 * (-2) = -sqrt(3)

making C+D the best options here

got to go get lunch

Answer 9

2

Answer 10

@Vocaloid

Answer 11

disclaimer: not my sol'n, but you would have to use the identity sin(x)sin(y) = (1/2)[(cosx - cosy) - (cosx+cosy)] so you would end up with "cos" in the blank

Answer 12

so 0.36

Answer 13

no

your problem is asking what goes into the blank of

(1/2)(cospi/12 - ___5pi/12)

notice how the solution given is

(1/2)(cospi/12 - cos5pi/12)

so cos would be the sol'n

Answer 14

oh okay i get it , i skipped the fact that there is a blank

Answer 15

A

Answer 16

well done

Answer 17

False

Answer 18

well done

Answer 19

C

Answer 20

it's the first identity so cos(a) should be in the blank

Answer 21

B

Answer 22

This is actually a wild guess

Answer 23

I dont know this one

Answer 24

sin(2 * theta) = 2 sin(theta) cos(theta)

so sin(theta)cos(theta) = (1/2)sin(2theta) so we can re-write the problem like:

h = v^2/4.9 * (1/2) * sin(2theta) = 150

plug v into the equation, leave sin(2theta) and 150 alone, see what you get

Answer 25

B

Answer 26

Actually tbh the calc i plugged it into is giving me error :/

Answer 27

v0 = 44 so

44^2/4.9 * (1/2) = ?

Answer 28

A =)

Answer 29

good

Answer 30

Cos?

Answer 31

first multiply the expression w/ foil

csc^2 - 1

now check, any identities that might be useful here?

Answer 32

cot?

Answer 33

good, cot(x) is your sol'n

idk how they want you to write it but "cot(x)" maybe? or

Answer 34

or maybe just cotx

Answer 35

hm

Answer 36

What do you think would be the best, its 12th grade calc

Answer 37

precalc*

Answer 38

maybe just cotx

Answer 39

I really don't know ;; since the original expression doesn't have parentheses then maybe just cotx

Answer 40

yeah thats what i thought too

Answer 41

T

Answer 42

good

Answer 43

C

Answer 44

I got something a little different

3sin(2x) = 5cos(x)

since sin(2x) = 2sin(x)cos(x) we can re-write the original as:

3[2sin(x)cos(x)] = 5cos(x)
or

6sin(x)cos(x) - 5cos(x) = 0

factor out cos, what do you get?

Answer 45

you don't need to solve for x

6sin(x)cos(x) - 5cos(x) = 0

factor out cos(x) from both terms, what do you get?

Answer 46

6sin(x)-5

Answer 47

good so the entire expression becomes

cos(x) * (6sin(x) - 5) = --> answer choice A

Answer 48

csc

Answer 49

or maybe sin ?

Answer 50

Im looking at the chart you provided

Answer 51

yeah cos is right

Answer 52

okay cos

Answer 53

not csc right

Answer 54

yes

Answer 55

cos?

Answer 56

if you let alpha = pi/3 and beta = pi/6 it should be the fourth identity, so sin not cos

Answer 57

F

Answer 58

this is the cos half angle identity re-stated so I'm leaning towards true

Answer 59

False

Answer 60

good

Answer 61

F

Answer 62

if you take 1 + cot^2(x) = csc^2(x), subtract csc^2(x) from both sides you get

1 + cot^2(x) - csc^2(x) = 0

subtracting 1 from each side gives us

cot^2(x) - csc^2(x) = -1

making the original statement true

Answer 63

A and D

Answer 64

hm

if we let A = x and B = pi/7 then our expression becomes

sin(A)cos(B) - cos(A)sin(B) = sqrt(2)/2

using the sin(A-B) identity we get

sin(A-B) = sqrt(2/2)

plugging the x and pi/7 back in we get

sin(x-pi/7) = sqrt(2/2)

take the arcsin of both sides and solve for x

Answer 65

B

Answer 66

good but there's one more value where sin(x) = sqrt(2)/2, check the unit circle again

Answer 67

C

Answer 68

good so B+C

can we close this question, it's getting a bit laggy