# How do you evaluate the inverse function by sketching a unit circle, locating the correct angle, and evaluating the ordered pair on the circle for: tan^-1 (0) and csc^-1 (2)?

Aug 4, 2014

The trigonometric functions ($\text{sin}$, $\text{cos}$, $\text{tan}$) all take angles as their arguments, and produce ratios. (remember SOHCAHTOA)

The inverse trigonometric functions ($\text{arcsin}$, $\text{arccos}$) take ratios as their arguments, and produce the corresponding angles.

Let us take a look at a unit circle diagram: $r$ is the radius of the circle, and it is also the hypotenuse of the right triangle.

We will start with $\arctan 0$. First, we know that the tangent of an angle equals the ratio between the opposite side and the adjacent side. And, we know that the arc tangent function takes a ratio of this form, and produces an angle. Since $0$ is our arc tangent's argument, then it must be equal to the ratio:

$\frac{y}{x} = 0$.

Clearly, this statement can only be true if $y = 0$. And if $y = 0$, then $\theta$ must also be $0$.

So,

$\arctan 0 = 0$.

Let us move on to $\text{arccsc} \left(2\right)$.

Well, the cosecant of an angle is the inverse of its sine. In other words,

$\csc \theta = \frac{1}{\sin} \theta$.

We know that sine gives a ratio between the opposite side and the hypotenuse. So, the cosecant function therefore gives a ratio between the hypotenuse and the opposite side. And, if the arc-cosecant takes this ratio as an argument, and gives the angle, then we know that $2$ must be the ratio between the hypotenuse and the opposite side.

$2 = \frac{r}{y}$

This is more conveniently written as:

$2 y = r$

Or, alternatively as:

$y = \frac{1}{2} r$

What this tells us is that for our angle $\theta$ to equal the $\text{arccsc}$ of $2$, we need a right triangle whose hypotenuse is twice the length of its opposite leg.

And, elementary geometry tells us that this is precisely what occurs in a 30-60-90 triangle.

If $r = 2 y$, then $x = y \sqrt{3}$. Therefore, $\theta$ is equal to $30$ degrees, or $\frac{\pi}{6}$.