# 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!

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What is the exact statement of the original problem? –  Jonas Meyer Jul 19 '11 at 1:28
@Jonas: Use the figure to find the values of the trig function, with the figure being a 5, 12, 13 right triangle, and the functions requested being cot(x/2), cos(2x), and csc(x/2), with x being the angle with sides 5 and 13 adjacent to it. –  Josh Jul 19 '11 at 1:29
Josh, I'm not sure I have the full context, but it sounds like $x$ is then assumed to be a positive acute angle. If you assume that $0<x<\frac{\pi}{2}$ there is no ambiguity. It sounds to me like it would be a good idea to ask an instructor (if this is for a class) to be 100% sure. –  Jonas Meyer Jul 19 '11 at 1:40
@Jonas: Great, thanks. I was thinking along those lines, but wasn't sure if there was any basic info I was just missing. –  Josh Jul 19 '11 at 1:48
@Josh, you probably know that $\sin\theta$ is always between $-1$ and $1$, and that $\csc\theta=1/\sin\theta$. From these it follows that $\csc\theta$ is always at least 1 in absolute value and so, as Andre notes in his answer, somehting must be wrong with what you got. –  Gerry Myerson Jul 19 '11 at 3:42

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.

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Thank you so much for the detailed explanation! I did indeed forget to put the 1 over my sin, hence the reciprocal. Thanks so much! –  Josh Jul 19 '11 at 11:26