# Graphing Inverse Trigonometric Functions

Inverse Sine - Graph and where it comes from

Tip: This isn't the place to ask a question because the teacher can't reply.

## Key Questions

• Since the graphs of $f \left(x\right)$ and $f ' \left(x\right)$ are symmetric about the line $y = x$, start with the graph of a trigonometric function with an appropriate restricted domain, then reflect it about the line $y = x$.

(Caution: Their domains must be restricted to an appropriate interval so that their inverses exist.)

Let us sketch the graph of $y = {\sin}^{- 1} x$.

The graph of $y = \sin x$ on $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ looks like:

By reflecting the graph above about the line $y = x$,

The curve in purple is the graph of $y = {\sin}^{- 1} x$.

The graphs of other inverse trigonometric functions can be obtained similarly.

I hope that this was helpful.

• Throughout the following answer, I will assume that you are asking about trigonometry restricted to real numbers.

Using Domain of $a r c \sin x$

Find $a r c \sin \left(3\right)$.

$3$ is not in the domain of $a r c \sin$, ($3$ is not in the range of $\sin$) so $a r c \sin \left(3\right)$ does not exist.

Using Range of $a r c \sin x$

Find $a r c \sin \left(\frac{1}{2}\right)$.

Although there are infinitely many $t$ with $\sin t = \frac{1}{2}$, the range of $a r c \sin$ restricts the value to those $t$ with $\frac{- \pi}{2} \le t \le \frac{\pi}{2}$, So the value we want is $\frac{\pi}{6}$.

This is the same as using the range, but it involves thinking about the problem more geometrically.

Find $a r c \sin \left(\frac{1}{2}\right)$.

$a r c \sin \left(\frac{1}{2}\right)$ is a number (an angle) in quadrant I or IV. It is the $t$ with smallest absolute value (the shortest path from the initial side).

Again, $a r c \sin \left(\frac{1}{2}\right) = \frac{\pi}{6}$.

• For a trig function, the range is called "Period"

For example, the function $f \left(x\right) = \cos x$ has a period of $2 \pi$; the function $f \left(x\right) = \tan x$ has a period of $\pi$. Solving or graphing a trig function must cover a whole period.

The range depends on each specific trig function.
For example, the inverse function $f \left(x\right) = \frac{1}{\cos x} = \sec x$ has as period $2 \pi$.

Its range varies from (+infinity) to Minimum $1$ then back to (+infinity), between ($- \frac{\pi}{2}$ and $\frac{\pi}{2}$).

Its range also varies from (-infinity) to Max -1 then back to to (-infinity), between ($\frac{\pi}{2}$ and $3 \frac{\pi}{2}$).

• It is best to start by graphing the parent inverse trig function. For example, if you graph the inverse tangent graph, you would have horizontal asymptotes where tangent has vertical asymptotes - which is where cosine is zero the first time in both directions - which is at - pi/2 and pi/2. You graph your horizontal asymptotes, plot the origin (because the tangent of zero is zero), then draw what looks pretty much like an ${x}^{3}$ graphing laying on its side.

Apply the transformations to the horizontal asymptotes and to the point at the origin, and draw the transformed graph the same way.

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