Question

http://prntscr.com/93aff4 I feel like I did something wrong.

Answer 1

cos (A/2) = sqrt (1-4/5 / 2) = sqrt (1/5 / 2) = sqrt (1/5 • 1/2) = sqrt (1/10)??

Answer 2

look at page 2 of the pdf. Notice how it's 1+cos(theta) and not 1-cos(theta)

http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

you should have cos(A) = 4/5 since angle A is in QIV

hopefully that helps you find cos(A/2)

Answer 3

oh so 1- 4/5

Answer 4

sorry +

Answer 5

it should be 1+4/5

Answer 6

yes

Answer 7

stupid typos

Answer 8

I put + gives me - ugh

Answer 9

ok so I got 1•4/5•1/2=4/10=2/5?

Answer 10

hmm strange

anyways, the original restrictions on A are that 270 < A < 360

divide EVERY side of that by 2 to get

270 < A < 360
270/2 < A/2 < 360/2
135 < A/2 < 180

So whatever the actual value of A/2 really is, we know that angle A/2 is somewhere in Q2 (between 135 and 180 degrees)

so the final result of cos(A/2) will be some negative number since cos is negative in Q2

Answer 11

what

Answer 12

do you see where it says `270 < A < 360` ?

Answer 13

um Astrophysics said do arcsin then use the angle found and divide by 2 to find the rest of the trig functions for A/2

Answer 14

I think

Answer 15

oh that will work too

Answer 16

Okay then thank you

Answer 17

tell me what you get

Answer 18

ok

Answer 19

I got -36.87 degrees

Answer 20

same here

-36.869897645844 which rounds to -36.87

so that's the approx value of A, but wait a minute, A is supposed to be between 270 and 360. How do we fix this?

Answer 21

http://prntscr.com/93aiy6 um idk o-o

Answer 22

so then um ... 360-(A/2)?

Answer 23

we add 360 degrees to find the coterminal angle

Answer 24

yes

Answer 25

so -36.869897645844 + 360 = 323.130102354157

Answer 26

A = 323.130102354157 roughly
A/2 = ??

Answer 27

323.130/2=161.565

Answer 28

and then compute the cosine of that to get what?

Answer 29

ummm http://prntscr.com/93ajzt

Answer 30

that result happens if you are in radian mode, but you were previously working in degree mode

Answer 31

I used Wolfram haha

Answer 32

here http://prntscr.com/93akg1

Answer 33

A = 323.130102354157
A/2 = 161.565051177079
cos(A/2) = -0.94868329805051 (this is the approximate answer)


It turns out that the exact answer is \[\Large \cos\left(\frac{A}{2}\right) = -\frac{3*\sqrt{10}}{10}\]
How did I get this? I used the identity off that reference sheet I posted above. Look at the attached PDF to see how I got this exact value


Notice how \[\Large -\frac{3*\sqrt{10}}{10} \approx -0.94868329805051\] (use a calculator to evaluate `-3*sqrt(10)/10`) so that confirms the answer in a way

Answer 34

wha

Answer 35

so -3/10 • sqrt 10

Answer 36

yes since \[\Large -\frac{3}{10}\sqrt{10} = \frac{-3\sqrt{10}}{10}\]

Answer 37

ok

Answer 38

so http://prntscr.com/93alc3

Answer 39

correct

Answer 40

okay

Answer 41

says it's wrong

Answer 42

why does it says A is in Q4, but then changes to say that A is in Q3? I just noticed that

Answer 43

oh wait, maybe the part about A being in Q4 is a hypothetical example. Nvm, let me fix

Answer 44

o-o

Answer 45

ok so A is in Q3 which means cos(A) is negative

cos(A) = -4/5 (not +4/5)

that leads to \(\Large \cos\left(\frac{A}{2}\right) = -\frac{\sqrt{10}}{10}\)

The steps are shown in the attached PDF

Answer 46

ok

Answer 47

so it's -sqr10/10?

Answer 48

yeah

Answer 49

Okay

Answer 50

wow that was very exhausting haha

Answer 51

yeah it's a lot but after a lot of practice, it should come easier

Answer 52

yes thanks :)

Answer 53

you're welcome

Answer 54

oh um @jim_thompson5910 I still need help haha -- http://prntscr.com/93aonf this one, I can't get a solid number to answer with ^^;

Answer 55

the degree I got was 5.77 for A/2

Answer 56

if B is in Q3, then what is the value of cos(B) ?

Answer 57

wha

Answer 58

hold on lol

Answer 59

270-11.54 degrees? then div by 2

Answer 60

Are you familiar with this identity ?

\[\Large \sin^2(B) + \cos^2(B) = 1\]

Answer 61

Yes

Answer 62

if sin(B) = -1/5, do you see how to find the value of cos(B) ?

Answer 63

OH
right yes... Forgot about tha

Answer 64

tell me what cos(B) is equal to

Answer 65

\[\sqrt{24}\]

Answer 66

or\[2\sqrt{6}\]

Answer 67

close

Answer 68

no? o_o\[\sqrt{5^{2}-1^{2}}=\sqrt{25-1}=\sqrt{24}=2\sqrt{6}\]

Answer 69

you should have

x^2 + (-1/5)^2 = 1^2
x^2 + 1/25 = 1
x^2 = 1 - 1/25
x^2 = 25/25 - 1/25
...
...
...
x = ???

Answer 70

what

Answer 71

For now, I'm letting cos(B) be unknown. So I'm just calling it x

Answer 72

sin(B) = -1/5

Answer 73

sorry multitasking multiple math questions

Answer 74

so that's how I got x^2 + (-1/5)^2 = 1

I plugged in sin(B) = -1/5 into that identity shown above

Answer 75

wait so did you put the answer or no because I'm not sure where you're going lol

Answer 76

no I haven't gotten to the answer yet

Answer 77

I was just showing how to find cos(B)

you'll then use cos(B) for the next step, but you need cos(B) first

Answer 78

Oh okay gotcha
I have to finish in 9 minutes btw

Answer 79

So I may not respond as quickly, I mean

Answer 80

well long story short, you should get \(\Large \cos\left(B\right) = -\frac{2\sqrt{6}}{5}\)

you'll plug that into the identity shown on the reference page (study that reference booklet, you'll use those identities a lot)

follow the steps shown on the attached pdf I'm sending and you'll get this

\[\Large \sin\left(\frac{B}{2}\right) = \sqrt{\frac{5+2\sqrt{6}}{10}}\]\

Answer 81

Oh, and I'm not sure why this is wrong but it won't accept the other non-theta symbol either so http://prntscr.com/93atzu

Answer 82

okay thank you

Answer 83

you have the right pattern (cos cos minus sin sin) but you have theta showing up twice

Answer 84

yeah but it won't accept the other symbol as a registered value in the system

Answer 85

should be \[\Large \cos(\theta + \phi) = \cos(\theta)\cos(\phi) - \sin(\theta)\sin(\phi)\]

Answer 86

yes but that symbol is unregistered

Answer 87

try typing it out like

cos(theta + phi) = cos(theta)cos(phi) - sin(theta)sin(phi)

kinda strange how they ask you a question and not have a proper way to answer

Answer 88

okay

Answer 89

WHA! I COPY PASTE AND IT DOESN'T WORK BUT IT WORKS LIKE THIS?

Answer 90

AUGH

Answer 91

computers can be really picky

Answer 92

well time's up, thanks for helping with what you did

Answer 93

:)

Answer 94

you're welcome