Trig - Determine Quadrant I'm almost done with a trig problem, and am at the point where I have to determine the quadrant and the signs of my answer. Here is what I have worked out, after solving the problems, and I feel like I don't have enough info to solve it, but here it is:
GIVEN:

Side lengths of a triangle, right: 5, 12, 13
I FOUND:

$cot( x/2 ) = +/- 3/2$
$cos( 2x ) = -119/169$
$csc( x/2 ) = +/- (2sqrt(13))/13$
I know it is in either 2nd or 3rd because cos is negative, but where do I go from there?
Thanks!
 A: There is enough material in what you wrote to reconstruct the original question.  But since full detail about the question is not supplied, there is the possibility of misinterpretation.  In the future, please think in terms of providing detail. That encourages a quick response from this site!
Since $5^2+12^2=13^2$, the Pythagorean Theorem tells us that we have a right-angled triangle.  If you did your calculation of $\cos(2x)$ correctly, the angle $x$ is the angle whose "adjacent" side has length $5$, and whose "opposite" side has length $12$.
You calculated $\cos(2x)$, using one of the standard identities for $\cos(2x)$, it doesn't matter which one, and got $-\dfrac{119}{169}$.  Since the angle $x$ is between $0$ and $90^\circ$, or equivalently between $0$ and $\pi/2$ (radians), and the $\cos(2x)$ is negative, $2x$ must be in the second quadrant.
As for $x/2$, since $x$ is between $0$ and $90^\circ$, the angle $x/2$ will be between $0$ and $45^\circ$, so all the trigonometric functions of $x/2$ will be positive.
I do not know how you computed $\cot(x/2)$. There are two ways to proceed, by calculator, and by "exact formula." Probably, since you are in the trigonometric identities section, exact calculation was expected.  I checked, the answer is exactly $3/2$. (As noted above, all trigonometric functions of $x/2$ are positive.)
As for $\csc(x/2)$, something went wrong in your calculation. I get $\sqrt{13}/2$.
That is the reciprocal of what you got, so it is easy to guess what might have gone wrong.
