Originally Posted by
tangerine_tomato ohhh ok i think get it now...so i just have to see what quadrant each is in??? AND look at the even odd properties???
You don't have to worry about even/odd, but it may be helpful. What you do need to realize is that you are being asked when the function is negative, not when the function's argument is negative. Thus, $\displaystyle 270^\circ$ is certainly positive, but its sine is negative since $\displaystyle \sin270^\circ = 1$.
For example, let's look at $\displaystyle \tan\left(193^\circ\right)$. Just because the 193 is negative doesn't mean the tangent will be. We need to look at the quadrant. Since this is a negative angle, we move clockwise from the xaxis. Go around 193 degrees, and you should see that the angle lies in quadrant II:
Code:
270°
QUAD 
II 
***  QUAD I
*** 
*** 
***
180° + 0°


QUAD III  QUAD IV



90°
So, since $\displaystyle \sin\left(193^\circ\right) > 0$ and $\displaystyle \cos\left(193^\circ < 0\right)$, we have
$\displaystyle \tan\left(193^\circ\right) = \frac{\sin\left(193^\circ\right)}{\cos\left(193^\circ\right)} < 0$
Or, you could use the fact that tangent is odd, so $\displaystyle \tan\left(193^\circ\right) = \tan193^\circ$, and since $\displaystyle 193^\circ$ is in quadrant III, the tangent will be positive, and then the negative in front makes it negative. Either method works.