Posted by Jon on Friday, February 15, 2008 at 5:53pm.
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

Trig  drwls, Friday, February 15, 2008 at 6:08pm
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.

Trig  Jon, Friday, February 15, 2008 at 6:13pm
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.

Trig  Damon, Friday, February 15, 2008 at 6:20pm
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

Trig TYPO  Damon, Friday, February 15, 2008 at 6:21pm
then in quadrant 1
sin T = y/1 so +
cos T = x/1 so +
tan T = y/x so +
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