# Thread: i need trig help - even/odd properties

1. ## i need trig help - even/odd properties

ok i dont understand this:

"which of the following trigonometric values are NEGATIVE?"

I. sin(-292°)
II. tan(-193°)
III. cos(-207°)
IV. cot222°

ok i have the answer key and apparently it's II and III

but i thought it was I and II

in my math book it says cos(-Θ) = cosΘ

so how the heck is III negative????

ok is this answer key wrong b/c that makes *no sense*!!!!!

2. Hello,

Originally Posted by tangerine_tomato
ok i dont understand this:

"which of the following trigonometric values are NEGATIVE?"

I. sin(-292°)
II. tan(-193°)
III. cos(-207°)
IV. cot222°

ok i have the answer key and apparently it's II and III

but i thought it was I and II

wtf??????? in my math book it says cos(-Θ) = cosΘ

so how the heck is III negative????

ok is this answer key wrong b/c that makes *no sense*!!!!!
$$\cos(-\theta)=\cos(\theta)$$

$\sin(-\theta)=-\sin(\theta)$

193, 207 and 222 are in the 3rd quadrant, that is to say negative cosine and negative sine.
292 is in the 4th quadrant, that is to say positive cosine and negative sine.

For example III, where you seem to have problems.

$\cos(-207)=\cos(207)<0$

3. Originally Posted by Moo
Hello,

$\cos(-\theta)=\cos(\theta)$

$\sin(-\theta)=-\sin(\theta)$

193, 207 and 222 are in the 3rd quadrant, that is to say negative cosine and negative sine.
292 is in the 4th quadrant, that is to say positive cosine and negative sine.

For example III, where you seem to have problems.

$\cos(-207)=\cos(207)<0$
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???

4. 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, $270^\circ$ is certainly positive, but its sine is negative since $\sin270^\circ = -1$.

For example, let's look at $\tan\left(-193^\circ\right)$. 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 x-axis. Go around 193 degrees, and you should see that the angle lies in quadrant II:

Code:
                -270°
II     |
***      |
***   |
***|
-180° ------------+-------------- 0°
|
|
|
|
|
-90°
So, since $\sin\left(-193^\circ\right) > 0$ and $\cos\left(-193^\circ < 0\right)$, we have

$\tan\left(-193^\circ\right) = \frac{\sin\left(-193^\circ\right)}{\cos\left(-193^\circ\right)} < 0$

Or, you could use the fact that tangent is odd, so $\tan\left(-193^\circ\right) = -\tan193^\circ$, and since $193^\circ$ is in quadrant III, the tangent will be positive, and then the negative in front makes it negative. Either method works.

5. ok that really confuses me

but i just devised this method of doing problems like that...lol

1. rewrite it using even-odd properties
2. draw a picture to see what quadrant it's in
3. apply the "all students take classes" rule