
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,
is certainly positive, but its sine is negative since
.
For example, let's look at
. 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 x-axis. 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
and
, we have
 = \frac{\sin\left(-193^\circ\right)}{\cos\left(-193^\circ\right)} < 0)
Or, you could use the fact that tangent is odd, so
, and since
is in quadrant III, the tangent will be positive, and then the negative in front makes it negative. Either method works.