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([1-cos(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 x-y 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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