Precalculus

I am trying to submit this homework in but i guess i'm not doing it in exact values because it is not accepting it. I know i'm supposed to be using half angle formulas but maybe the quadrants are messing me up. Please help!

Find sin x/2 , cos x/2 , and tan x/2 from the given information. tan x = square root (15) , 0° < x < 90°

Find sin x/2 , cos x/2 , and tan x/2 from the given information.
cos x = (−12/13) 180° < x < 270°

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1. if tanx = √15 or √15/1 , and x is in quad I
make a sketch of a right-angled triangle with opposite = √15
r^2 = (√15)^2 + 1^2 = 16
r = 4
then sinx = √15/4 and cosx = 1/4

if x is in quad I, then certainly x/2 is in quad I and everybody stays nicely positive

we know:
cosx = 1 - 2sin^2 (x/2)
1/4 = 1 - 2sin^2 (x/2)
2sin^2 (x/2) = 3/4
sin^2 x/2 = 3/8 = 6/16
sin (x/2) = √6/4
also sin^2 (x/2) + cos^2 (x/2) = 1
cos^2 (x/2) = 1 - 6/16 = 10/16
cos (x/2) = √10/4

also tan(x/2) = sin(x/2) / cos(x/2)
= (√6/4) / (√10/4)
= √6/√10
or √60/10 = 2√15/10 = √15/5

for the second, since x is in the third,
x/2 is in II
(e.g. if x = 182°, x = 91° ; if x = 278 , x/2 = 139°

given: cosx = - 12/13, other side is 5
sinx = -5/13

same as before ...
cos x = 1 - 2sin^2 (x/2)
-12/13 = 1 - 2sin^2 (x/2)
2sin^2 (x/2) = 1 + 12/13 = 25/13
sin^2 (x/2) = 25/26
sin (x/2) = ± 5/√26 , but x/2 is in II, so
sin (x/2) = 5/√26 or 5√26/26

again, as before:
sin^2 (x/2) + cos^2 (x/2) = 1
25/26 + cos^2(x/2) = 1
cos^2(x/2) = 1-25/26 = 1/26
cos(x/2) = ±1/√26, but in II cos(x/2) = -1/√26

finally, tan(x/2) = (5/√26) / (-1/√26) = -5

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posted by Reiny

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