Unit 3 · Topic 3.9 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Precalculus 3.9: Inverse Trigonometric Functions

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

On the left, the unit circle. Its right half, from -pi/2 at the bottom through 0 to pi/2 at the top, is drawn in purple and labeled arcsine outputs, and its top half, from 0 to pi, is drawn in green just outside the circle and labeled arccosine outputs. A dashed line at height 1/2 meets the circle at pi/6, on the purple half, and at 5pi/6, off it. On the right, the graph of y = sin x for x from -pi/2 to pi/2, rising from (-pi/2, -1) to (pi/2, 1), and its mirror image across the dashed line y = x, the graph of y = arcsin x, rising from (-1, -pi/2) to (1, pi/2).

Part 1 Think first How far has the wheel turned?

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

Your seat on a Ferris wheel started level with the hub, on the right, and the wheel turns counterclockwise. A friend who looked away tells you only one thing: right now your seat is half a radius above the hub. You want to tell them how far the wheel has turned since the start. What can the height alone tell you?

In short. An inverse trigonometric function runs sine, cosine or tangent backward: it takes a value of the function as its input and gives back the one angle in a fixed interval that produces it.

Sine, cosine and tangent are periodic, so each output comes from many angles. Each has an inverse only on a restricted domain: $[- \frac{π}{2}, \frac{π}{2}]$ for sine, $[0, π]$ for cosine and $(- \frac{π}{2}, \frac{π}{2})$ for tangent.
The inverses are arcsine, arccosine and arctangent, also written $\sin^{- 1} x$ , $\cos^{- 1} x$ and $\tan^{- 1} x$ . The $- 1$ means inverse, not reciprocal.
Inputs and outputs switch, so the output is an angle, and it always lies in the restricted domain. An angle outside it never comes back out: $\arcsin (\sin \frac{5 π}{6}) = \frac{π}{6}$ .

Part 2 Learn it Running the trig functions backward

3a. Too many angles, so restrict the domain

In 2.8 you saw that a function has an inverse only where each output comes from exactly one input. Sine fails that badly. $\sin \frac{π}{6} = \frac{1}{2}$ , and so does $\sin \frac{5 π}{6}$ , and so does the sine of every angle a whole number of turns away from either. Sine is periodic, so every value it takes comes back again and again. Cosine and tangent are periodic too.

So none of the three has an inverse on its full domain. Each one is invertible only on a restricted domain: a stretch of angles on which the function takes each of its values exactly once. On that stretch it is one-to-one, and running it backward gives back one angle.

The three inverse functions are called arcsine, arccosine and arctangent. You will also see them written $\sin^{- 1} x$ , $\cos^{- 1} x$ and $\tan^{- 1} x$ . That $- 1$ means inverse, exactly as it does in $f^{- 1}$ . It is not an exponent. The reciprocal of sine is written $(\sin x)^{- 1}$ or $\frac{1}{\sin x}$ , and it is a different function altogether.

Quick check · Multiple choice

No calculator. The function $f (θ) = \sin θ$ has domain all real numbers. Which statement explains why $f$ has no inverse function on that domain?

3b. Where each function is fenced in

These three restricted domains are the standard ones, so arcsine, arccosine and arctangent each give one agreed value for every input they accept.

Sine is restricted to $[- \frac{π}{2}, \frac{π}{2}]$ , the right half of the unit circle. As P climbs that half from the bottom to the top, its height goes from $- 1$ to $1$ and passes every height once.
Cosine is restricted to $[0, π]$ , the top half. As P goes over the top from $(1, 0)$ to $(- 1, 0)$ , its $x$ -coordinate goes from $1$ to $- 1$ , each value once.
Tangent is restricted to $(- \frac{π}{2}, \frac{π}{2})$ , the right half with its two ends left open. At $\pm \frac{π}{2}$ the terminal ray is vertical, so tangent is not defined there. In between, the slope of the ray takes every real value once.

An inverse switches inputs and outputs, so each restricted domain becomes the set of outputs of the inverse. Arcsine and arctangent give angles on the right half of the circle. A negative input gives a negative angle, such as $- \frac{π}{4}$ , never $\frac{7 π}{4}$ . Arccosine gives angles from $0$ to $π$ , so a negative input gives an angle in Quadrant II.

Switching inputs and outputs also reflects each graph across the line $y = x$ : the point $(a, b)$ becomes $(b, a)$ . So $y = \arcsin x$ rises from $(- 1, - \frac{π}{2})$ to $(1, \frac{π}{2})$ , as the diagram shows, and stops there. Arctangent's graph never reaches $\pm \frac{π}{2}$ ; those are its horizontal asymptotes.

Quick check · Sort it

No calculator. Each card is an angle in radians. Tap a card, then tap which of arcsine and arccosine can give that angle as an output.

Both can

Only arcsine

Only arccosine

Neither

3c. Reading an inverse value

Sine takes an angle and gives a number from $- 1$ to $1$ . Arcsine runs the other way: it takes that number and gives back the angle. So $\sin \frac{π}{6} = \frac{1}{2}$ becomes $\arcsin \frac{1}{2} = \frac{π}{6}$ . The input of an inverse trig function is a value of sine, cosine or tangent, and the output is an angle.

That also tells you which inputs make sense. Sine and cosine only produce values from $- 1$ to $1$ , so $\arcsin 2$ and $\arccos (- 3)$ do not exist. Tangent produces every real number, so arctangent accepts any input at all.

To find a value without a calculator, ask the question, then check the fence. For $\arccos (- \frac{1}{2})$ , ask which angle has cosine $- \frac{1}{2}$ . Several do, such as $\frac{2 π}{3}$ and $\frac{4 π}{3}$ . Only $\frac{2 π}{3}$ lies in $[0, π]$ , so $\arccos (- \frac{1}{2}) = \frac{2 π}{3}$ .

Quick check · Fill the statement

No calculator. Each value is an angle in radians. Tap a blank, then tap an answer.

$\arcsin (- \frac{\sqrt{2}}{2}) =$ , $\arccos (- \frac{\sqrt{3}}{2}) =$ and $\arctan \sqrt{3} =$ .

Part 3 Watch it done An inverse wrapped around its own function

Watch each step, then do the last one yourself.

Find $\arcsin (\sin \frac{4 π}{3})$ without a calculator.

Step 1 · Work the inside first
$\frac{4 π}{3}$ is in Quadrant III, $\frac{π}{3}$ past the negative $x$ -axis. There P is $(- \frac{1}{2}, - \frac{\sqrt{3}}{2})$ , so $\sin \frac{4 π}{3} = - \frac{\sqrt{3}}{2}$ . The problem is now $\arcsin (- \frac{\sqrt{3}}{2})$ .
Why this step?
Why is the answer not simply $\frac{4 π}{3}$ ? Arcsine undoes sine, after all.
Arcsine only ever outputs an angle in $[- \frac{π}{2}, \frac{π}{2}]$ , and $\frac{4 π}{3}$ is outside that interval. An inverse undoes its function only on the restricted domain, so $\arcsin (\sin θ) = θ$ holds only when $θ$ is already between $- \frac{π}{2}$ and $\frac{π}{2}$ .
Step 2 · Look on the right half of the circle
The answer is an angle on the right half of the unit circle whose sine is $- \frac{\sqrt{3}}{2}$ . A negative sine puts P below the $x$ -axis, so the angle is between $- \frac{π}{2}$ and $0$ : a clockwise turn from the positive $x$ -axis. Find the point on that quarter of the circle at height $- \frac{\sqrt{3}}{2}$ .
Your step · Calculate

No calculator. Finish it. The value of $\arcsin (\sin \frac{4 π}{3})$ , in radians, is $k π$ . What is $k$ ? Give the exact value; a fraction such as 7/4 is fine.

Your answer

Part 4 Try it Many angles, one answer

The lab draws P on the unit circle with the sine graph underneath. Set $θ$ to $\frac{π}{6}$ and read $\sin θ$ . Now hunt for other angles with the same sine: move the slider, or type angles, until P sits at exactly the same height again. Find at least three more between $- 2 π$ and $4 π$ , and watch where each one lands on the graph.

Every one of those angles has sine $\frac{1}{2}$ , yet $\arcsin \frac{1}{2}$ is a single number. Which of your angles lies in $[- \frac{π}{2}, \frac{π}{2}]$ ? That one is the answer, and it is the only one. Then close the lab and answer the two questions on paper.

Lab

Unit Circle

The radius is fixed at 1 here, so the height of P is $\sin θ$ . Type an angle such as 13pi/6 in the box to land on it exactly.

Open the full Unit Circle lab

Try it · Calculate

Graphing calculator required. The terminal ray of an angle $θ$ , with $- \frac{π}{2} < θ < \frac{π}{2}$ , passes through the point $(4, - 3)$ . Use arctangent to find $θ$ in radians. Give a decimal to three places.

Your answer

Try it · Decide and justify

No calculator. A student writes $\arccos (\cos (- \frac{π}{3})) = - \frac{π}{3}$ , because arccosine undoes cosine. Is the student right?

Which reason supports your answer?

Part 5 Explain it Write it the way the exam asks

Nothing you type leaves this page.

Free response is 37.5% of the exam score. This one is shaped like Question 4, Symbolic Manipulations.

Free-response practice · Question 4 style

No calculator. Let $k (x) = \arctan x$ . (i) Give the range of $k$ . Explain why tangent has to be restricted before it has an inverse, and why the range leaves out $- \frac{π}{2}$ and $\frac{π}{2}$ . (ii) Explain why $k$ accepts every real number as an input, while $\arcsin x$ does not. (iii) A classmate writes $\arctan (\tan \frac{3 π}{4}) = \frac{3 π}{4}$ . Explain the error and find the correct value.

Your answer stays on this page. It is never sent or saved.

Model answer

(i) The range of $k$ is $(- \frac{π}{2}, \frac{π}{2})$ . Tangent is periodic, so each of its values comes from infinitely many angles, and it has an inverse only on a domain where each value comes from one angle. That domain is $(- \frac{π}{2}, \frac{π}{2})$ . Its ends are left out because the terminal ray is vertical at $\pm \frac{π}{2}$ , so tangent is not defined there. (ii) The inputs of $k$ are outputs of tangent. As $θ$ goes from $- \frac{π}{2}$ to $\frac{π}{2}$ , the slope of the ray takes every real value, so $k$ accepts any real number. Sine only gives values from $- 1$ to $1$ , so $\arcsin x$ needs $- 1 \leq x \leq 1$ . (iii) $\frac{3 π}{4}$ is not in $(- \frac{π}{2}, \frac{π}{2})$ , so arctangent can never output it. $\tan \frac{3 π}{4} = - 1$ , and the angle in $(- \frac{π}{2}, \frac{π}{2})$ with tangent $- 1$ is $- \frac{π}{4}$ . So $\arctan (\tan \frac{3 π}{4}) = - \frac{π}{4}$ .

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Gives the range $(- \frac{π}{2}, \frac{π}{2})$ and says tangent is not defined at $\pm \frac{π}{2}$ , so the ends are left out.
Explains that tangent is periodic, so it is one-to-one only on a restricted domain, and only there does it have an inverse.
Explains that the inputs of $k$ are outputs of tangent, which can be any real number, while sine only gives values from $- 1$ to $1$ .
Finds $\tan \frac{3 π}{4} = - 1$ and gives $- \frac{π}{4}$ , saying that $\frac{3 π}{4}$ is outside the range of arctangent.

Part 6 Check yourself One table, run backward

The two case questions use the same table of $Q$ . Answer them from it and the unit circle.

A function $Q$ is defined by $Q (θ) = \cos θ$ for $0 \leq θ \leq π$ only, with $θ$ in radians. On that domain $Q$ is decreasing, so each output comes from exactly one input and $Q$ has an inverse function, $Q^{- 1}$ . The table lists some values of $Q$ .

Values of $Q (θ) = \cos θ$ on $0 \leq θ \leq π$
$θ$	$Q (θ)$
$0$	$1$
$\frac{π}{6}$	$\frac{\sqrt{3}}{2}$
$\frac{π}{4}$	$\frac{\sqrt{2}}{2}$
$\frac{π}{2}$	$0$
$\frac{2 π}{3}$	$- \frac{1}{2}$
$\frac{3 π}{4}$	$- \frac{\sqrt{2}}{2}$
$π$	$- 1$

Case question 1 · Calculate

No calculator. The function $Q (θ) = \cos θ$ is defined for $0 \leq θ \leq π$ . The value of $Q^{- 1} (- \frac{\sqrt{2}}{2})$ , in radians, is $k π$ . What is $k$ ? Give a fraction or a decimal.

Your answer

Case question 2 · Multiple choice

No calculator. The function $Q (θ) = \cos θ$ is defined for $0 \leq θ \leq π$ . Reading from left to right, which describes the graph of $y = Q^{- 1} (x)$ ?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student evaluates $\arccos \frac{1}{2} + \arcsin (- \frac{1}{2})$ . Tap the first line that goes wrong.

Check yourself · Match

No calculator. Tap each expression, then its value. Angles are in radians.

Part 7 Keep it Bring back earlier ideas

Pulling an older idea back up is what makes it last.

From 3.3 Sine and Cosine Function Values · Multiple choice

No calculator. For which angle $θ$ , in radians, is $\cos θ = - \frac{\sqrt{3}}{2}$ and $\sin θ = \frac{1}{2}$ ?

From 2.8 Inverse Functions · Fill the statement

No calculator. A function $g$ has $g (- 2) = 6$ , $g (0) = 4$ , $g (1) = - 1$ , $g (3) = 0$ and $g (5) = - 3$ . Tap a blank, then tap an answer.

$g^{- 1} (4)$ = and $g^{- 1} (- 3)$ = . The inputs of $g^{- 1}$ are the of $g$ .

In three lines

Sine, cosine and tangent are periodic, so each has an inverse only on a restricted domain: $[- \frac{π}{2}, \frac{π}{2}]$ , $[0, π]$ and $(- \frac{π}{2}, \frac{π}{2})$ .
Arcsine, arccosine and arctangent, written $\sin^{- 1} x$ , $\cos^{- 1} x$ and $\tan^{- 1} x$ , take a trig value and give back an angle in that interval.
Always check the interval: $\arcsin (- \frac{1}{2}) = - \frac{π}{6}$ , $\arccos (- \frac{1}{2}) = \frac{2 π}{3}$ , and an angle outside it never comes back out.

Vocabulary

Inverse trigonometric function: A function that switches the inputs and outputs of sine, cosine or tangent, so its input is a value of that function and its output is an angle.
Arcsine: The inverse of sine on $[- \frac{π}{2}, \frac{π}{2}]$ , also written $\sin^{- 1} x$ ; it takes a value from $- 1$ to $1$ and gives an angle in that interval.
Arccosine: The inverse of cosine on $[0, π]$ , also written $\cos^{- 1} x$ ; it takes a value from $- 1$ to $1$ and gives an angle in that interval.
Arctangent: The inverse of tangent on $(- \frac{π}{2}, \frac{π}{2})$ , also written $\tan^{- 1} x$ ; it takes any real number and gives an angle in that interval.
Restricted domain: A part of a function's domain on which each output comes from exactly one input, so the function is invertible there.

Course alignment, for teachers

AP Precalculus topic 3.9, Unit 3: Trigonometric and Polar Functions.

Skill 1.C: Construct new functions, using transformations, compositions, inverses, or regressions, that may be useful in modeling contexts, criteria, or data, with and without technology.
Skill 2.B: Construct equivalent graphical, numerical, analytical, and verbal representations of functions that are useful in a given mathematical or applied context, with and without technology.

This lesson maps all 3 essential knowledge statements of the topic to the part that teaches each one: 3 are checked by a scored question. The map is in the lesson file.