Question 691480
Find sin x/2, cos x/2, and tan x/2
from the given information.
sec x = 10/9 ,270° < x < 360°
sin x/2=	
cos x/2=
tan x/2=
**
sec x=10/9=hypotenuse/adjacent side
hypotenuse=10
adjacent side=9
opposite side=&#8730;(10^2-9^2)=&#8730;(100-81)=&#8730;19
sin x=-&#8730;19/10 (in quadrant IV where sin<0)
cos x=9/10
..
Identities:
sin x/2=±&#8730;[(1-cos x)/2]
choose positive root because x/2 is in quadrant II where sin>0
sin x/2=&#8730;[(1-cos x)/2]=&#8730;[(1-9/10/2]=&#8730;.1/2=&#8730;.05
..
cos x/2=±&#8730;[(1+cos x)/2]
choose negative root because x/2 is in quadrant II where cos<0
cos x/2=-&#8730;[(1+cos x)/2]=-&#8730;[(1+9/10/2]=-&#8730;1.9/2=-&#8730;.95
.. 
tan x/2=sin x/(1+cos x)=-(&#8730;19/10)/(1+9/10)=-(&#8730;19/10)/1.9
..
How to check answers with calculator:
sec x=10/9
cos x=9/10
cos^-1=(9/10)&#8776;25.84º (reference angle in specified quadrant IV)
standard position of angle=360-25.84=334.16
x/2=334.16/2=167.08
reference angle=180-167.08=12.92º
..
sin x/2=sin 12.92&#8776;0.2236..(in quadrant II where sin>0)
&#8730;.05=0.2236..
..
you can check cos x/2 and tan x/2 in the same way