Trig
posted by Jon .
Find the exact value of sinx/2 if cosx = 2/3 and 270 < x < 360.
A)1/3
B)1/3
C)sqrt 6/6
D)sqrt 6/6
C, since I KNOW cosx is always positive but I don't know the work involved. I know the half angle formula

First of all, x/2 will be in the second quadrant, since x is in the fourth quadrant. The sine of x/2 will therefore be positive.
Use the formula for sin (x/2) in terms of cos x.
sin(x/2) = sqrt([1cos(x)]/2) = sqrt (1/6) = sqrt6/6
You got the right answer, but you it ssmes to have been a lucky guess.
Cos x is NOT always positive, but it is in this case. 
Thank you. It wasn't really a guess it was either C or D and then I just knew it was positive so that just leaves C.

Jon,
Perhaps it would help if you drew an xy axis system with a unit radius vector in each of the four quadrants.
then in quadrant 1
sin T = y/1 so +
cos T = x/1 so 
tan T = y/x so +
then in quadrant 2
sin T = y/1 so +
cos T = x/1 so  because x is  in q 2
tan T = y/x so 
then in quadrant 3
sin T = y/1 so 
cos T = x/1 so 
tan T = y/x so + because top and bottom both 
then in quadrant 4
sin T = y/1 so 
cos T = x/1 so +
tan T = y/x so 
sin has same sign as its inverse csc
cos has same sign as its inverse sec
tan has same sign as its inverse ctan 
then in quadrant 1
sin T = y/1 so +
cos T = x/1 so +
tan T = y/x so +
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