$$\definecolor{flatred}{RGB}{192, 57, 43} \definecolor{flatblue}{RGB}{41, 128, 185} \definecolor{flatgreen}{RGB}{39, 174, 96} \definecolor{flatpurple}{RGB}{142, 68, 173} \definecolor{flatlightblue}{RGB}{52, 152, 219} \definecolor{flatlightred}{RGB}{231, 76, 60} \definecolor{flatlightgreen}{RGB}{46, 204, 113} \definecolor{flatteal}{RGB}{22, 160, 133} \definecolor{flatpink}{RGB}{243, 104, 224}$$

## March 3, 2021

### Inverse Circular Functions

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Textbook Reference: Section 6.3 Inverse Trigonometric Functions

As we've been solving equations this week, perhaps you've noticed that something seems a little off...

All of the equations have been too "nice."

What do we do with an equation like this? $\sin(\theta)=\dfrac13$ This equation should have a solution, since $$\dfrac13$$ is well within the range of heights on the unit circle — it's between $$-1$$ and $$1$$. But there's no "nice" angle with a height of $$\dfrac13$$. Are we just out of luck?

Of course not.

We need to dive a little deeper and figure out how to "undo" a circular function ... using inverse functions.

## Inverse functions

If a function is a "machine" that turns inputs into outputs, then an inverse function is a corresponding "machine" that turns the outputs back into inputs.

For example, if $$f(x)=3x+1$$, then $$f$$ is a function that first triples a number and then adds $$1$$. Then, its inverse function $$f^{-1}$$ needs to "undo" these steps — first subtract $$1$$, and then divide by $$3$$ — so $$f^{-1}(x)=\dfrac{x-1}3$$.

Many of the functions you've learned about so far are inverses of each other:

 Addition $$\leftrightarrow$$ Subtraction Multiplication $$\leftrightarrow$$ Division Powers* $$\leftrightarrow$$ Roots Exponentials $$\leftrightarrow$$ Logarithms

There's a catch though.

It becomes much more difficult to specify an inverse when your original function isn't one-to-one.

For example, did you notice that asterisk on "Powers" above? Say we have $$f(x)=x^2$$, and I tell you that $$f(a)=4$$. Then you don't have quite enough information to figure out what $$a$$ is, because there are two possibilities: $$a$$ could be $$2$$, or it could be $$-2$$. This happens because $$f$$ isn't a one-to-one function — it doesn't pass the horizontal line test — so, given an output, you can't always uniquely determine the original input.

Drag the black point to move the horizontal line.
Notice how most outputs have two inputs.

We do have a trick up our sleeve though: we can restrict ourselves to only looking at part of the original graph of $$f(x)$$, and decide that our inverse function should always give values on that branch.

That's what happens when we define $$\sqrt{x}$$ to always give the positive square root of $$x$$, so that $$\sqrt{4}=2$$ rather than being undecided between $$2$$ and $$-2$$.

What's more, if you look at the graph of our newly-defined inverse function, you should notice that it's a reflection of our original graph, over the line $$y=x$$ ... which makes sense, because we're swapping our inputs and outputs, that is, $$x$$ and $$y$$ coordinates.

Drag either the red or blue points.
What do you notice about the coordinates?

## Creating inverses of circular functions

To solve our equation $$\sin(\theta)=\dfrac13$$ above, we'd like to create an inverse of the sine function, but again we have the problem that it's not one-to-one. In fact, it's very not one-to-one.$$\DeclareMathOperator{\Sin}{Sin} \DeclareMathOperator{\Cos}{Cos} \DeclareMathOperator{\Tan}{Tan} \DeclareMathOperator{\Csc}{Csc} \DeclareMathOperator{\Sec}{Sec} \DeclareMathOperator{\Cot}{Cot} \DeclareMathOperator{\Arcsin}{Arcsin} \DeclareMathOperator{\Arccos}{Arccos} \DeclareMathOperator{\Arctan}{Arctan} \DeclareMathOperator{\Arccsc}{Arccsc} \DeclareMathOperator{\Arcsec}{Arcsec} \DeclareMathOperator{\Arccot}{Arccot}$$

Drag the black point. Notice how many intersections there are!

But we can use the same trick from earlier, where we only focus on part of the graph of $$y=\sin(x)$$. We could use any part we want as long as it passes the horizontal line test, but for convenience's sake we'll choose the "simplest" part of the graph: the part containing the origin, from $$-\dfrac\pi2$$ to $$\dfrac\pi2$$.

Drag the black point, and notice that this part of the graph passes the horizontal line test.

If we reflect this graph over the line $$y=x$$, swapping all the input and outputs, we get a new graph that should serve as our inverse function:

We'll call this the inverse sine function, and we'll write it as $$\color{flatpurple}\Sin^{-1}(x)$$.

(By the way, I usually write inverse circular functions with a capital letter at the beginning. You'll also see them with a lowercase letter instead; for example, Desmos uses all-lowercase.)

To put it more succinctly:

$$\color{flatpurple}\Sin^{-1}(x)$$ is the angle
between $$-\dfrac\pi2$$ and $$\dfrac\pi2$$
whose sine is $$x$$.

For example, notice that $$\color{flatblue}\sin\left(\dfrac\pi6\right)=\sin\left(\dfrac{5\pi}6\right)=\dfrac12$$, so $$\color{flatpurple}\Sin^{-1}\left(\dfrac12\right)=\dfrac\pi6$$, since $$\color{flatpurple}\dfrac\pi6$$ is the unique angle that falls within the range from $$-\dfrac\pi2$$ to $$\dfrac\pi2$$.

By restricting the other circular functions similarly by defining the ranges we want for their inverses:

• $$\color{flatpurple}\Sin^{-1}(x)$$, $$\color{flatpurple}\Csc^{-1}(x)$$, and $$\color{flatpurple}\Tan^{-1}(x)$$ always give angles between $$-\dfrac\pi2$$ and $$\dfrac\pi2$$.
• $$\color{flatpurple}\Cos^{-1}(x)$$, $$\color{flatpurple}\Sec^{-1}(x)$$, and $$\color{flatpurple}\Cot^{-1}(x)$$ always give angles between $$0$$ and $$\pi$$.

Here's the big take-away:

Inverse circular functions tell us angles.

# Preview Activity 10

1. Fill in the following table of values for $$\color{flatpurple}\Sin^{-1}(x)$$.
 $$x$$ $$-1$$ $$-\dfrac{\sqrt3}2$$ $$-\dfrac{\sqrt2}2$$ $$-\dfrac12$$ $$0$$ $$\dfrac12$$ $$\dfrac{\sqrt2}2$$ $$\dfrac{\sqrt3}2$$ $$1$$ $$\color{flatpurple}\Sin^{-1}(x)$$

Make sure all values are between $$-\dfrac\pi2$$ and $$\dfrac\pi2$$.
2. Fill in the following table of values for $$\color{flatpurple}\Cos^{-1}(x)$$.
 $$x$$ $$-1$$ $$-\dfrac{\sqrt3}2$$ $$-\dfrac{\sqrt2}2$$ $$-\dfrac12$$ $$0$$ $$\dfrac12$$ $$\dfrac{\sqrt2}2$$ $$\dfrac{\sqrt3}2$$ $$1$$ $$\color{flatpurple}\Cos^{-1}(x)$$

Make sure all values are between $$0$$ and $$\pi$$.
3. What's the difference between $$\color{flatpurple}\Sin^{-1}(x)$$ and $$\color{flatlightblue}\csc(x)$$? Explain the best you can.
4. Use your calculator or Desmos to solve the equation from the beginning of the section: $$\sin(x)=\dfrac13$$. Are there other solutions?
5. Answer AT LEAST one of the following questions:
2. What was an "a-ha" moment you had while doing this reading?
3. What was the muddiest point of this reading for you?