4.9 — Inverse Trig Functions

Section 4.9

Inverse Trig
Functions

Evaluate and graph inverse sine functions.
Evaluate and graph other inverse trigonometric functions.
Evaluate compositions of trigonometric functions.

Concept

Recall about inverses…

Trig Function

angle

→

ratio

Concept

Inverse Sine Function

\(y = \sin x\) does not pass the Horizontal Line Test

However, when you restrict the domain to the interval \(\left[-\dfrac{\pi}{2},\;\dfrac{\pi}{2}\right]\), the following properties hold:

1. The function \(y = \sin x\) is increasing.

2. \(y = \sin x\) takes on its full range of values, \(-1 \le \sin x \le 1\).

3. \(y = \sin x\) is one-to-one.

\(y = \arcsin(x)\) or \(y = \sin^{-1}(x)\)

Concept

Inverse Sine Function

\(\arcsin x\) means the angle (or arc) whose sine is \(x\).

Both notations, \(\arcsin(x)\) and \(\sin^{-1}(x)\), are commonly used.

Remember that \(\sin^{-1}(x)\) denotes the inverse sine function; not \(\dfrac{1}{\sin(x)}\).

The values of \(\arcsin(x)\) lie in the interval \(\displaystyle -\dfrac{\pi}{2} \le \arcsin(x) \le \dfrac{\pi}{2}\).

Example 1

Sketch a graph of \(y = \arcsin(x)\) by hand.

Concept

Inverse Cosine

The cosine function has an inverse function on the restricted domain \([0, \pi]\) called the inverse cosine function.

\(y = \arccos(x)\) or \(y = \cos^{-1}(x)\)

Concept

Inverse Tangent

The tangent function has an inverse function on the restricted domain \(\left(-\dfrac{\pi}{2},\;\dfrac{\pi}{2}\right)\) called the inverse tangent function.

\(y = \arctan(x)\) or \(y = \tan^{-1}(x)\)

Reference

Inverse Trigonometric Functions

Function	Domain	Range
\(y = \arcsin x\)	\([-1,\;1]\)	\(\left[-\dfrac{\pi}{2},\;\dfrac{\pi}{2}\right]\)
\(y = \arccos x\)	\([-1,\;1]\)	\([0,\;\pi]\)
\(y = \arctan x\)	\((-\infty,\;\infty)\)	\(\left(-\dfrac{\pi}{2},\;\dfrac{\pi}{2}\right)\)

FYI

Transformations of Inverse Trig Functions

Just like with all functions, you can transform inverse trig functions, too!

\(y = \arcsin(2x)\) \(y = \arcsin(x)\)

Examples 2 & 3

If possible, find the exact value.

\(\displaystyle \text{2) } \arcsin\!\left(-\tfrac{1}{2}\right)\)

What angle gives a sine of \(-\tfrac{1}{2}\)?

The range of inverse sine is \(\left[-\dfrac{\pi}{2},\;\dfrac{\pi}{2}\right]\).
Sine is negative in QIV. The reference angle is \(\dfrac{\pi}{6}\).

\(= \displaystyle -\dfrac{\pi}{6}\)

\(\text{3) } \sin^{-1}(2)\)

What angle gives a sine of 2?

The sine of any angle can never be greater than 1!

Not possible
because the domain of inverse sine is \([-1,\;1]\)

Example 2

Graphical View: \(\arcsin\!\left(-\tfrac{1}{2}\right) = -\dfrac{\pi}{6}\)

We can see this exact answer dynamically on both the Sine graph and the Inverse Sine graph.

\(y = \sin(x)\) on \(\left[-\dfrac{\pi}{2},\;\dfrac{\pi}{2}\right]\)

Where does \(y = \sin(x)\) cross the line \(y = -\tfrac{1}{2}\)?
At \(x = -\dfrac{\pi}{6}\)

\(y = \arcsin(x)\)

What is the \(y\)-value on the \(y = \arcsin(x)\) graph when \(x = -\tfrac{1}{2}\)?
\(y = -\dfrac{\pi}{6}\)

Examples 4–6

Use a calculator to approximate the value (if possible).

4) \(\arctan(-8.45)\)

Check: Is your calculator in Radian mode?

\(\approx -1.4530\)

or \(\approx -83°15'\)

5) \(\arccos(2)\)

Try it in your calculator — you will get an error!
Not possible: domain of inverse cosine is \([-1,\;1]\)

6) \(\sec^{-1}(3.4)\)

Calculators don't have a \(\sec^{-1}\) key. Use the reciprocal identity:
\(\sec\theta = 3.4\) \(\Rightarrow\) \(\cos\theta = \dfrac{1}{3.4}\)

\(= \cos^{-1}\!\left(\dfrac{1}{3.4}\right)\)

\(\approx 1.2720\)

Concept

Compositions of Functions

Inverse functions have the properties… \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\)

\(\cos^{-1}(\cos 30°)\)

\(= \cos^{-1}\!\left(\dfrac{\sqrt{3}}{2}\right)\)

\(= 30°\)

\(\cos\!\left(\cos^{-1}\!\left(\dfrac{\sqrt{3}}{2}\right)\right)\)

\(= \cos(30°)\)

\(= \dfrac{\sqrt{3}}{2}\)

Concept

Compositions — Tricky Cases

You can NOT always just “cancel” a function and its inverse!
The answer must be in the restricted range of the outer inverse function.

\(\cos^{-1}(\cos 210°)\)

1. Evaluate the inner function first:

\(= \cos^{-1}\!\left(-\dfrac{\sqrt{3}}{2}\right)\)

2. What angle in \([0°,\;180°]\) has a cosine of \(-\dfrac{\sqrt{3}}{2}\)?
It must be in Quadrant II.

\(= 150°\)

NOT 210°! (210° is outside \([0°,\;180°]\))

\(\arcsin\!\left(\sin\dfrac{3\pi}{2}\right)\)

1. Evaluate the inner function first:

\(= \arcsin(-1)\)

2. What angle in \(\left[-\dfrac{\pi}{2},\;\dfrac{\pi}{2}\right]\) has a sine of \(-1\)?
It must be at the bottom of the unit circle.

\(= -\dfrac{\pi}{2}\)

NOT \(\dfrac{3\pi}{2}\)! (\(\dfrac{3\pi}{2}\) is outside \(\left[-\dfrac{\pi}{2},\;\dfrac{\pi}{2}\right]\))

Concept

Tips on working with Composition

Always work from the inside out.

Do NOT just “cancel out” a function and its inverse.

If the question gives an angle in radians, give the answer in radians.

If the question gives an angle in degrees, give the answer in degrees.

Examples 7 & 8

If possible, find the exact value.

\(\displaystyle \text{7) } \arcsin\!\left(\sin\dfrac{5\pi}{3}\right)\)

Step 1: \(\sin\!\left(\dfrac{5\pi}{3}\right) = -\dfrac{\sqrt{3}}{2}\) (Quadrant IV)

\(= \arcsin\!\left(-\dfrac{\sqrt{3}}{2}\right)\)

Step 2: What angle in \(\left[-\dfrac{\pi}{2},\;\dfrac{\pi}{2}\right]\) gives a sine of \(-\dfrac{\sqrt{3}}{2}\)?

\(= -\dfrac{\pi}{3}\)

\(\text{8) } \tan[\arctan(-5)]\)

Step 1: Is \(-5\) in the domain of arctan?

Yes, domain of arctan is \((-\infty,\;\infty)\).
Therefore, the inverse properties apply perfectly here.

\(= -5\)

tan and arctan cancel algebraically.

Example 9

Find the exact value of \(\tan\!\left(\arccos\dfrac{2}{3}\right)\).

Honors

Step 1 — Set up the inner function as an angle

Let \(\theta = \arccos\!\left(\dfrac{2}{3}\right)\)
Then \(\cos\theta = \dfrac{2}{3} = \dfrac{\text{adj}}{\text{hyp}}\)

Step 2 — Find the missing side

By Pythagorean Theorem: \(\text{opp} = \sqrt{3^2 - 2^2} = \sqrt{5}\)

Step 3 — Evaluate the outer function

We want \(\tan(\theta)\). \(\tan\theta = \dfrac{\text{opp}}{\text{adj}}\)

Answer

\[\tan\!\left(\arccos\dfrac{2}{3}\right) = \dfrac{\sqrt{5}}{2}\]

Example 10

Write as an algebraic expression: \(\cot(\arccos(3x))\), \(0 \le x \le \tfrac{1}{3}\)

Honors

Step 1 — Set up the inner function as an angle

Let \(\theta = \arccos(3x)\)
Then \(\cos\theta = 3x = \dfrac{3x}{1} = \dfrac{\text{adj}}{\text{hyp}}\)

Step 2 — Find the missing side

By Pythagorean Theorem: \(\text{opp} = \sqrt{1^2 - (3x)^2} = \sqrt{1 - 9x^2}\)

Step 3 — Evaluate the outer function

We want \(\cot(\theta)\). \(\cot\theta = \dfrac{\text{adj}}{\text{opp}}\)

Answer

\[\cot(\arccos(3x)) = \dfrac{3x}{\sqrt{1 - 9x^2}}\]

What are your questions?