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: ${tan}^{- 1} (0)$ $tan^-1 (0)$ and ${csc}^{- 1} (2)$ $csc^-1 (2)$ ?

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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: ${tan}^{- 1} (0)$ $tan^-1 (0)$ and ${csc}^{- 1} (2)$ $csc^-1 (2)$ ?

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Answer 1 · 2014-08-04T02:43:13

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

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

Let us take a look at a unit circle diagram:
Unit circle diagram with radius

$r$ $r$ is the radius of the circle, and it is also the hypotenuse of the right triangle.

We will start with $arctan 0$ $arctan 0$ . 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 $0$ $0$ is our arc tangent's argument, then it must be equal to the ratio:

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

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

So,

$arctan 0 = 0$ $arctan 0 = 0$ .

Let us move on to $arccsc (2)$ $"arccsc"(2)$ .

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

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

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 $2$ $2$ must be the ratio between the hypotenuse and the opposite side.

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

This is more conveniently written as:

$2 y = r$ $2y = r$

Or, alternatively as:

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

What this tells us is that for our angle $θ$ $theta$ to equal the $arccsc$ $"arccsc"$ of $2$ $2$ , 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 $r = 2 y$ $r = 2y$ , then $x = y \sqrt{3}$ $x = ysqrt(3)$ . Therefore, $θ$ $theta$ is equal to $30$ $30$ degrees, or $\frac{π}{6}$ $pi/6$ .

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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: ${tan}^{- 1} (0)$ $tan^-1 (0)$ and ${csc}^{- 1} (2)$ $csc^-1 (2)$ ?

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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)tan^-1 (0) and csc−1(2)csc^-1 (2)?

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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: ${tan}^{- 1} (0)$ $tan^-1 (0)$ and ${csc}^{- 1} (2)$ $csc^-1 (2)$ ?