Solving a Trig Question in 3rd Quadrant

[zaddyzad]

Homework Statement

We have cosx = -5/6 in the third quadrant, and I solved for sin(x/2).

I did sin^2(x/2) = 1-cos(x)/2 --> I get to sin^2(x/2)= 11/12, and here's my question.

When rooting I should choose the negative value because sine is negative in the third quadrant right? But in my webwork, the answer is the positive one. Can anyone explain to me why this is?

 

[Gridvvk]

Your basis for the answer being negative is that sin(x) < 0 in the third quadrant, but what does this tell you about the sign of sin(x/2)?

It may help to think angles in the third quadrant are in the range 180 < x < 270.

 

[zaddyzad]

is that the only basis of choosing the right sign? it being 90<x<135 ?

 

[Gridvvk]

is that the only basis of choosing the right sign? it being 90<x<135 ?

Yes, x/2 is located in the second quadrant. And the sign of sine is positive in quadrant II.

Also, you should be careful with notation. We have 90 < x/2 < 135, not 90 < x < 135.