# How to know if the angle is positive or negative in inverse trigonometry

I have been doing this problem this problem:

$$Cos[Tan^{-1}(-\frac{2}{3})]$$ So I was instructed to draw a triangle to guide me so I did

Now once I drew my triangle I found the hypotenuse, which is $$\sqrt{13}$$

And then I was able to obtain the answer to this expression which I got:

$$\frac{2\sqrt{13}}{13}$$

However, I am told I drew the triangle wrong, it is actually -2 (negative) and (3) positive. Why is it that the triangle is wrong? I was told the actual answer is

$$\frac{3\sqrt{13}}{13}$$

Arctan has a range of $\frac{-\pi}{2}\le{y}\le\frac{\pi}{2}$. Now let $arctan\frac{-2}{3}=y$. This implies $tany=\frac{-2}{3}$. Because tan is negative, we know y must lie in quadrant IV. It cannot lie in quadrant I because tan is positive in quadrant I. Therefore we draw our angle as you have above. Now, note that $tan\theta=\frac{opposite}{adjacent}$. Therefore, you should have $-2$ where $3$ is in your picture, and you should have $3$ where $-2$ is. Your line should be drawn to the coordinate $(3,-2)$. Now, we find the hypotenuse as you have already done using the Pythagorean Theorem. We find that it is $\sqrt{13}$ as you've noted. Now, because we are finding $cos(arctan\frac{-2}{3})$, we will use the fact that $cos\theta=\frac{adjacent}{hypotenuse}$. So, this gives $\frac{3}{\sqrt{13}}$. Rationalizing we have $\frac{3\sqrt{13}}{13}$
• Yes, tangent is negative in quadrant IV and II, but the range for arctan is limited to $\frac{-\pi}{2}\le{y}\le\frac{\pi}{2}$ so our angle cannot lie in quadrant II. It must lie in either quadrant I or IV. And since it is negative, it must lie in quadrant IV. – MathGuy Jan 27 '17 at 3:04
• Yes, the ranges are different though depending on the inverse trig function. For arcsin the range is $\frac{-\pi}{2}\le{y}\le\frac{\pi}{2}$, for arccos the range is $0\le{y}\le\pi$ and for arctan the range is $\frac{-\pi}{2}\le{y}\le\frac{\pi}{2}$ – MathGuy Jan 27 '17 at 3:12