SOLUTION: If sin(x)=-5/13 and x is in quadrant III, with 0° ≤ x < 360°, find the exact values of the expressions without solving for x. sin(x/2) cos(x/2) tan(x/2)

So for angle x, we have a 5-12-13 right triangle in quadrant III,
where x=-12, y=-5, and r=+13.

There is often a conflict of notation when x is used for an angle, and also
for values of the adjacent side of the defining right triangle. There is a
problem here. But I think you won't get confused.  Teachers aren't always
careful to point out this conflict, which happens a lot. (just like the problem
of how to say "the sign of the sine". J )

So cos(x)=x/r=(-12)/(+13)=-12/13 and tan(x)=y/x=(-5)/(-12)=+5/12


 so x/2 is in quadrant II (the upper half of quadrant II).

So sin(x/2) is positive, cos(x/2) is negative, and tan(x/2) is negative.

So we just use the half-angle formulas:

, 

We know to use the + because x/2 is in quadrant II, where sine is positive.



We know to use the - because x/2 is in quadrant II, where cosine is negative.



Now we could use a formula for tan(x/2), but now all we need is 



Edwin