Question

need help answering math problem regarding sin x/2 cosx/2 and tan x/2 (attached file)
Will give best response and possibly become a fan

Answer 1

all half-angle identities, use a root variation of cos(x), sin(x/2), cos(x/2) and tan(x/2) as well

so if csc(x) = 9 that means that 1/sin(x) = 9,
solving that for "sin(x)" then sin(x) = 1/9

\(\bf sin(x) = \cfrac{1}{9} \implies\cfrac{\text{opposite}}{\text{hypotenuse}} \implies\cfrac{b}{c}\\
c^2 = a^2 + b^2 \implies b = \sqrt{9^2 - 1^2}\\
\color{blue}{cos(x) = \cfrac{b}{c}}\)

Answer 2

well, actually we want "a" not "b" one sec

Answer 3

\(\bf c^2 = a^2 + b^2 \implies a = \sqrt{9^2 - 1^2}\\
\color{blue}{cos(x) = \cfrac{a}{c}}\)

Answer 4

one thing to keep in mind, you're working on the 2nd Quadrant, thus the cosine is negative

Answer 5

okay so what are the ansewrs

Answer 6

I am confused becasue its all over 2?

Answer 7

well, you'd need to use the half-angle identities, do you have them?

Answer 8

I think so

Answer 9

this is what I did

Answer 10

then for all the half-angle identities, as you can see them, they all use cos(x)

Answer 11

well, I don't think the box there was expecting the whole equation, so much as the answer alone

Answer 12

sow aht do we do?

Answer 13

well, we find cos(x) first

\(\bf sin(x) = \cfrac{1}{9} \implies\cfrac{\text{opposite}}{\text{hypotenuse}} \implies\cfrac{b}{c}\\
c^2 = a^2 + b^2 \implies a = \sqrt{9^2 - 1^2}\\
cos(x) = \cfrac{a}{c}\)

Answer 14

what would that give you for cosine?

Answer 15

ok so cosx is = sqrt 80

Answer 16

so sqrt 80/9

Answer 17

\(\bf cos(x) = \cfrac{a}{c}\\
cos(x) = \cfrac{\sqrt{80}}{9}\\
\sqrt{80} \implies\sqrt{4^2\times 5}\implies 4\sqrt{5}\\
cos(x) = \cfrac{ 4\sqrt{5}}{9}\)