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

March 3, 2021

Inverse Circular Functions

Return to MAT 130 main page

Go to Preview Activity

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

Answer these questions and submit your answers as a document on Moodle. (Please submit as .docx or .pdf if possible.)

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

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

    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:
    1. What was something you found interesting about this reading?
    2. What was an "a-ha" moment you had while doing this reading?
    3. What was the muddiest point of this reading for you?
    4. What question(s) do you have about anything you've read?


Return to MAT 130 main page

No comments :

Post a Comment

Contact Form


Email *

Message *