Question

Find sin (a/2) if cos(a)= 12/13 for 0 degrees≤ a≤ 90 degrees

Answer 1

hmmm if the angle is between \(\bf 0^o < a < 90^o\ that'd be 1st Quadrant

what signs are sine and cosine? is sine + or -? is cosine + or - anyway?

Answer 2

hmmm well.... just notice.. you just need the half-angle... hmmm ok

Answer 3

It does't say so i assume they are both positive

Answer 4

well just bear in mind that -> \(\bf sin\left(\cfrac{{\color{red}{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1-cos({\color{red}{ \theta}})}{2}}\)

and you're given the cosine already

keep in mind that you're in the 1st Quadrant, so sine has a sign there, and that's the root you'd keep

Answer 5

If it's in the first quadrant, that means its positive, right?

Answer 6

yeap, which means \(\bf sin\left(\cfrac{{\color{red}{ \theta}}}{2}\right)=+ \sqrt{\cfrac{1-cos({\color{red}{ \theta}})}{2}}\)

Answer 7

So how do i find theta?

Answer 8

well.. you're asked to find sin(a/2)
you're already given the cosine of angle "a" anyhow
and that's all you'd need to get sin(a/2)

Answer 9

So all i have to do is input 12/13 for theta into the equation and solve?

Answer 10

@jdoe0001

Answer 11

yeap pretty much, yes

Answer 12

do you need help

Answer 13

I got -6, but thats not an option for me

Answer 14

\(\bf sin\left(\cfrac{{\color{red}{ \theta}}}{2}\right)=+ \sqrt{\cfrac{1-\left({\color{brown}{ \frac{12}{13}}}\right)}{2}}\)

Answer 15

-6? from a square root...... ?

Answer 16

keep in mind that 1 = 13/13
thus one could say \(\large {
sin\left(\cfrac{{\color{red}{ a}}}{2}\right)=+ \sqrt{\cfrac{\frac{13}{13}-\left({\color{brown}{ \frac{12}{13}}}\right)}{2}} }\)

Answer 17

On the top where I have 12/13, wouldn't i have to get a common denominator?

Answer 18

yues

Answer 19

anyhow.. yes.. but doesn't end up with -6 though

Answer 20

then I'm left with 1/13

Answer 21

yeap

Answer 22

and then divide that by 2 and find the square root?

Answer 23

I got it! thank you so much(:

Answer 24

yeap

Answer 25

yw