Exact Value of an Inverse Trig Function

[Jason76]

$\displaystyle \sin(\tan^{-1}\dfrac{1}{2})$

Figuring out the sin is no problem (use Pythagorean theorem), but how do you determine the sign of the answer. According to ASTC (acronym), tan is positive in the 1st and 3rd quadrants, yet the answer to this problem $\displaystyle \dfrac{\sqrt{5}}{5}$ is positive, and is shown to be totally in the 1st quadrant, not the 3rd. How do we know it is in the 1st and not the 3rd?

 

[HallsofIvy]

The tangent is a periodic function, with period $\displaystyle \pi$ and so there are many value of x such that tan(x)= 1/2. Strictly speaking "tan(x)" does not have an inverse. HOWEVER, in order to be abe to have an inverse function, we restict its domain. We define "Tan(x)" (some texts use capital t to make the distinction clearer) to have domain $\displaystyle -\pi/2$ to $\displaystyle \pi/2$ and then $\displaystyle Tan^{-1}(x)$ (which we then write "$\displaystyle tan^{-1}(x)$" just because we are lazy!) has value between $\displaystyle -\pi/2$ and $\displaystyle \pi/2$. In particular, if x is positive then $\displaystyle tan^{-1}(x)$ is between 0 and $\displaystyle /pi/2$

In particuar, $\displaystyle \theta= tan^{-1}(1/2)$ is in the first quadrant, since 1/2 is positive. And we can then visualize it as a right triangle with "opposite side" 1 and "near side" 2. By the Pythagorean theorem the length of the hypotenuse is $\displaystyle \sqrt{1^2+ 2^2}= \sqrt{5}$ and so $\displaystyle sin(\theta)= \frac{1}{\sqrt{5}}= \frac{\sqrt{5}}{5}$.

 

[Jason76]

The tangent is a periodic function, with period $\displaystyle \pi$ and so there are many value of x such that tan(x)= 1/2. Strictly speaking "tan(x)" does not have an inverse. HOWEVER, in order to be abe to have an inverse function, we restict its domain. We define "Tan(x)" (some texts use capital t to make the distinction clearer) to have domain $\displaystyle -\pi/2$ to $\displaystyle \pi/2$ and then $\displaystyle Tan^{-1}(x)$ (which we then write "$\displaystyle tan^{-1}(x)$" just because we are lazy!) has value between $\displaystyle -\pi/2$ and $\displaystyle \pi/2$. In particular, if x is positive then $\displaystyle tan^{-1}(x)$ is between 0 and $\displaystyle /pi/2$

In particuar, $\displaystyle \theta= tan^{-1}(1/2)$ is in the first quadrant, since 1/2 is positive. And we can then visualize it as a right triangle with "opposite side" 1 and "near side" 2. By the Pythagorean theorem the length of the hypotenuse is $\displaystyle \sqrt{1^2+ 2^2}= \sqrt{5}$ and so $\displaystyle sin(\theta)= \frac{1}{\sqrt{5}}= \frac{\sqrt{5}}{5}$.

So the answer $\displaystyle \dfrac{1}{2}$ being between $\displaystyle + \dfrac{\pi}{2}$ and $\displaystyle - \dfrac{\pi}{2}$ restricts our answer to the 1st and 4th quadrant. So the answer, being positive, must be in the 1st. Cause the other positive answer would be in the 3rd quadrant. Is this right?