Question 1138396
 

If theta is a third quadrant angle and {{{cos( theta) = -2/3}}}, find {{{sin (theta)}}}. 

From the sign on the cosine value, I only know that the angle is in QII or QIII. That's why they had to give me that additional specification: so I'd know which of those quadrants I'm in.


 {{{cos( theta) = -2/3}}}=>{{{x/r}}}

The Pythagorean Theorem gives me:

{{{r^2 = x^2 + y^2 }}}

{{{3^2 = (-2)^2 + y^2 }}}

{{{9 = 4 + y^2 }}}

{{{9-4 = y^2 }}}

{{{y}}}=±{{{sqrt(5)}}}

Since  in QIII,  below the x-axis, so y is negative. 
I'll take the negative solution to the equation, and {{{y= -sqrt(5)}}}

 {{{sin (theta)=y/r}}}

 {{{sin (theta)=-sqrt(5)/3}}}