Contents

Preface
1Functions
2Linear Functions
3Polynomial and Rational Functions
4Exponential and Logarithmic Functions
5Trigonometric Functions
6Periodic Functions
7Trigonometric Identities and Equations
8Further Applications of Trigonometry
9Systems of Equations and Inequalities
10Analytic Geometry
11Sequences, Probability and Counting Theory
12Introduction to Calculus
ABasic Functions and Identities
Answer Key
Index

Inverse Trigonometric Functions

Learning Objectives

In this section, you will:

Understand and use the inverse sine, cosine, and tangent functions.
Find the exact value of expressions involving the inverse sine, cosine, and tangent functions.
Use a calculator to evaluate inverse trigonometric functions.
Find exact values of composite functions with inverse trigonometric functions.

For any right triangle, given one other angle and the length of one side, we can figure out what the other angles and sides are. But what if we are given only two sides of a right triangle? We need a procedure that leads us from a ratio of sides to an angle. This is where the notion of an inverse to a trigonometric function comes into play. In this section, we will explore the inverse trigonometric functions.

Understanding and Using the Inverse Sine, Cosine, and Tangent Functions

In order to use inverse trigonometric functions, we need to understand that an inverse trigonometric function “undoes” what the original trigonometric function “does,” as is the case with any other function and its inverse. In other words, the domain of the inverse function is the range of the original function, and vice versa, as summarized in Figure 1.

A chart that says “Trig Functinos”, “Inverse Trig Functions”, “Domain: Measure of an angle”, “Domain: Ratio”, “Range: Ratio”, and “Range: Measure of an angle”.

Figure 1

For example, if $f (x) = \sin x, f (x) = \sin$ then we would write $f^{- 1} (x) = \sin^{- 1} x . f^{- 1} (x) = \sin^{- 1} x .$ Be aware that $\sin^{- 1} x \sin^{- 1} x$ does not mean $\frac{1}{\sin x} . \frac{1}{\sin x} .$ The following examples illustrate the inverse trigonometric functions:

Since $sin (\frac{π}{6}) = \frac{1}{2}, sin (\frac{π}{6}) = \frac{1}{2},$ then $\frac{π}{6} = {sin}^{- 1} (\frac{1}{2}) . \frac{π}{6} = {sin}^{- 1} (\frac{1}{2}) .$
Since $\cos (π) = - 1, \cos (π) = - 1,$ then $π = \cos^{- 1} (- 1) . π = \cos^{- 1} (- 1) .$
Since $\tan (\frac{π}{4}) = 1, \tan (\frac{π}{4}) = 1,$ then $\frac{π}{4} = \tan^{- 1} (1) . \frac{π}{4} = \tan^{- 1} (1) .$

In previous sections, we evaluated the trigonometric functions at various angles, but at times we need to know what angle would yield a specific sine, cosine, or tangent value. For this, we need inverse functions. Recall that, for a one-to-one function, if $f (a) = b, f (a) = b,$ then an inverse function would satisfy $f^{- 1} (b) = a . f^{- 1} (b) = a .$

Bear in mind that the sine, cosine, and tangent functions are not one-to-one functions. The graph of each function would fail the horizontal line test. In fact, no periodic function can be one-to-one because each output in its range corresponds to at least one input in every period, and there are an infinite number of periods. As with other functions that are not one-to-one, we will need to restrict the domain of each function to yield a new function that is one-to-one. We choose a domain for each function that includes the number 0. Figure 2 shows the graph of the sine function limited to $[- \frac{π}{2}, \frac{π}{2}] [- \frac{π}{2}, \frac{π}{2}]$ and the graph of the cosine function limited to $[0, π] . [0, π] .$

Two side-by-side graphs. The first graph, graph A, shows half of a period of the function sine of x. The second graph, graph B, shows half a period of the function cosine of x.

Figure 2 (a) Sine function on a restricted domain of

[- \frac{π}{2}, \frac{π}{2}]; [- \frac{π}{2}, \frac{π}{2}];

(b) Cosine function on a restricted domain of

[0, π] [0, π]

Figure 3 shows the graph of the tangent function limited to $(- \frac{π}{2}, \frac{π}{2}) . (- \frac{π}{2}, \frac{π}{2}) .$

A graph of one period of tangent of x, from -pi/2 to pi/2.

Figure 3 Tangent function on a restricted domain of

(- \frac{π}{2}, \frac{π}{2}) (- \frac{π}{2}, \frac{π}{2})

These conventional choices for the restricted domain are somewhat arbitrary, but they have important, helpful characteristics. Each domain includes the origin and some positive values, and most importantly, each results in a one-to-one function that is invertible. The conventional choice for the restricted domain of the tangent function also has the useful property that it extends from one vertical asymptote to the next instead of being divided into two parts by an asymptote.

On these restricted domains, we can define the inverse trigonometric functions.

The inverse sine function $y = \sin^{- 1} x y = \sin^{- 1} x$ means $x = \sin y . x = \sin$ The inverse sine function is sometimes called the arcsine function, and notated $\arcsin x . \arcsin x .$
$y = \sin^{- 1} x has domain [−1, 1] and range [- \frac{π}{2}, \frac{π}{2}] y = \sin^{- 1} x$
The inverse cosine function $y = \cos^{- 1} x y = \cos^{- 1} x$ means $x = \cos y . x = \cos$ The inverse cosine function is sometimes called the arccosine function, and notated $\arccos x . \arccos$
$y = \cos^{- 1} x has domain [−1, 1] and range [0, π] y = \cos^{- 1} x$
The inverse tangent function $y = \tan^{- 1} x y = \tan^{- 1} x$ means $x = \tan y . x = \tan$ The inverse tangent function is sometimes called the arctangent function, and notated $\arctan x . \arctan$
$y = \tan^{- 1} x has domain (−∞, \infty) and range (- \frac{π}{2}, \frac{π}{2}) y = \tan^{- 1} x$

The graphs of the inverse functions are shown in Figure 4, Figure 5, and Figure 6. Notice that the output of each of these inverse functions is a number, an angle in radian measure. We see that $\sin^{- 1} x \sin^{- 1} x$ has domain $[−1, 1] [−1, 1]$ and range $[- \frac{π}{2}, \frac{π}{2}], [- \frac{π}{2}, \frac{π}{2}],$ $\cos^{- 1} x \cos^{- 1} x$ has domain $[−1,1] [−1,1]$ and range $[0, π], [0, π],$ and $\tan^{- 1} x \tan^{- 1} x$ has domain of all real numbers and range $(- \frac{π}{2}, \frac{π}{2}) . (- \frac{π}{2}, \frac{π}{2}) .$ To find the domain and range of inverse trigonometric functions, switch the domain and range of the original functions. Each graph of the inverse trigonometric function is a reflection of the graph of the original function about the line $y = x . y = x .$

A graph of the functions of sine of x and arc sine of x. There is a dotted line y=x between the two graphs, to show inverse nature of the two functions

Figure 4 The sine function and inverse sine (or arcsine) function

A graph of the functions of cosine of x and arc cosine of x. There is a dotted line at y=x to show the inverse nature of the two functions.

Figure 5 The cosine function and inverse cosine (or arccosine) function

A graph of the functions of tangent of x and arc tangent of x. There is a dotted line at y=x to show the inverse nature of the two functions.

Figure 6 The tangent function and inverse tangent (or arctangent) function

Relations for Inverse Sine, Cosine, and Tangent Functions

For angles in the interval $[- \frac{π}{2}, \frac{π}{2}], [- \frac{π}{2}, \frac{π}{2}],$ if $\sin y = x, \sin$ then $\sin^{- 1} x = y . \sin^{- 1} x = y .$

For angles in the interval $[0, π], [0, π],$ if $\cos y = x, \cos$ then $\cos^{- 1} x = y . \cos^{- 1} x = y .$

For angles in the interval $(- \frac{π}{2}, \frac{π}{2}), (- \frac{π}{2}, \frac{π}{2}),$ if $\tan y = x, \tan$ then $\tan^{- 1} x = y . \tan^{- 1} x = y .$

Example 1

Writing a Relation for an Inverse Function

Given $\sin (\frac{5 π}{12}) \approx 0.96593, \sin (\frac{5 π}{12}) \approx 0.96593,$ write a relation involving the inverse sine.

Show solution

Solution

Use the relation for the inverse sine. If $\sin y = x, \sin$ then $\sin^{- 1} x = y \sin^{- 1} x = y$ .

In this problem, $x = 0.96593, x = 0.96593,$ and $y = \frac{5 π}{12} . y = \frac{5 π}{12} .$

\sin^{- 1} (0.96593) \approx \frac{5 π}{12} \sin^{- 1} (0.96593) \approx \frac{5 π}{12}

Try It #1

Given $\cos (0.5) \approx 0.8776, \cos (0.5) \approx 0.8776,$ write a relation involving the inverse cosine.

Finding the Exact Value of Expressions Involving the Inverse Sine, Cosine, and Tangent Functions

Now that we can identify inverse functions, we will learn to evaluate them. For most values in their domains, we must evaluate the inverse trigonometric functions by using a calculator, interpolating from a table, or using some other numerical technique. Just as we did with the original trigonometric functions, we can give exact values for the inverse functions when we are using the special angles, specifically $\frac{π}{6} \frac{π}{6}$ (30°), $\frac{π}{4} \frac{π}{4}$ (45°), and $\frac{π}{3} \frac{π}{3}$ (60°), and their reflections into other quadrants.

How To

Given a “special” input value, evaluate an inverse trigonometric function.

Find angle $x x$ for which the original trigonometric function has an output equal to the given input for the inverse trigonometric function.
If $x x$ is not in the defined range of the inverse, find another angle $y y$ that is in the defined range and has the same sine, cosine, or tangent as $x, x,$ depending on which corresponds to the given inverse function.

Example 2

Evaluating Inverse Trigonometric Functions for Special Input Values

Evaluate each of the following.

ⓐ ${sin}^{- 1} (\frac{1}{2}) {sin}^{- 1} (\frac{1}{2})$
ⓑ ${sin}^{- 1} (- \frac{\sqrt{2}}{2}) {sin}^{- 1} (- \frac{\sqrt{2}}{2})$
ⓒ $\cos^{- 1} (- \frac{\sqrt{3}}{2}) \cos^{- 1} (- \frac{\sqrt{3}}{2})$
ⓓ $\tan^{- 1} (1) \tan^{- 1} (1)$

Show solution

Solution

ⓐ Evaluating $\sin^{- 1} (\frac{1}{2}) \sin^{- 1} (\frac{1}{2})$ is the same as determining the angle that would have a sine value of $\frac{1}{2} . \frac{1}{2} .$ In other words, what angle $x x$ would satisfy $\sin (x) = \frac{1}{2} ? \sin (x) = \frac{1}{2} ?$ There are multiple values that would satisfy this relationship, such as $\frac{π}{6} \frac{π}{6}$ and $\frac{5 π}{6}, \frac{5 π}{6},$ but we know we need the angle in the interval $[- \frac{π}{2}, \frac{π}{2}], [- \frac{π}{2}, \frac{π}{2}],$ so the answer will be $\sin^{- 1} (\frac{1}{2}) = \frac{π}{6} . \sin^{- 1} (\frac{1}{2}) = \frac{π}{6} .$ Remember that the inverse is a function, so for each input, we will get exactly one output.
ⓑ To evaluate $\sin^{- 1} (- \frac{\sqrt{2}}{2}), \sin^{- 1} (- \frac{\sqrt{2}}{2}),$ we know that $\frac{5 π}{4} \frac{5 π}{4}$ and $\frac{7 π}{4} \frac{7 π}{4}$ both have a sine value of $- \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2},$ but neither is in the interval $[- \frac{π}{2}, \frac{π}{2}] . [- \frac{π}{2}, \frac{π}{2}] .$ For that, we need the negative angle coterminal with $\frac{7 π}{4} : \frac{7 π}{4} :$ ${sin}^{- 1} (- \frac{\sqrt{2}}{2}) = - \frac{π}{4} . {sin}^{- 1} (- \frac{\sqrt{2}}{2}) = - \frac{π}{4} .$
ⓒTo evaluate $\cos^{- 1} (- \frac{\sqrt{3}}{2}), \cos^{- 1} (- \frac{\sqrt{3}}{2}),$ we are looking for an angle in the interval $[0, π] [0, π]$ with a cosine value of $- \frac{\sqrt{3}}{2} . - \frac{\sqrt{3}}{2} .$ The angle that satisfies this is $\cos^{- 1} (- \frac{\sqrt{3}}{2}) = \frac{5 π}{6} . \cos^{- 1} (- \frac{\sqrt{3}}{2}) = \frac{5 π}{6} .$
ⓓ Evaluating $\tan^{- 1} (1), \tan^{- 1} (1),$ we are looking for an angle in the interval $(- \frac{π}{2}, \frac{π}{2}) (- \frac{π}{2}, \frac{π}{2})$ with a tangent value of 1. The correct angle is $\tan^{- 1} (1) = \frac{π}{4} . \tan^{- 1} (1) = \frac{π}{4} .$

Try It #2

Evaluate each of the following.

ⓐ ${sin}^{−1} (−1) {sin}^{−1} (−1)$
ⓑ $\tan^{−1} (−1) \tan^{−1} (−1)$
ⓒ $\cos^{−1} (−1) \cos^{−1} (−1)$
ⓓ $\cos^{−1} (\frac{1}{2}) \cos^{−1} (\frac{1}{2})$

Using a Calculator to Evaluate Inverse Trigonometric Functions

To evaluate inverse trigonometric functions that do not involve the special angles discussed previously, we will need to use a calculator or other type of technology. Most scientific calculators and calculator-emulating applications have specific keys or buttons for the inverse sine, cosine, and tangent functions. These may be labeled, for example, SIN $^{−1}^{−1}$ , ARCSIN, or ASIN.

In the previous chapter, we worked with trigonometry on a right triangle to solve for the sides of a triangle given one side and an additional angle. Using the inverse trigonometric functions, we can solve for the angles of a right triangle given two sides, and we can use a calculator to find the values to several decimal places.

In these examples and exercises, the answers will be interpreted as angles and we will use $θ θ$ as the independent variable. The value displayed on the calculator may be in degrees or radians, so be sure to set the mode appropriate to the application.

Example 3

Evaluating the Inverse Sine on a Calculator

Evaluate $\sin^{- 1} (0.97) \sin^{- 1} (0.97)$ using a calculator.

Show solution

Solution

Because the output of the inverse function is an angle, the calculator will give us a degree value if in degree mode and a radian value if in radian mode. Calculators also use the same domain restrictions on the angles as we are using.

In radian mode, $\sin^{- 1} (0.97) \approx 1.3252. \sin^{- 1} (0.97) \approx 1.3252.$ In degree mode, $\sin^{- 1} (0.97) \approx 75.93°. \sin^{- 1} (0.97) \approx 75.93°.$ Note that in calculus and beyond we will use radians in almost all cases.

Try It #3

Evaluate $\cos^{- 1} (- 0.4) \cos^{- 1} (- 0.4)$ using a calculator.

How To

Given two sides of a right triangle like the one shown in Figure 7, find an angle.

An illustration of a right triangle with an angle theta. Adjacent to theta is the side a, opposite theta is the side p, and the hypoteneuse is side h.

Figure 7

If one given side is the hypotenuse of length $h h$ and the side of length $a a$ adjacent to the desired angle is given, use the equation $θ = \cos^{- 1} (\frac{a}{h}) . θ = \cos^{- 1} (\frac{a}{h}) .$
If one given side is the hypotenuse of length $h h$ and the side of length $p p$ opposite to the desired angle is given, use the equation $θ = \sin^{- 1} (\frac{p}{h}) . θ = \sin^{- 1} (\frac{p}{h}) .$
If the two legs (the sides adjacent to the right angle) are given, then use the equation $θ = \tan^{- 1} (\frac{p}{a}) . θ = \tan^{- 1} (\frac{p}{a}) .$

Example 4

Applying the Inverse Cosine to a Right Triangle

Solve the triangle in Figure 8 for the angle $θ . θ .$

An illustration of a right triangle with the angle theta. Adjacent to the angle theta is a side with a length of 9 and a hypoteneuse of length 12.

Figure 8

Show solution

Solution

Because we know the hypotenuse and the side adjacent to the angle, it makes sense for us to use the cosine function.

\begin{array}{l} \cos θ = \frac{9}{12} & \begin{matrix}  \end{matrix} \\ θ = \cos^{- 1} (\frac{9}{12}) & \begin{matrix}  \end{matrix} Apply definition of the inverse . \\ θ \approx 0.7227 or about 41.4096° & \begin{matrix}  \end{matrix} Evaluate . \end{array} \begin{array}{l} \cos & \begin{matrix}  \end{matrix} \\ θ = \cos^{- 1} (\frac{9}{12}) & \begin{matrix}  \end{matrix} Apply definition of the inverse . \\ θ \approx 0.7227 or about 41.4096° & \begin{matrix}  \end{matrix} Evaluate . \end{array}

Try It #4

Solve the triangle in Figure 9 for the angle $θ . θ .$

An illustration of a right triangle with the angle theta. Opposite to the angle theta is a side with a length of 6 and a hypoteneuse of length 10.

Figure 9

Finding Exact Values of Composite Functions with Inverse Trigonometric Functions

There are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can usually find exact values for the resulting expressions without resorting to a calculator. Even when the input to the composite function is a variable or an expression, we can often find an expression for the output. To help sort out different cases, let $f (x) f (x)$ and $g (x) g (x)$ be two different trigonometric functions belonging to the set ${\sin (x), \cos (x), \tan (x)} {\sin (x), \cos (x), \tan (x)}$ and let $f^{- 1} (y) f^{- 1} (y)$ and $g^{- 1} (y) g^{- 1} (y)$ be their inverses.

Evaluating Compositions of the Form f(f⁻¹(y)) and f⁻¹(f(x))

For any trigonometric function, $f (f^{- 1} (y)) = y f (f^{- 1} (y)) = y$ for all $y y$ in the proper domain for the given function. This follows from the definition of the inverse and from the fact that the range of $f f$ was defined to be identical to the domain of $f^{- 1} . f^{- 1} .$ However, we have to be a little more careful with expressions of the form $f^{- 1} (f (x)) . f^{- 1} (f (x)) .$

Compositions of a trigonometric function and its inverse

\begin{array}{l} \sin (\sin^{- 1} x) = x for - 1 \leq x \leq 1 \\ \cos (\cos^{- 1} x) = x for - 1 \leq x \leq 1 \\ \tan (\tan^{- 1} x) = x for - \infty < x < \infty \end{array} \begin{array}{l} \sin (\sin^{- 1} x) = x \\ \cos (\cos^{- 1} x) = x \\ \tan (\tan^{- 1} x) = x \end{array}

\begin{array}{l} \sin^{- 1} (\sin x) = x only for - \frac{π}{2} \leq x \leq \frac{π}{2} \\ \cos^{- 1} (\cos x) = x only for 0 \leq x \leq π \\ \tan^{- 1} (\tan x) = x only for - \frac{π}{2} < x < \frac{π}{2} \end{array} \begin{array}{l} \sin^{- 1} (\sin \\ \cos^{- 1} (\cos \\ \tan^{- 1} (\tan \end{array}

Q&A

Is it correct that $\sin^{- 1} (\sin x) = x ? \sin^{- 1} (\sin$

No. This equation is correct if $x x$ belongs to the restricted domain $[- \frac{π}{2}, \frac{π}{2}], [- \frac{π}{2}, \frac{π}{2}],$ but sine is defined for all real input values, and for $x x$ outside the restricted interval, the equation is not correct because its inverse always returns a value in $[- \frac{π}{2}, \frac{π}{2}] . [- \frac{π}{2}, \frac{π}{2}] .$ The situation is similar for cosine and tangent and their inverses. For example, $\sin^{- 1} (\sin (\frac{3 π}{4})) = \frac{π}{4} . \sin^{- 1} (\sin (\frac{3 π}{4})) = \frac{π}{4} .$

How To

Given an expression of the form f⁻¹(f(θ)) where $f (θ) = \sin θ, \cos θ, or \tan θ, f (θ) = \sin$ evaluate.

If $θ θ$ is in the restricted domain of $f, then f^{- 1} (f (θ)) = θ . f, then f^{- 1} (f (θ)) = θ .$
If not, then find an angle $ϕ ϕ$ within the restricted domain of $f f$ such that $f (ϕ) = f (θ) . f (ϕ) = f (θ) .$ Then $f^{- 1} (f (θ)) = ϕ . f^{- 1} (f (θ)) = ϕ .$

Example 5

Using Inverse Trigonometric Functions

Evaluate the following:

ⓐ $\sin^{- 1} (\sin (\frac{π}{3})) \sin^{- 1} (\sin (\frac{π}{3}))$
ⓑ $\sin^{- 1} (\sin (\frac{2 π}{3})) \sin^{- 1} (\sin (\frac{2 π}{3}))$
ⓒ $\cos^{- 1} (\cos (\frac{2 π}{3})) \cos^{- 1} (\cos (\frac{2 π}{3}))$
ⓓ $\cos^{- 1} (\cos (- \frac{π}{3})) \cos^{- 1} (\cos (- \frac{π}{3}))$

Show solution

Solution

ⓐ $\frac{π}{3} is in [- \frac{π}{2}, \frac{π}{2}], \frac{π}{3} is in [- \frac{π}{2}, \frac{π}{2}],$ so $\sin^{- 1} (\sin (\frac{π}{3})) = \frac{π}{3} . \sin^{- 1} (\sin (\frac{π}{3})) = \frac{π}{3} .$
ⓑ $\frac{2 π}{3} is not in [- \frac{π}{2}, \frac{π}{2}], \frac{2 π}{3} is not in [- \frac{π}{2}, \frac{π}{2}],$ but $\sin (\frac{2 π}{3}) = \sin (\frac{π}{3}), \sin (\frac{2 π}{3}) = \sin (\frac{π}{3}),$ so $\sin^{- 1} (\sin (\frac{2 π}{3})) = \frac{π}{3} . \sin^{- 1} (\sin (\frac{2 π}{3})) = \frac{π}{3} .$
ⓒ $\frac{2 π}{3} is in [0, π], \frac{2 π}{3} is in [0, π],$ so $\cos^{- 1} (\cos (\frac{2 π}{3})) = \frac{2 π}{3} . \cos^{- 1} (\cos (\frac{2 π}{3})) = \frac{2 π}{3} .$
ⓓ $- \frac{π}{3} is not in [0, π], - \frac{π}{3} is not in [0, π],$ but $\cos (- \frac{π}{3}) = \cos (\frac{π}{3}) \cos (- \frac{π}{3}) = \cos (\frac{π}{3})$ because cosine is an even function. $\frac{π}{3} is in [0, π], \frac{π}{3} is in [0, π],$ so $\cos^{- 1} (\cos (- \frac{π}{3})) = \frac{π}{3} . \cos^{- 1} (\cos (- \frac{π}{3})) = \frac{π}{3} .$

Try It #5

Evaluate $\tan^{- 1} (\tan (\frac{π}{8})) and \tan^{- 1} (\tan (\frac{11 π}{9})) . \tan^{- 1} (\tan (\frac{π}{8}))$

Evaluating Compositions of the Form f⁻¹(g(x))

Now that we can compose a trigonometric function with its inverse, we can explore how to evaluate a composition of a trigonometric function and the inverse of another trigonometric function. We will begin with compositions of the form $f^{- 1} (g (x)) . f^{- 1} (g (x)) .$ For special values of $x, x,$ we can exactly evaluate the inner function and then the outer, inverse function. However, we can find a more general approach by considering the relation between the two acute angles of a right triangle where one is $θ, θ,$ making the other $\frac{π}{2} - θ . \frac{π}{2} - θ .$ Consider the sine and cosine of each angle of the right triangle in Figure 10.

An illustration of a right triangle with angles theta and pi/2 - theta. Opposite the angle theta and adjacent the angle pi/2-theta is the side a. Adjacent the angle theta and opposite the angle pi/2 - theta is the side b. The hypoteneuse is labeled c.

Figure 10 Right triangle illustrating the cofunction relationships

Because $\cos θ = \frac{b}{c} = \sin (\frac{π}{2} - θ), \cos$ we have $\sin^{- 1} (\cos θ) = \frac{π}{2} - θ \sin^{- 1} (\cos) = \frac{π}{2} - θ$ if $0 \leq θ \leq π . 0 \leq θ \leq π .$ If $θ θ$ is not in this domain, then we need to find another angle that has the same cosine as $θ θ$ and does belong to the restricted domain; we then subtract this angle from $\frac{π}{2} . \frac{π}{2} .$ Similarly, $\sin θ = \frac{a}{c} = \cos (\frac{π}{2} - θ), \sin$ so $\cos^{- 1} (\sin θ) = \frac{π}{2} - θ \cos^{- 1} (\sin) = \frac{π}{2} - θ$ if $- \frac{π}{2} \leq θ \leq \frac{π}{2} . - \frac{π}{2} \leq θ \leq \frac{π}{2} .$ These are just the function-cofunction relationships presented in another way.

How To

Given functions of the form $\sin^{- 1} (\cos x) \sin^{- 1} (\cos)$ and $\cos^{- 1} (\sin x), \cos^{- 1} (\sin),$ evaluate them.

If $x is in [0, π], x is in [0, π],$ then $\sin^{- 1} (\cos x) = \frac{π}{2} - x . \sin^{- 1} (\cos) = \frac{π}{2} - x .$
If $x is not in [0, π], x is not in [0, π],$ then find another angle $y in [0, π] y in [0, π]$ such that $\cos y = \cos x . \cos$
$\sin^{- 1} (\cos x) = \frac{π}{2} - y \sin^{- 1} (\cos) = \frac{π}{2} - y$
If $x is in [- \frac{π}{2}, \frac{π}{2}], x is in [- \frac{π}{2}, \frac{π}{2}],$ then $\cos^{- 1} (\sin x) = \frac{π}{2} - x . \cos^{- 1} (\sin) = \frac{π}{2} - x .$
If $x is not in [- \frac{π}{2}, \frac{π}{2}], x is not in,$ then find another angle $y in [- \frac{π}{2}, \frac{π}{2}] y in [- \frac{π}{2}, \frac{π}{2}]$ such that $\sin y = \sin x . \sin$
$\cos^{- 1} (\sin x) = \frac{π}{2} - y \cos^{- 1} (\sin) = \frac{π}{2} - y$

Example 6

Evaluating the Composition of an Inverse Sine with a Cosine

Evaluate $\sin^{- 1} (\cos (\frac{13 π}{6})) \sin^{- 1} (\cos (\frac{13 π}{6}))$

ⓐby direct evaluation.
ⓑ by the method described previously.

Show solution

Solution

ⓐ Here, we can directly evaluate the inside of the composition.
$\begin{array}{l} \begin{array}{l} \cos (\frac{13 π}{6}) = \cos (\frac{π}{6} + 2 π) \\ = \cos (\frac{π}{6}) \\ = \frac{\sqrt{3}}{2} \end{array} \end{array} \begin{array}{l} \begin{array}{l} \cos (\frac{13 π}{6}) = \cos (\frac{π}{6} + 2 π) \\ = \cos (\frac{π}{6}) \\ = \frac{\sqrt{3}}{2} \end{array} \end{array}$
Now, we can evaluate the inverse function as we did earlier.

$\sin^{- 1} (\frac{\sqrt{3}}{2}) = \frac{π}{3} \sin^{- 1} (\frac{\sqrt{3}}{2}) = \frac{π}{3}$
ⓑ We have $x = \frac{13 π}{6}, y = \frac{π}{6}, x = \frac{13 π}{6},$ and
$\begin{array}{r} \sin^{- 1} (\cos (\frac{13 π}{6})) = \frac{π}{2} - \frac{π}{6} \\ = \frac{π}{3} \end{array} \begin{array}{r} \sin^{- 1} (\cos (\frac{13 π}{6})) = \frac{π}{2} - \frac{π}{6} \\ = \frac{π}{3} \end{array}$

Try It #6

Evaluate $\cos^{- 1} (\sin (- \frac{11 π}{4})) . \cos^{- 1} (\sin (- \frac{11 π}{4})) .$

Evaluating Compositions of the Form f(g⁻¹(x))

To evaluate compositions of the form $f (g^{- 1} (x)), f (g^{- 1} (x)),$ where $f f$ and $g g$ are any two of the functions sine, cosine, or tangent and $x x$ is any input in the domain of $g^{- 1}, g^{- 1},$ we have exact formulas, such as $\sin (\cos^{- 1} x) = \sqrt{1 - x^{2}} . \sin (\cos^{- 1} x) = \sqrt{1 - x^{2}} .$ When we need to use them, we can derive these formulas by using the trigonometric relations between the angles and sides of a right triangle, together with the use of Pythagoras’s relation between the lengths of the sides. We can use the Pythagorean identity, $\sin^{2} x + \cos^{2} x = 1, \sin^{2} x + \cos^{2} x = 1,$ to solve for one when given the other. We can also use the inverse trigonometric functions to find compositions involving algebraic expressions.

Example 7

Evaluating the Composition of a Sine with an Inverse Cosine

Find an exact value for $\sin (\cos^{- 1} (\frac{4}{5})) . \sin (\cos^{- 1} (\frac{4}{5})) .$

Show solution

Solution

Beginning with the inside, we can say there is some angle such that $θ = \cos^{- 1} (\frac{4}{5}), θ = \cos^{- 1} (\frac{4}{5}),$ which means $\cos θ = \frac{4}{5}, \cos$ and we are looking for $\sin θ . \sin$ We can use the Pythagorean identity to do this.

\begin{array}{l} \sin^{2} θ + \cos^{2} θ = 1 & Use our known value for cosine . \\ \sin^{2} θ + {(\frac{4}{5})}^{2} = 1 & Solve for sine . \\ \sin^{2} θ = 1 - \frac{16}{25} \\ \sin θ = \pm \sqrt{\frac{9}{25}} = \pm \frac{3}{5} \end{array} \begin{array}{l} \sin^{2} θ + \cos^{2} θ = 1 & Use our known value for cosine . \\ \sin^{2} θ + {(\frac{4}{5})}^{2} = 1 & Solve for sine . \\ \sin^{2} θ = 1 - \frac{16}{25} \\ \sin \end{array}

Since $θ = \cos^{- 1} (\frac{4}{5}) θ = \cos^{- 1} (\frac{4}{5})$ is in quadrant I, $\sin θ \sin$ must be positive, so the solution is $\frac{3}{5} . \frac{3}{5} .$ See Figure 11.

An illustration of a right triangle with an angle theta. Oppostie the angle theta is a side with length 3. Adjacent the angle theta is a side with length 4. The hypoteneuse has angle of length 5.

Figure 11 Right triangle illustrating that if

\cos θ = \frac{4}{5}, \cos

then

\sin θ = \frac{3}{5} \sin

We know that the inverse cosine always gives an angle on the interval $[0, π], [0, π],$ so we know that the sine of that angle must be positive; therefore $\sin (\cos^{- 1} (\frac{4}{5})) = \sin θ = \frac{3}{5} . \sin (\cos^{- 1} (\frac{4}{5})) = \sin$

Try It #7

Evaluate $\cos (\tan^{- 1} (\frac{5}{12})) . \cos (\tan^{- 1} (\frac{5}{12})) .$

Example 8

Evaluating the Composition of a Sine with an Inverse Tangent

Find an exact value for $\sin (\tan^{- 1} (\frac{7}{4})) . \sin (\tan^{- 1} (\frac{7}{4})) .$

Show solution

Solution

While we could use a similar technique as in Example 6, we will demonstrate a different technique here. From the inside, we know there is an angle such that $\tan θ = \frac{7}{4} . \tan$ We can envision this as the opposite and adjacent sides on a right triangle, as shown in Figure 12.

An illustration of a right triangle with angle theta. Adjacent the angle theta is a side with length 4. Opposite the angle theta is a side with length 7.

Figure 12 A right triangle with two sides known

Using the Pythagorean Theorem, we can find the hypotenuse of this triangle.

\begin{array}{l} \begin{array}{l} 4^{2} + 7^{2} = {hypotenuse}^{2} \end{array} \\ hypotenuse = \sqrt{65} \end{array} \begin{array}{l} \begin{array}{l} 4^{2} + 7^{2} = {hypotenuse}^{2} \end{array} \\ hypotenuse = \sqrt{65} \end{array}

Now, we can evaluate the sine of the angle as the opposite side divided by the hypotenuse.

\sin θ = \frac{7}{\sqrt{65}} \sin

This gives us our desired composition.

\begin{array}{l} \sin (\tan^{- 1} (\frac{7}{4})) = \sin θ \\ = \frac{7}{\sqrt{65}} \\ = \frac{7 \sqrt{65}}{65} \end{array} \begin{array}{l} \sin (\tan^{- 1} (\frac{7}{4})) = \sin \\ = \frac{7}{\sqrt{65}} \\ = \frac{7 \sqrt{65}}{65} \end{array}

Try It #8

Evaluate $\cos (\sin^{- 1} (\frac{7}{9})) . \cos (\sin^{- 1} (\frac{7}{9})) .$

Example 9

Finding the Cosine of the Inverse Sine of an Algebraic Expression

Find a simplified expression for $\cos (\sin^{- 1} (\frac{x}{3})) \cos (\sin^{- 1} (\frac{x}{3}))$ for $- 3 \leq x \leq 3. - 3 \leq x \leq 3.$

Show solution

Solution

We know there is an angle $θ θ$ such that $\sin θ = \frac{x}{3} . \sin$

\begin{array}{l} \sin^{2} θ + \cos^{2} θ = 1 & Use the Pythagorean Theorem . \\ {(\frac{x}{3})}^{2} + \cos^{2} θ = 1 & Solve for cosine . \\ \cos^{2} θ = 1 - \frac{x^{2}}{9} \\ \cos θ = \pm \sqrt{\frac{9 - x^{2}}{9}} = \pm \frac{\sqrt{9 - x^{2}}}{3} \end{array} \begin{array}{l} \sin^{2} θ + \cos^{2} θ = 1 & Use the Pythagorean Theorem . \\ Solve for cosine . \\ \cos^{2} θ = 1 - \frac{x^{2}}{9} \\ \cos θ = \pm \sqrt{\frac{9 - x^{2}}{9}} = \pm \frac{\sqrt{9 - x^{2}}}{3} \end{array}

Because we know that the inverse sine must give an angle on the interval $[- \frac{π}{2}, \frac{π}{2}], [- \frac{π}{2}, \frac{π}{2}],$ we can deduce that the cosine of that angle must be positive.

\cos (\sin^{- 1} (\frac{x}{3})) = \frac{\sqrt{9 - x^{2}}}{3} \cos (\sin^{- 1} (\frac{x}{3})) = \frac{\sqrt{9 - x^{2}}}{3}

Try It #9

Find a simplified expression for $\sin (\tan^{- 1} (4 x)) \sin (\tan^{- 1} (4 x))$ for $- \frac{1}{4} \leq x \leq \frac{1}{4} . - \frac{1}{4} \leq x \leq \frac{1}{4} .$

Media

Access this online resource for additional instruction and practice with inverse trigonometric functions.

Evaluate Expressions Involving Inverse Trigonometric Functions

Verbal

Why do the functions $f (x) = \sin^{- 1} x f (x) = \sin^{- 1} x$ and $g (x) = \cos^{- 1} x g (x) = \cos^{- 1} x$ have different ranges?

Since the functions $y = \cos x y = \cos$ and $y = \cos^{- 1} x y = \cos^{- 1} x$ are inverse functions, why is $\cos^{- 1} (\cos (- \frac{π}{6})) \cos^{- 1} (\cos (- \frac{π}{6}))$ not equal to $- \frac{π}{6} ? - \frac{π}{6} ?$

Explain the meaning of $\frac{π}{6} = \arcsin (0.5) . \frac{π}{6} = \arcsin (0.5) .$

Most calculators do not have a key to evaluate $\sec^{- 1} (2) . \sec^{- 1} (2) .$ Explain how this can be done using the cosine function or the inverse cosine function.

Why must the domain of the sine function, $\sin x, \sin$ be restricted to $[- \frac{π}{2}, \frac{π}{2}] [- \frac{π}{2}, \frac{π}{2}]$ for the inverse sine function to exist?

Discuss why this statement is incorrect: $\arccos (\cos x) = x \arccos (\cos) = x$ for all $x . x .$

Determine whether the following statement is true or false and explain your answer: $\arccos (- x) = π - \arccos x . \arccos (- x) = π - \arccos$

Algebraic

For the following exercises, evaluate the expressions.

$\sin^{- 1} (\frac{\sqrt{2}}{2}) \sin^{- 1} (\frac{\sqrt{2}}{2})$

$\sin^{- 1} (- \frac{1}{2}) \sin^{- 1} (- \frac{1}{2})$

10.

$\cos^{- 1} (\frac{1}{2}) \cos^{- 1} (\frac{1}{2})$

11.

$\cos^{- 1} (- \frac{\sqrt{2}}{2}) \cos^{- 1} (- \frac{\sqrt{2}}{2})$

12.

$\tan^{- 1} (1) \tan^{- 1} (1)$

13.

$\tan^{- 1} (- \sqrt{3}) \tan^{- 1} (- \sqrt{3})$

14.

$\tan^{- 1} (- 1) \tan^{- 1} (- 1)$

15.

$\tan^{- 1} (\sqrt{3}) \tan^{- 1} (\sqrt{3})$

16.

$\tan^{- 1} (\frac{- 1}{\sqrt{3}}) \tan^{- 1} (\frac{- 1}{\sqrt{3}})$

For the following exercises, use a calculator to evaluate each expression. Express answers to the nearest hundredth.

17.

$\cos^{- 1} (- 0.4) \cos^{- 1} (- 0.4)$

18.

$\arcsin (0.23) \arcsin (0.23)$

19.

$\arccos (\frac{3}{5}) \arccos (\frac{3}{5})$

20.

$\cos^{- 1} (0.8) \cos^{- 1} (0.8)$

21.

$\tan^{- 1} (6) \tan^{- 1} (6)$

For the following exercises, find the angle $θ θ$ in the given right triangle. Round answers to the nearest hundredth.

22.

An illustration of a right triangle with angle theta. Opposite the angle theta is a side with length of 7. The hypotenuse has a lngeth of 10.

23.

An illustration of a right triangle with angle theta. Adjacent the angle theta is a side of length 19. Opposite the angle theta is a side with length 12.

For the following exercises, find the exact value, if possible, without a calculator. If it is not possible, explain why.

24.

$\sin^{- 1} (\cos (π)) \sin^{- 1} (\cos (π))$

25.

$\tan^{- 1} (\sin (π)) \tan^{- 1} (\sin (π))$

26.

$\cos^{- 1} (\sin (\frac{π}{3})) \cos^{- 1} (\sin (\frac{π}{3}))$

27.

$\tan^{- 1} (\sin (\frac{π}{3})) \tan^{- 1} (\sin (\frac{π}{3}))$

28.

$\sin^{- 1} (\cos (\frac{- π}{2})) \sin^{- 1} (\cos (\frac{- π}{2}))$

29.

$\tan^{- 1} (\sin (\frac{4 π}{3})) \tan^{- 1} (\sin (\frac{4 π}{3}))$

30.

$\sin^{- 1} (\sin (\frac{5 π}{6})) \sin^{- 1} (\sin (\frac{5 π}{6}))$

31.

$\tan^{- 1} (\sin (\frac{- 5 π}{2})) \tan^{- 1} (\sin (\frac{- 5 π}{2}))$

32.

$\cos (\sin^{- 1} (\frac{4}{5})) \cos (\sin^{- 1} (\frac{4}{5}))$

33.

$\sin (\cos^{- 1} (\frac{3}{5})) \sin (\cos^{- 1} (\frac{3}{5}))$

34.

$\sin (\tan^{- 1} (\frac{4}{3})) \sin (\tan^{- 1} (\frac{4}{3}))$

35.

$\cos (\tan^{- 1} (\frac{12}{5})) \cos (\tan^{- 1} (\frac{12}{5}))$

36.

$\cos (\sin^{- 1} (\frac{1}{2})) \cos (\sin^{- 1} (\frac{1}{2}))$

For the following exercises, find the exact value of the expression in terms of $x x$ with the help of a reference triangle.

37.

$\tan (\sin^{- 1} (x - 1)) \tan (\sin^{- 1} (x - 1))$

38.

$\sin (\cos^{- 1} (1 - x)) \sin (\cos^{- 1} (1 - x))$

39.

$\cos (\sin^{- 1} (\frac{1}{x})) \cos (\sin^{- 1} (\frac{1}{x}))$

40.

$\cos (\tan^{- 1} (3 x - 1)) \cos (\tan^{- 1} (3 x - 1))$

41.

$\tan (\sin^{- 1} (x + \frac{1}{2})) \tan (\sin^{- 1} (x + \frac{1}{2}))$

Extensions

For the following exercises, evaluate the expression without using a calculator. Give the exact value.

42.

$\frac{\sin^{- 1} (\frac{1}{2}) - \cos^{- 1} (\frac{\sqrt{2}}{2}) + \sin^{- 1} (\frac{\sqrt{3}}{2}) - \cos^{- 1} (1)}{\cos^{- 1} (\frac{\sqrt{3}}{2}) - \sin^{- 1} (\frac{\sqrt{2}}{2}) + \cos^{- 1} (\frac{1}{2}) - \sin^{- 1} (0)} \frac{\sin^{- 1} (\frac{1}{2}) - \cos^{- 1} (\frac{\sqrt{2}}{2}) + \sin^{- 1} (\frac{\sqrt{3}}{2}) - \cos^{- 1} (1)}{\cos^{- 1} (\frac{\sqrt{3}}{2}) - \sin^{- 1} (\frac{\sqrt{2}}{2}) + \cos^{- 1} (\frac{1}{2}) - \sin^{- 1} (0)}$

For the following exercises, find the function if $\sin t = \frac{x}{x + 1} . \sin$

43.

$\cos t \cos$

44.

$\sec t \sec$

45.

$\cot t \cot$

46.

$\cos (\sin^{- 1} (\frac{x}{x + 1})) \cos (\sin^{- 1} (\frac{x}{x + 1}))$

47.

$\tan^{- 1} (\frac{x}{\sqrt{2 x + 1}}) \tan^{- 1} (\frac{x}{\sqrt{2 x + 1}})$

Graphical

48.

Graph $y = \sin^{- 1} x y = \sin^{- 1} x$ and state the domain and range of the function.

49.

Graph $y = \arccos x y = \arccos$ and state the domain and range of the function.

50.

Graph one cycle of $y = \tan^{- 1} x y = \tan^{- 1} x$ and state the domain and range of the function.

51.

For what value of $x x$ does $\sin x = \sin^{- 1} x ? \sin$ Use a graphing calculator to approximate the answer.

52.

For what value of $x x$ does $\cos x = \cos^{- 1} x ? \cos$ Use a graphing calculator to approximate the answer.

Real-World Applications

53.

Suppose a 13-foot ladder is leaning against a building, reaching to the bottom of a second-ﬂoor window 12 feet above the ground. What angle, in radians, does the ladder make with the building?

54.

Suppose you drive 0.6 miles on a road so that the vertical distance changes from 0 to 150 feet. What is the angle of elevation of the road?

55.

An isosceles triangle has two congruent sides of length 9 inches. The remaining side has a length of 8 inches. Find the angle that a side of 9 inches makes with the 8-inch side.

56.

Without using a calculator, approximate the value of $\arctan (10, 000) . \arctan (10, 000) .$ Explain why your answer is reasonable.

57.

A truss for the roof of a house is constructed from two identical right triangles. Each has a base of 12 feet and height of 4 feet. Find the measure of the acute angle adjacent to the 4-foot side.

58.

The line $y = \frac{3}{5} x y = \frac{3}{5} x$ passes through the origin in the x,y-plane. What is the measure of the angle that the line makes with the positive x-axis?

59.

The line $y = \frac{- 3}{7} x y = \frac{- 3}{7} x$ passes through the origin in the x,y-plane. What is the measure of the angle that the line makes with the negative x-axis?

60.

What percentage grade should a road have if the angle of elevation of the road is 4 degrees? (The percentage grade is defined as the change in the altitude of the road over a 100-foot horizontal distance. For example a 5% grade means that the road rises 5 feet for every 100 feet of horizontal distance.)

61.

A 20-foot ladder leans up against the side of a building so that the foot of the ladder is 10 feet from the base of the building. If specifications call for the ladder's angle of elevation to be between 35 and 45 degrees, does the placement of this ladder satisfy safety specifications?

62.

Suppose a 15-foot ladder leans against the side of a house so that the angle of elevation of the ladder is 42 degrees. How far is the foot of the ladder from the side of the house?

Learning Objectives

Understanding and Using the Inverse Sine, Cosine, and Tangent Functions

Relations for Inverse Sine, Cosine, and Tangent Functions

Example 1

Writing a Relation for an Inverse Function

Solution

Try It #1

Finding the Exact Value of Expressions Involving the Inverse Sine, Cosine, and Tangent Functions

How To

Example 2

Evaluating Inverse Trigonometric Functions for Special Input Values

Solution

Try It #2

Using a Calculator to Evaluate Inverse Trigonometric Functions

Example 3

Evaluating the Inverse Sine on a Calculator

Solution

Try It #3

How To

Example 4

Applying the Inverse Cosine to a Right Triangle

Solution

Try It #4

Finding Exact Values of Composite Functions with Inverse Trigonometric Functions

Evaluating Compositions of the Form f(f−1(y)) and f−1(f(x))

Compositions of a trigonometric function and its inverse

Q&A

How To

Example 5

Using Inverse Trigonometric Functions

Solution

Try It #5

Evaluating Compositions of the Form f−1(g(x))

How To

Example 6

Evaluating the Composition of an Inverse Sine with a Cosine

Solution

Try It #6

Evaluating Compositions of the Form f(g−1(x))

Example 7

Evaluating the Composition of a Sine with an Inverse Cosine

Solution

Try It #7

Example 8

Evaluating the Composition of a Sine with an Inverse Tangent

Solution

Try It #8

Example 9

Finding the Cosine of the Inverse Sine of an Algebraic Expression

Solution

Try It #9

Media

Verbal

Algebraic

Extensions

Graphical

Real-World Applications

Evaluating Compositions of the Form f(f⁻¹(y)) and f⁻¹(f(x))

Evaluating Compositions of the Form f⁻¹(g(x))

Evaluating Compositions of the Form f(g⁻¹(x))