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.

Figure 1

For example, if f(x)=sinx,𝑓(𝑥)=sin𝑥, then we would write f−1(x)=sin−1x.𝑓−1(𝑥)=sin−1𝑥. Be aware that sin−1xsin−1𝑥 does not mean 1sinx.1sin𝑥. The following examples illustrate the inverse trigonometric functions:

• Since sin(π6)=12,sin(𝜋6)=12, then π6=sin−1(12).𝜋6=sin−1(12).
• Since cos(π)=−1,cos(𝜋)=−1, then π=cos−1(−1).𝜋=cos−1(−1).
• Since tan(π4)=1,tan(𝜋4)=1, then π4=tan−1(1).𝜋4=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 [−π2,π2][−𝜋2,𝜋2] and the graph of the cosine function limited to [0,π].[0,𝜋].

Figure 2 (a) Sine function on a restricted domain of [−π2,π2];[−𝜋2,𝜋2]; (b) Cosine function on a restricted domain of [0,π][0,𝜋]

Figure 3 shows the graph of the tangent function limited to (−π2,π2).(−𝜋2,𝜋2).

Figure 3 Tangent function on a restricted domain of (−π2,π2)(−𝜋2,𝜋2)

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−1x𝑦=sin−1𝑥 means x=siny.𝑥=sin𝑦. The inverse sine function is sometimes called the arcsine function, and notated arcsinx.arcsin𝑥.y=sin−1xhas domain[−1,1]and range[−π2,π2]𝑦=sin−1𝑥has domain[−1,1]and range[−𝜋2,𝜋2]
• The inverse cosine function y=cos−1x𝑦=cos−1𝑥 means x=cosy.𝑥=cos𝑦. The inverse cosine function is sometimes called the arccosine function, and notated arccosx.arccos𝑥.y=cos−1xhas domain[−1,1]and range[0,π]𝑦=cos−1𝑥has domain[−1,1]and range[0,𝜋]
• The inverse tangent function y=tan−1x𝑦=tan−1𝑥 means x=tany.𝑥=tan𝑦. The inverse tangent function is sometimes called the arctangent function, and notated arctanx.arctan𝑥.y=tan−1xhas domain(−∞,∞)and range(−π2,π2)𝑦=tan−1𝑥has domain(−∞,∞)and range(−𝜋2,𝜋2)

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−1xsin−1𝑥 has domain [−1,1][−1,1] and range [−π2,π2],[−𝜋2,𝜋2], cos−1xcos−1𝑥 has domain [−1,1][−1,1] and range [0,π],[0,𝜋], and tan−1xtan−1𝑥 has domain of all real numbers and range (−π2,π2).(−𝜋2,𝜋2). 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.

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

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

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

RELATIONS FOR INVERSE SINE, COSINE, AND TANGENT FUNCTIONS

For angles in the interval [−π2,π2],[−𝜋2,𝜋2], if siny=x,sin𝑦=𝑥, then sin−1x=y.sin−1𝑥=𝑦.

For angles in the interval [0,π],[0,𝜋], if cosy=x,cos𝑦=𝑥, then cos−1x=y.cos−1𝑥=𝑦.

For angles in the interval (−π2,π2),(−𝜋2,𝜋2), if tany=x,tan𝑦=𝑥, then tan−1x=y.tan−1𝑥=𝑦.

EXAMPLE 1

Writing a Relation for an Inverse Function

Given sin(5π12)≈0.96593,sin(5𝜋12)≈0.96593, write a relation involving the inverse sine.

Solution

Use the relation for the inverse sine. If siny=x,sin𝑦=𝑥, then sin−1x=ysin−1𝑥=𝑦.

In this problem, x=0.96593,𝑥=0.96593, and y=5π12.𝑦=5𝜋12.

sin−1(0.96593)≈5π12sin−1(0.96593)≈5𝜋12

TRY IT #1

Given cos(0.5)≈0.8776,cos(0.5)≈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 π6𝜋6 (30°), π4𝜋4 (45°), and π3𝜋3 (60°), and their reflections into other quadrants.

HOW TO

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

1. Find angle x𝑥 for which the original trigonometric function has an output equal to the given input for the inverse trigonometric function.
2. If x𝑥 is not in the defined range of the inverse, find another angle y𝑦 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.

1. ⓐ sin−1(12)sin−1(12)
2. ⓑ sin−1(−2√2)sin−1(−22)
3. ⓒ cos−1(−3√2)cos−1(−32)
4. ⓓ tan−1(1)tan−1(1)

Solution

1. ⓐ Evaluating sin−1(12)sin−1(12) is the same as determining the angle that would have a sine value of 12.12. In other words, what angle x𝑥 would satisfy sin(x)=12?sin(𝑥)=12? There are multiple values that would satisfy this relationship, such as π6𝜋6 and 5π6,5𝜋6, but we know we need the angle in the interval [−π2,π2],[−𝜋2,𝜋2], so the answer will be sin−1(12)=π6.sin−1(12)=𝜋6. Remember that the inverse is a function, so for each input, we will get exactly one output.
2. ⓑ To evaluate sin−1(−2√2),sin−1(−22), we know that 5π45𝜋4 and 7π47𝜋4 both have a sine value of −2√2,−22, but neither is in the interval [−π2,π2].[−𝜋2,𝜋2]. For that, we need the negative angle coterminal with 7π4:7𝜋4: sin−1(−2√2)=−π4.sin−1(−22)=−𝜋4.
3. ⓒTo evaluate cos−1(−3√2),cos−1(−32), we are looking for an angle in the interval [0,π][0,𝜋] with a cosine value of −3√2.−32. The angle that satisfies this is cos−1(−3√2)=5π6.cos−1(−32)=5𝜋6.
4. ⓓ Evaluating tan−1(1),tan−1(1), we are looking for an angle in the interval (−π2,π2)(−𝜋2,𝜋2) with a tangent value of 1. The correct angle is tan−1(1)=π4.tan−1(1)=𝜋4.

TRY IT #2

Evaluate each of the following.

1. ⓐ sin−1(−1)sin−1(−1)
2. ⓑ tan−1(−1)tan−1(−1)
3. ⓒ cos−1(−1)cos−1(−1)
4. ⓓ cos−1(12)cos−1(12)

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 θ𝜃 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.

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)≈1.3252.sin−1(0.97)≈1.3252. In degree mode, sin−1(0.97)≈75.93°.sin−1(0.97)≈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.

Figure 7

1. If one given side is the hypotenuse of length hℎ and the side of length a𝑎 adjacent to the desired angle is given, use the equation θ=cos−1(ah).𝜃=cos−1(𝑎ℎ).
2. If one given side is the hypotenuse of length hℎ and the side of length p𝑝 opposite to the desired angle is given, use the equation θ=sin−1(ph).𝜃=sin−1(𝑝ℎ).
3. If the two legs (the sides adjacent to the right angle) are given, then use the equation θ=tan−1(pa).𝜃=tan−1(𝑝𝑎).

EXAMPLE 4

Applying the Inverse Cosine to a Right Triangle

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

Figure 8

Solution

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

cosθ=912θ=cos−1(912)θ≈0.7227 or about 41.4096°Apply definition of the inverse.Evaluate.cos𝜃=912𝜃=cos−1(912)Apply definition of the inverse.𝜃≈0.7227 or about 41.4096°Evaluate.

TRY IT #4

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

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))=y𝑓(𝑓−1(𝑦))=𝑦 for all y𝑦 in the proper domain for the given function. This follows from the definition of the inverse and from the fact that the range of f𝑓 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

sin(sin−1x)=xfor−1≤x≤1cos(cos−1x)=xfor−1≤x≤1tan(tan−1x)=xfor−∞<x<∞sin(sin−1𝑥)=𝑥for−1≤𝑥≤1cos(cos−1𝑥)=𝑥for−1≤𝑥≤1tan(tan−1𝑥)=𝑥for−∞<𝑥<∞

sin−1(sinx)=xonly for −π2≤x≤π2cos−1(cosx)=xonly for 0≤x≤πtan−1(tanx)=xonly for −π2<x<π2sin−1(sin𝑥)=𝑥only for −𝜋2≤𝑥≤𝜋2cos−1(cos𝑥)=𝑥only for 0≤𝑥≤𝜋tan−1(tan𝑥)=𝑥only for −𝜋2<𝑥<𝜋2

Q&A

Is it correct that sin−1(sinx)=x?sin−1(sin𝑥)=𝑥?

No. This equation is correct if x𝑥 belongs to the restricted domain [−π2,π2],[−𝜋2,𝜋2], but sine is defined for all real input values, and for x𝑥 outside the restricted interval, the equation is not correct because its inverse always returns a value in [−π2,π2].[−𝜋2,𝜋2]. The situation is similar for cosine and tangent and their inverses. For example, sin−1(sin(3π4))=π4.sin−1(sin(3𝜋4))=𝜋4.

HOW TO

Given an expression of the form f⁻¹(f(θ)) where f(θ)=sinθ,cosθ, or tanθ,𝑓(𝜃)=sin𝜃,cos𝜃, or tan𝜃, evaluate.

1. If θ𝜃 is in the restricted domain of f, then f−1(f(θ))=θ.𝑓, then 𝑓−1(𝑓(𝜃))=𝜃.
2. If not, then find an angle ϕ𝜙 within the restricted domain of f𝑓 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:

1. ⓐ sin−1(sin(π3))sin−1(sin(𝜋3))
2. ⓑ sin−1(sin(2π3))sin−1(sin(2𝜋3))
3. ⓒ cos−1(cos(2π3))cos−1(cos(2𝜋3))
4. ⓓ cos−1(cos(−π3))cos−1(cos(−𝜋3))

Solution

1. ⓐ π3 is in [−π2,π2],𝜋3 is in [−𝜋2,𝜋2], so sin−1(sin(π3))=π3.sin−1(sin(𝜋3))=𝜋3.
2. ⓑ 2π3 is not in [−π2,π2],2𝜋3 is not in [−𝜋2,𝜋2], but sin(2π3)=sin(π3),sin(2𝜋3)=sin(𝜋3), so sin−1(sin(2π3))=π3.sin−1(sin(2𝜋3))=𝜋3.
3. ⓒ 2π3 is in [0,π],2𝜋3 is in [0,𝜋], so cos−1(cos(2π3))=2π3.cos−1(cos(2𝜋3))=2𝜋3.
4. ⓓ −π3 is not in [0,π],−𝜋3 is not in [0,𝜋], but cos(−π3)=cos(π3)cos(−𝜋3)=cos(𝜋3) because cosine is an even function. π3 is in [0,π],𝜋3 is in [0,𝜋], so cos−1(cos(−π3))=π3.cos−1(cos(−𝜋3))=𝜋3.

TRY IT #5

Evaluate tan−1(tan(π8))andtan−1(tan(11π9)).tan−1(tan(𝜋8))andtan−1(tan(11𝜋9)).

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 π2−θ.𝜋2−𝜃. Consider the sine and cosine of each angle of the right triangle in Figure 10.

Figure 10 Right triangle illustrating the cofunction relationships

Because cosθ=bc=sin(π2−θ),cos𝜃=𝑏𝑐=sin(𝜋2−𝜃), we have sin−1(cosθ)=π2−θsin−1(cos𝜃)=𝜋2−𝜃 if 0≤θ≤π.0≤𝜃≤𝜋. If θ𝜃 is not in this domain, then we need to find another angle that has the same cosine as θ𝜃 and does belong to the restricted domain; we then subtract this angle from π2.𝜋2. Similarly, sinθ=ac=cos(π2−θ),sin𝜃=𝑎𝑐=cos(𝜋2−𝜃), so cos−1(sinθ)=π2−θcos−1(sin𝜃)=𝜋2−𝜃 if −π2≤θ≤π2.−𝜋2≤𝜃≤𝜋2. These are just the function-cofunction relationships presented in another way.

HOW TO

Given functions of the form sin−1(cosx)sin−1(cos𝑥) and cos−1(sinx),cos−1(sin𝑥), evaluate them.

1. If x is in [0,π],𝑥 is in [0,𝜋], then sin−1(cosx)=π2−x.sin−1(cos𝑥)=𝜋2−𝑥.
2. 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 cosy=cosx.cos𝑦=cos𝑥.sin−1(cosx)=π2−ysin−1(cos𝑥)=𝜋2−𝑦
3. If x is in [−π2,π2],𝑥 is in [−𝜋2,𝜋2], then cos−1(sinx)=π2−x.cos−1(sin𝑥)=𝜋2−𝑥.
4. If x is not in[−π2,π2],𝑥 is not in[−𝜋2,𝜋2], then find another angle y in [−π2,π2]𝑦 in [−𝜋2,𝜋2] such that siny=sinx.sin𝑦=sin𝑥.cos−1(sinx)=π2−ycos−1(sin𝑥)=𝜋2−𝑦

EXAMPLE 6

Evaluating the Composition of an Inverse Sine with a Cosine

Evaluate sin−1(cos(13π6))sin−1(cos(13𝜋6))

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

Solution

1. ⓐ Here, we can directly evaluate the inside of the composition.cos(13π6)=cos(π6+2π)               =cos(π6)               =3√2cos(13𝜋6)=cos(𝜋6+2𝜋)               =cos(𝜋6)               =32Now, we can evaluate the inverse function as we did earlier.sin−1(3–√2)=π3sin−1(32)=𝜋3
2. ⓑ We have x=13π6,y=π6,𝑥=13𝜋6,𝑦=𝜋6, andsin−1(cos(13π6))=π2−π6=π3       sin−1(cos(13𝜋6))=𝜋2−𝜋6=𝜋3       

TRY IT #6

Evaluate cos−1(sin(−11π4)).cos−1(sin(−11𝜋4)).

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

To evaluate compositions of the form f(g−1(x)),𝑓(𝑔−1(𝑥)), where f𝑓 and g𝑔 are any two of the functions sine, cosine, or tangent and x𝑥 is any input in the domain of g−1,𝑔−1, we have exact formulas, such as sin(cos−1x)=1−x2−−−−−√.sin(cos−1𝑥)=1−𝑥2. 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, sin2x+cos2x=1,sin2𝑥+cos2𝑥=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(45)).sin(cos−1(45)).

Solution

Beginning with the inside, we can say there is some angle such that θ=cos−1(45),𝜃=cos−1(45), which means cosθ=45,cos𝜃=45, and we are looking for sinθ.sin𝜃. We can use the Pythagorean identity to do this.

sin2θ+cos2θ=1sin2θ+(45)2=1sin2θ=1−1625sinθ=±925−−√=±35Use our known value for cosine.Solve for sine.sin2𝜃+cos2𝜃=1Use our known value for cosine.sin2𝜃+(45)2=1Solve for sine.sin2𝜃=1−1625sin𝜃=±925=±35

Since θ=cos−1(45)𝜃=cos−1(45) is in quadrant I, sinθsin𝜃 must be positive, so the solution is 35.35. See Figure 11.

Figure 11 Right triangle illustrating that if cosθ=45,cos𝜃=45, then sinθ=35sin𝜃=35

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(45))=sinθ=35.sin(cos−1(45))=sin𝜃=35.

TRY IT #7

Evaluate cos(tan−1(512)).cos(tan−1(512)).

EXAMPLE 8

Evaluating the Composition of a Sine with an Inverse Tangent

Find an exact value for sin(tan−1(74)).sin(tan−1(74)).

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 tanθ=74.tan𝜃=74. We can envision this as the opposite and adjacent sides on a right triangle, as shown in Figure 12.

Figure 12 A right triangle with two sides known

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

      42+72=hypotenuse2hypotenuse=65−−√      42+72=hypotenuse2hypotenuse=65

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

sinθ=765−−√sin𝜃=765

This gives us our desired composition.

sin(tan−1(74))=sinθ                      =765√                      =765√65sin(tan−1(74))=sin𝜃                      =765                      =76565

TRY IT #8

Evaluate cos(sin−1(79)).cos(sin−1(79)).

EXAMPLE 9

Finding the Cosine of the Inverse Sine of an Algebraic Expression

Find a simplified expression for cos(sin−1(x3))cos(sin−1(𝑥3)) for −3≤x≤3.−3≤𝑥≤3.

Solution

We know there is an angle θ𝜃 such that sinθ=x3.sin𝜃=𝑥3.

sin2θ+cos2θ=1(x3)2+cos2θ=1cos2θ=1−x29cosθ=±9−x29−−−−√=±9−x2√3Use the Pythagorean Theorem.Solve for cosine.sin2𝜃+cos2𝜃=1Use the Pythagorean Theorem.(𝑥3)2+cos2𝜃=1Solve for cosine.cos2𝜃=1−𝑥29cos𝜃=±9−𝑥29=±9−𝑥23

Because we know that the inverse sine must give an angle on the interval [−π2,π2],[−𝜋2,𝜋2], we can deduce that the cosine of that angle must be positive.

cos(sin−1(x3))=9−x2−−−−−√3cos(sin−1(𝑥3))=9−𝑥23

TRY IT #9

Find a simplified expression for sin(tan−1(4x))sin(tan−1(4𝑥)) for −14≤x≤14.