inverse trig functions problem solving

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Precalculus

Course: precalculus   >   unit 2.

• Intro to arcsine
• Intro to arctangent
• Intro to arccosine
• Evaluate inverse trig functions
• Restricting domains of functions to make them invertible
• Domain & range of inverse tangent function
• Using inverse trig functions with a calculator

Inverse trigonometric functions review

What are the inverse trigonometric functions?

Range of the inverse trig functions.

Check your understanding

• (Choice A)   − 11.20 ‍   A − 11.20 ‍  
• (Choice B)   − 4.91 ‍   B − 4.91 ‍  
• (Choice C)   1.37 ‍   C 1.37 ‍  
• (Choice D)   7.65 ‍   D 7.65 ‍  

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8.3 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 .

For example, if f ( x ) = sin x , f ( x ) = sin x , 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 1 sin x . 1 sin x . The following examples illustrate the inverse trigonometric functions:

• Since sin ( π 6 ) = 1 2 , sin ( π 6 ) = 1 2 , then π 6 = sin − 1 ( 1 2 ) . π 6 = sin − 1 ( 1 2 ) .
• Since cos ( π ) = − 1 , cos ( π ) = − 1 , then π = cos − 1 ( − 1 ) . π = cos − 1 ( − 1 ) .
• Since tan ( π 4 ) = 1 , tan ( π 4 ) = 1 , then π 4 = tan − 1 ( 1 ) . 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 [ − π 2 , π 2 ] [ − π 2 , π 2 ] and the graph of the cosine function limited to [ 0 , π ] . [ 0 , π ] .

Figure 3 shows the graph of the tangent function limited to ( − π 2 , π 2 ) . ( − π 2 , 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 y . 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 [ − π 2 , π 2 ] y = sin − 1 x has domain [ −1 , 1 ] and range [ − π 2 , π 2 ]
• The inverse cosine function y = cos − 1 x y = cos − 1 x means x = cos y . x = cos y . The inverse cosine function is sometimes called the arccosine function, and notated arccos x . arccos x . y = cos − 1 x has domain [ −1 , 1 ] and range [ 0 , π ] y = cos − 1 x has domain [ −1 , 1 ] and range [ 0 , π ]
• The inverse tangent function y = tan − 1 x y = tan − 1 x means x = tan y . x = tan y . The inverse tangent function is sometimes called the arctangent function, and notated arctan x . arctan x . y = tan − 1 x has domain ( −∞ , ∞ ) and range ( − π 2 , π 2 ) y = tan − 1 x has domain ( −∞ , ∞ ) and range ( − π 2 , π 2 )

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 [ − π 2 , π 2 ] , [ − π 2 , 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 ( − π 2 , π 2 ) . ( − π 2 , 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 .

Relations for Inverse Sine, Cosine, and Tangent Functions

For angles in the interval [ − π 2 , π 2 ] , [ − π 2 , π 2 ] , if sin y = x , sin y = x , then sin − 1 x = y . sin − 1 x = y .

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

For angles in the interval ( − π 2 , π 2 ) , ( − π 2 , π 2 ) , if tan y = x , tan y = x , then tan − 1 x = y . tan − 1 x = y .

Writing a Relation for an Inverse Function

Given sin ( 5 π 12 ) ≈ 0.96593 , sin ( 5 π 12 ) ≈ 0.96593 , write a relation involving the inverse sine.

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

In this problem, x = 0.96593 , x = 0.96593 , and y = 5 π 12 . y = 5 π 12 .

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 π 6 π 6 (30°), π 4 π 4 (45°), and π 3 π 3 (60°), and their reflections into other quadrants.

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.

Evaluating Inverse Trigonometric Functions for Special Input Values

Evaluate each of the following.

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

Evaluating the Inverse Sine on a Calculator

Evaluate sin − 1 ( 0.97 ) sin − 1 ( 0.97 ) using a calculator.

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.

Evaluate cos − 1 ( − 0.4 ) cos − 1 ( − 0.4 ) using a calculator.

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

• 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 ( a h ) . θ = cos − 1 ( 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 ( p h ) . θ = sin − 1 ( p h ) .
• If the two legs (the sides adjacent to the right angle) are given, then use the equation θ = tan − 1 ( p a ) . θ = tan − 1 ( p a ) .

Applying the Inverse Cosine to a Right Triangle

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

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

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

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

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

No. This equation is correct if x x belongs to the restricted domain [ − π 2 , π 2 ] , [ − π 2 , 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 [ − π 2 , π 2 ] . [ − π 2 , π 2 ] . The situation is similar for cosine and tangent and their inverses. For example, sin − 1 ( sin ( 3 π 4 ) ) = π 4 . sin − 1 ( sin ( 3 π 4 ) ) = π 4 .

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

Using Inverse Trigonometric Functions

Evaluate the following:

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

Evaluate tan − 1 ( tan ( π 8 ) ) and tan − 1 ( tan ( 11 π 9 ) ) . tan − 1 ( tan ( π 8 ) ) and tan − 1 ( tan ( 11 π 9 ) ) .

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 π 2 − θ . π 2 − θ . Consider the sine and cosine of each angle of the right triangle in Figure 10 .

Because cos θ = b c = sin ( π 2 − θ ) , cos θ = b c = sin ( π 2 − θ ) , we have sin − 1 ( cos θ ) = π 2 − θ sin − 1 ( cos θ ) = π 2 − θ if 0 ≤ θ ≤ π . 0 ≤ θ ≤ 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 π 2 . π 2 . Similarly, sin θ = a c = cos ( π 2 − θ ) , sin θ = a c = cos ( π 2 − θ ) , so cos − 1 ( sin θ ) = π 2 − θ cos − 1 ( sin θ ) = π 2 − θ if − π 2 ≤ θ ≤ π 2 . − π 2 ≤ θ ≤ π 2 . These are just the function-cofunction relationships presented in another way.

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

• If x  is in  [ 0 , π ] , x  is in  [ 0 , π ] , then sin − 1 ( cos x ) = π 2 − x . sin − 1 ( cos x ) = 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 y = cos x . sin − 1 ( cos x ) = π 2 − y sin − 1 ( cos x ) = π 2 − y
• If x  is in  [ − π 2 , π 2 ] , x  is in  [ − π 2 , π 2 ] , then cos − 1 ( sin x ) = π 2 − x . cos − 1 ( sin x ) = π 2 − x .
• If x  is not in [ − π 2 , π 2 ] , x  is not in [ − π 2 , π 2 ] , then find another angle y  in  [ − π 2 , π 2 ] y  in  [ − π 2 , π 2 ] such that sin y = sin x . sin y = sin x . cos − 1 ( sin x ) = π 2 − y cos − 1 ( sin x ) = π 2 − y

Evaluating the Composition of an Inverse Sine with a Cosine

Evaluate sin − 1 ( cos ( 13 π 6 ) ) sin − 1 ( cos ( 13 π 6 ) )

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

Now, we can evaluate the inverse function as we did earlier.

• ⓑ We have x = 13 π 6 , y = π 6 , x = 13 π 6 , y = π 6 , and sin − 1 ( cos ( 13 π 6 ) ) = π 2 − π 6 = π 3         sin − 1 ( cos ( 13 π 6 ) ) = π 2 − π 6 = π 3        

Evaluate cos − 1 ( sin ( − 11 π 4 ) ) . cos − 1 ( sin ( − 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 ) = 1 − x 2 . sin ( cos − 1 x ) = 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.

Evaluating the Composition of a Sine with an Inverse Cosine

Find an exact value for sin ( cos − 1 ( 4 5 ) ) . sin ( cos − 1 ( 4 5 ) ) .

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

Since θ = cos − 1 ( 4 5 ) θ = cos − 1 ( 4 5 ) is in quadrant I, sin θ sin θ must be positive, so the solution is 3 5 . 3 5 . See Figure 11 .

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 ( 4 5 ) ) = sin θ = 3 5 . sin ( cos − 1 ( 4 5 ) ) = sin θ = 3 5 .

Evaluate cos ( tan − 1 ( 5 12 ) ) . cos ( tan − 1 ( 5 12 ) ) .

Evaluating the Composition of a Sine with an Inverse Tangent

Find an exact value for sin ( tan − 1 ( 7 4 ) ) . sin ( tan − 1 ( 7 4 ) ) .

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 = 7 4 . tan θ = 7 4 . We can envision this as the opposite and adjacent sides on a right triangle, as shown in Figure 12 .

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

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

This gives us our desired composition.

Evaluate cos ( sin − 1 ( 7 9 ) ) . cos ( sin − 1 ( 7 9 ) ) .

Finding the Cosine of the Inverse Sine of an Algebraic Expression

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

We know there is an angle θ θ such that sin θ = x 3 . sin θ = x 3 .

Because we know that the inverse sine must give an angle on the interval [ − π 2 , π 2 ] , [ − π 2 , π 2 ] , we can deduce that the cosine of that angle must be positive.

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

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

• Evaluate Expressions Involving Inverse Trigonometric Functions

8.3 Section Exercises

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 x and y = cos − 1 x y = cos − 1 x are inverse functions, why is cos − 1 ( cos ( − π 6 ) ) cos − 1 ( cos ( − π 6 ) ) not equal to − π 6 ? − π 6 ?

Explain the meaning of π 6 = arcsin ( 0.5 ) . 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 x , be restricted to [ − π 2 , π 2 ] [ − π 2 , π 2 ] for the inverse sine function to exist?

Discuss why this statement is incorrect: arccos ( cos x ) = x arccos ( cos x ) = 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 x .

For the following exercises, evaluate the expressions.

sin − 1 ( 2 2 ) sin − 1 ( 2 2 )

sin − 1 ( − 1 2 ) sin − 1 ( − 1 2 )

cos − 1 ( 1 2 ) cos − 1 ( 1 2 )

cos − 1 ( − 2 2 ) cos − 1 ( − 2 2 )

tan − 1 ( 1 ) tan − 1 ( 1 )

tan − 1 ( − 3 ) tan − 1 ( − 3 )

tan − 1 ( − 1 ) tan − 1 ( − 1 )

tan − 1 ( 3 ) tan − 1 ( 3 )

tan − 1 ( − 1 3 ) tan − 1 ( − 1 3 )

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

cos − 1 ( − 0.4 ) cos − 1 ( − 0.4 )

arcsin ( 0.23 ) arcsin ( 0.23 )

arccos ( 3 5 ) arccos ( 3 5 )

cos − 1 ( 0.8 ) cos − 1 ( 0.8 )

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.

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

sin − 1 ( cos ( π ) ) sin − 1 ( cos ( π ) )

tan − 1 ( sin ( π ) ) tan − 1 ( sin ( π ) )

cos − 1 ( sin ( π 3 ) ) cos − 1 ( sin ( π 3 ) )

tan − 1 ( sin ( π 3 ) ) tan − 1 ( sin ( π 3 ) )

sin − 1 ( cos ( − π 2 ) ) sin − 1 ( cos ( − π 2 ) )

tan − 1 ( sin ( 4 π 3 ) ) tan − 1 ( sin ( 4 π 3 ) )

sin − 1 ( sin ( 5 π 6 ) ) sin − 1 ( sin ( 5 π 6 ) )

tan − 1 ( sin ( − 5 π 2 ) ) tan − 1 ( sin ( − 5 π 2 ) )

cos ( sin − 1 ( 4 5 ) ) cos ( sin − 1 ( 4 5 ) )

sin ( cos − 1 ( 3 5 ) ) sin ( cos − 1 ( 3 5 ) )

sin ( tan − 1 ( 4 3 ) ) sin ( tan − 1 ( 4 3 ) )

cos ( tan − 1 ( 12 5 ) ) cos ( tan − 1 ( 12 5 ) )

cos ( sin − 1 ( 1 2 ) ) cos ( sin − 1 ( 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.

tan ( sin − 1 ( x − 1 ) ) tan ( sin − 1 ( x − 1 ) )

sin ( cos − 1 ( 1 − x ) ) sin ( cos − 1 ( 1 − x ) )

cos ( sin − 1 ( 1 x ) ) cos ( sin − 1 ( 1 x ) )

cos ( tan − 1 ( 3 x − 1 ) ) cos ( tan − 1 ( 3 x − 1 ) )

tan ( sin − 1 ( x + 1 2 ) ) tan ( sin − 1 ( x + 1 2 ) )

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

sin − 1 ( 1 2 ) − cos − 1 ( 2 2 ) + sin − 1 ( 3 2 ) − cos − 1 ( 1 ) cos − 1 ( 3 2 ) − sin − 1 ( 2 2 ) + cos − 1 ( 1 2 ) − sin − 1 ( 0 ) sin − 1 ( 1 2 ) − cos − 1 ( 2 2 ) + sin − 1 ( 3 2 ) − cos − 1 ( 1 ) cos − 1 ( 3 2 ) − sin − 1 ( 2 2 ) + cos − 1 ( 1 2 ) − sin − 1 ( 0 )

For the following exercises, find the function if sin t = x x + 1 . sin t = x x + 1 .

cos t cos t

sec t sec t

cot t cot t

cos ( sin − 1 ( x x + 1 ) ) cos ( sin − 1 ( x x + 1 ) )

tan − 1 ( x 2 x + 1 ) tan − 1 ( x 2 x + 1 )

Graph y = sin − 1 x y = sin − 1 x and state the domain and range of the function.

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

Graph one cycle of y = tan − 1 x y = tan − 1 x and state the domain and range of the function.

For what value of x x does sin x = sin − 1 x ? sin x = sin − 1 x ? Use a graphing calculator to approximate the answer.

For what value of x x does cos x = cos − 1 x ? cos x = cos − 1 x ? Use a graphing calculator to approximate the answer.

Real-World Applications

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?

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?

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.

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

A truss (interior beam structure) 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.

The line y = 3 5 x y = 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?

The line y = − 3 7 x y = − 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?

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.)

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?

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?

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Chapter 8: More Functions and Identities

Exercises: 8.2 Inverse Trigonometric Functions Exercises

Practice each skill in the Homework Problems listed.

• Decide whether a function has an inverse function #1–8
• Evaluate the inverse trig functions #9–20
• Model problems with inverse trig functions #21–24
• Solve formulas #25–30
• Simplify expressions involving the inverse trig functions #31–42, 51–68
• Graph the inverse trig functions #43–50, 69 and 70

Suggested Homework

Exercises for 8.2 Inverse Trigonometric Functions

Exercise group.

In Problems 1–4, which functions have an inverse function? Explain your answer.

For Problems 5–8, graph the function and decide if it has an inverse function.

[latex]f(x)=\sin 2x - \cos x[/latex]

[latex]g(x)=4e^{-(x/4)^2}[/latex]

[latex]G(x)=\sqrt{25-x^2}[/latex]

[latex]F(x)=\ln(x^3+8)[/latex]

For Problems 9–14, use a calculator to evaluate. Round your answers to the nearest tenth of a degree.

[latex]\sin^{-1}(0.2838)[/latex]

[latex]\tan^{-1}(4.8972)[/latex]

[latex]\cos^{-1}(0.6894)[/latex]

[latex]\arccos(-0.8134)[/latex]

[latex]\arctan(-1.2765)[/latex]

[latex]\arcsin(-07493)[/latex]

For Problems 15–20, give exact values in radians.

[latex]\cos^{-1} \dfrac{-1}{\sqrt{2}}[/latex]

[latex]\tan^{-1}(-1)[/latex]

[latex]\sin^{-1}\dfrac{-1}{2}[/latex]

[latex]\arccos\dfrac{\sqrt{3}}{2}[/latex]

[latex]\arctan \dfrac{1}{\sqrt{3}}[/latex]

[latex]\arcsin (-1)[/latex]

For Problems 21–26, sketch a figure to help you model each problem.

Delbert is watching the launch of a satellite at Cape Canaveral. The viewing area is 500 yards from the launch site. The angle of elevation, [latex]\theta{,}[/latex] of Delbert’s line of sight increases as the booster rocket rises.

• Write a formula for the height, [latex]h{,}[/latex] of the rocket as a function of [latex]\theta{.}[/latex]
• Write a formula for [latex]\theta[/latex] as a function of [latex]h{.}[/latex]
• Evaluate the formula in part (b) for [latex]h=1000{,}[/latex] and interpret the result.

Francine’s house lies under the flight path from the city airport, and commercial airliners pass overhead at an altitude of 35,000 feet. As Francine watches an airplane recede, its angle of elevation, [latex]\theta{,}[/latex] decreases.

• Write a formula for the horizontal distance, [latex]d{,}[/latex] to the airplane as a function of [latex]\theta{.}[/latex]
• Write a formula for [latex]\theta[/latex] as a function of [latex]d{.}[/latex]
• Evaluate the formula in part (b) for [latex]d=20,000{,}[/latex] and interpret the result.

While driving along the interstate, you approach an enormous 50-foot-wide billboard that sits just beside the road. Your viewing angle, [latex]\theta{,}[/latex] increases as you get closer to the billboard.

• Write a formula for your distance, [latex]d{,}[/latex] from the billboard as a function of [latex]\theta{.}[/latex]
• Evaluate the formula in part (b) for [latex]d=200{,}[/latex] and interpret the result.

Emma is walking along the bank of a straight river toward a 20-meter-long bridge over the river. Let [latex]\theta[/latex] be the angle subtended horizontally by Emma’s view of the bridge.

• Write a formula for Emma’s distance from the bridge, [latex]d{,}[/latex] as a function of [latex]\theta{.}[/latex]
• Evaluate the formula in part (b) for [latex]d=500{,}[/latex] and interpret the result.

Martin is viewing a 4-meter-tall painting whose base is 1 meter above his eye level.

• Write a formula for [latex]\alpha{,}[/latex] the angle subtended from Martin’s eye level to the bottom of the painting, when he stands [latex]x[/latex] meters from the wall.
• Write a formula for [latex]\beta{,}[/latex] the angle subtended by the painting, in terms of [latex]x{.}[/latex]
• Evaluate the formula in part (b) for [latex]x=5{,}[/latex] and interpret the result.

A 5-foot mirror is positioned so that its bottom is 1.5 feet below Jane’s eye level.

• Write a formula for [latex]\alpha{,}[/latex] the angle subtended by the section of mirror below Jane’s eye level, when she stands [latex]x[/latex] feet from the mirror.
• Write a formula for [latex]\theta{,}[/latex] the angle subtended by the entire mirror, in terms of [latex]x{.}[/latex]
• Evaluate the formula in part (b) for [latex]x=10{,}[/latex] and interpret the result.

For Problems 27–32, solve the formula for the given variable.

[latex]V=V_0 \sin(2\pi\omega t+\phi){,}[/latex] for [latex]t[/latex]

[latex]R=\dfrac{1}{32} v_0^2\sin (2\theta){,}[/latex] for [latex]\theta[/latex]

[latex]\dfrac{a}{\sin A}= \dfrac{b}{\sin B}{,}[/latex] for [latex]A[/latex]

[latex]c^2=a^2 + b^2 - 2ab\cos C{,}[/latex] for [latex]C[/latex]

[latex]P=\dfrac{k}{R^4\cos \theta}[/latex] for [latex]\theta[/latex]

[latex]\dfrac{r}{z}=\dfrac{1}{\tan (\alpha + \beta)}{,}[/latex] for [latex]\alpha[/latex]

For Problems 33–38, find exact values without using a calculator.

[latex]\tan\left(\sin^{-1}\left(\dfrac{2}{3}\right)\right)[/latex]

[latex]\tan\left(\cos^{-1}\left(\dfrac{3}{4}\right)\right)[/latex]

[latex]\cos\left[\tan^{-1}(-2)\right][/latex]

[latex]\sin \left[\tan^{-1}\left(\dfrac{-3}{\sqrt{5}}\right)\right][/latex]

[latex]\sin\left[\cos^{-1}\left(\dfrac{-2\sqrt{6}}{7}\right)\right][/latex]

[latex]\cos\left[\sin^{-1}\left(\dfrac{-2}{7}\right)\right][/latex]

For Problems 39–44, simplify the expression.

[latex]\tan(\cos^{-1}x)[/latex]

[latex]\cos(\tan^{-1}a)[/latex]

[latex]\cos(\sin^{-1}h)[/latex]

[latex]\sin(\cos^{-1}v)[/latex]

[latex]\sin(\tan^{-1}2t)[/latex]

[latex]\tan(\sin^{-1}3b)[/latex]

For Problems 45–47, complete the table of values and sketch the function.

Use a graphing calculator to answer each of the following questions. Then explain the results.

• Does [latex]\cos^{-1}x=\dfrac{1}{\cos x}{?}[/latex]
• Does [latex]\sin^{-1}x=\dfrac{1}{\sin x}{?}[/latex]
• Does [latex]\tan^{-1}x=\dfrac{1}{\tan x}{?}[/latex]
• Sketch a graph of [latex]y=\cos^{-1}x{,}[/latex] and label the scales on the axes.
• Use transformations to sketch graphs of [latex]y=2\cos^{-1}x[/latex] and [latex]y=\cos^{-1}(2x){.}[/latex]

Does [latex]2\cos^{-1}x=\cos^{-1}(2x){?}[/latex]

• Sketch a graph of [latex]y=\sin^{-1}x{,}[/latex] and label the scales on the axes.
• Use transformations to sketch graphs of [latex]y=\dfrac{1}{2}\sin^{-1}x[/latex] and [latex]y=\cos^{-1}\left(\dfrac{1}{2}x\right){.}[/latex]
• Does [latex]\dfrac{1}{2}\sin^{-1}x=\cos^{-1}\left(\dfrac{1}{2}x\right){?}[/latex]
• Sketch a graph of [latex]y=\tan^{-1}x{,}[/latex] and label the scales on the axes.
• Use your calculator to graph [latex]y=\dfrac{\sin^{-1}x}{\cos^{-1}x}[/latex] on a suitable domain.
• Does [latex]\tan^{-1}x=\dfrac{\sin^{-1}x}{\cos^{-1}x}{?}[/latex]
• Use your calculator to sketch [latex]y=\sqrt[3]{x}[/latex] and [latex]\tan^{-1}x[/latex] on [latex][-10, 10]{.}[/latex]
• Describe the similarities and differences in the two graphs.

Use the identities from Section 8.1 to help you find exact values for the expressions in Problems 53–58.

[latex]\sin (2\tan^{-1}4)[/latex]

[latex]\cos\left(2\sin^{-1}\left(\dfrac{5}{13}\right)\right)[/latex]

[latex]\tan\left(2\cos^{-1}\left(\dfrac{3}{4}\right)\right)[/latex]

[latex]\sin\left(2\cos^{-1}\left(\dfrac{-4}{5}\right)\right)[/latex]

[latex]\tan\left(2\sin^{-1}\left(\dfrac{1}{3}\right)\right)[/latex]

[latex]\cos\left(2\tan^{-1}\left(\dfrac{3}{2}\right)\right)[/latex]

Let [latex]\alpha= \cos^{-1}\left(\dfrac{-4}{5}\right),~\beta=\sin^{-1}\left(\dfrac{5}{13}\right){.}[/latex] Find exact values for the following:

[latex]\displaystyle \sin (\alpha+\beta)[/latex]

[latex]\displaystyle \cos (\alpha-\beta)[/latex]

[latex]\displaystyle \sin (\alpha-\beta)[/latex]

Let [latex]\alpha= \sin^{-1}\left(\dfrac{-15}{17}\right),~\beta=\tan^{-1}\left(\dfrac{4}{3}\right){.}[/latex] Find exact values for the following:

Find an exact value for [latex]\sin\left(\tan^{-1}\left(\dfrac{3}{4}\right)-\sin^{-1}\left(\dfrac{-4}{5}\right)\right){.}[/latex]

Find an exact value for [latex]\cos\left(\tan^{-1}\left(\dfrac{5}{12}\right)+\sin^{-1}\left(\dfrac{-3}{5}\right)\right){.}[/latex]

Express in terms of [latex]x[/latex] without trigonometric functions.

[latex]\displaystyle \cos(2\sin^{-1}x)[/latex]

Express in terms of [latex]w[/latex] without trigonometric functions.

[latex]\displaystyle \cos(2\tan^{-1}w)[/latex]

If [latex]x=5\sin \theta,~0° \lt \theta \lt 90°{,}[/latex] express [latex]\sin 2\theta[/latex] and [latex]\cos 2\theta[/latex] in terms of [latex]x{.}[/latex]

If [latex]x-1=2\cos \theta,~0° \lt \theta \lt 90°{,}[/latex] express [latex]\sin 2\theta[/latex] and [latex]\cos 2\theta[/latex] in terms of [latex]x{.}[/latex]

If [latex]x=3\tan \theta{,}[/latex] write [latex]\theta +\dfrac{1}{4}\sin 2\theta[/latex] in terms of [latex]x{.}[/latex]

If [latex]x=5\cos \theta{,}[/latex] write [latex]\dfrac{\theta}{2}-\cos 2\theta[/latex] in terms of [latex]x{.}[/latex]

• For what values of [latex]x[/latex] is the function [latex]f(x)=\sin (\arcsin x)[/latex] defined?
• Is [latex]\sin (\arcsin x)=x[/latex] for all [latex]x[/latex] where it is defined? If not, for what values of [latex]x[/latex] is the equation false?
• For what values of [latex]x[/latex] is the function [latex]g(x)=\arcsin(\sin x)[/latex] defined?
• Is [latex]\arcsin (\sin x)=x[/latex] for all [latex]x[/latex] where it is defined? If not, for what values of [latex]x[/latex] is the equation false?
• For what values of [latex]x[/latex] is the function [latex]f(x)=\cos (\arccos x)[/latex] defined?
• Is [latex]\cos (\arccos x)=x[/latex] for all [latex]x[/latex] where it is defined? If not, for what values of [latex]x[/latex] is the equation false?
• For what values of [latex]x[/latex] is the function [latex]g(x)=\arccos(\cos x)[/latex] defined?
• Is [latex]\arccos (\cos x)=x[/latex] for all [latex]x[/latex] where it is defined? If not, for what values of [latex]x[/latex] is the equation false?

Use your calculator to graph [latex]y=\sin^{-1}x+\cos^{-1}x{.}[/latex]

• State the domain and range of the graph.
• Explain why the graph looks as it does.

Use your calculator to graph [latex]y=\tan^{-1}x+\tan^{-1}(\dfrac{1}{x}){.}[/latex]

In Problems 73–74, we find a formula for the area under part of a semicircle.

• Write an expression for the area of the shaded sector in terms of [latex]\theta{.}[/latex]
• How are [latex]\theta[/latex] and [latex]t[/latex] related in the figure? (Hint: Write an expression for [latex]\sin \theta{.}[/latex])
• Combine your answers to (a) and (b) to write an expression for the area of the sector in terms of [latex]t{.}[/latex]
• Write an expression for the height of the shaded triangle in terms of [latex]t{.}[/latex] (Hint: Use the Pythagorean theorem.)
• Write an expression for the area of the triangle in terms of [latex]t{.}[/latex]
• Combine your answers to (b) and to Problem 73 to write an expression for the area bounded above by the upper semicircle, below by the [latex]x[/latex]-axis, on the left by the [latex]y[/latex]-axis, and on the right by [latex]x=t{,}[/latex] when [latex]0 \le t \le 1{.}[/latex]

Trigonometry Copyright © 2024 by Bimal Kunwor; Donna Densmore; Jared Eusea; and Yi Zhen. All Rights Reserved.

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Math 112 (Online) Walkthrough Text

Walkthrough of unit 5: inverse trig functions and solving trig equations, learning outcomes.

• 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.
• Use inverse trigonometric functions to solve right triangles.
• Find exact values of composite functions with 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.

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

• Since [latex]\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}[/latex], then [latex]\frac{\pi}{6}=\sin^{−1}(\frac{1}{2})[/latex].
• Since [latex]\cos(\pi)=−1[/latex], then [latex]\pi=\cos^{−1}(−1)[/latex].
• Since [latex]\tan\left(\frac{\pi}{4}\right)=1[/latex], then [latex]\frac{\pi}{4}=\tan^{−1}(1)[/latex].

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 [latex]f(a)=b[/latex], then an inverse function would satisfy [latex]f^{−1}(b)=a[/latex].

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 [latex]\left[\frac{−\pi}{2}\text{, }\frac{\pi}{2}\right][/latex] and the graph of the cosine function limited to [0, π].

Figure 2. (a) Sine function on a restricted domain of [latex]\left[−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right][/latex]; (b) Cosine function on a restricted domain of [0, π]

Figure 3 shows the graph of the tangent function limited to [latex]\left(−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right)[/latex].

Figure 3.  Tangent function on a restricted domain of [latex]\left(−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right)[/latex]

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

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 [latex]\sin^{−1}x[/latex] has domain [−1, 1] and range [latex]\left[−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right][/latex], [latex]\cos^{−1}x[/latex] has domain [−1, 1] and range [0, π], and [latex]\tan^{−1}x[/latex] has domain of all real numbers and range [latex]\left(−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right)[/latex]. 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 [latex]y=x[/latex].

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

A General Note: Relations for Inverse Sine, Cosine, and Tangent Functions

For angles in the interval [latex]\left[−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right][/latex], if [latex]\sin y=x[/latex], then [latex]\sin^{−1}x=y[/latex].

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

For angles in the interval [latex]\left(−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right)[/latex], if [latex]\tan y=x[/latex], then [latex]\tan^{−1}x=y[/latex].

Example 1: Writing a Relation for an Inverse Function

Given [latex]\sin\left(\frac{5\pi}{12}\right)\approx 0.96593[/latex], write a relation involving the inverse sine.

Use the relation for the inverse sine. If [latex]\sin y=x[/latex], then [latex]\sin^{−1}x=y[/latex].

In this problem, [latex]x=0.96593[/latex], and [latex]y=\frac{5\pi}{12}[/latex].

[latex]\sin^{−1}(0.96593)\approx \frac{5\pi}{12}[/latex]

Given [latex]\cos(0.5)\approx 0.8776[/latex], write a relation involving the inverse cosine.

[latex]\arccos(0.8776)\approx0.5[/latex]

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 [latex]\frac{\pi}{ 6} (30^\circ)\text{, }\frac{\pi}{ 4} (45^\circ),\text{ and } \frac{\pi}{ 3} (60^\circ)[/latex], and their reflections into other quadrants.

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

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

a. [latex]\sin−1\left(\frac{1}{2}\right)[/latex]

b. [latex]\sin−1\left(−\frac{2}{\sqrt{2}}\right)[/latex]

c. [latex]\cos−1\left(−\frac{3}{\sqrt{2}}\right)[/latex]

d. [latex]\tan^{− 1}(1)[/latex]

a. Evaluating [latex]\sin^{−1}(\frac{1}{2})[/latex] is the same as determining the angle that would have a sine value of [latex]\frac{1}{2}[/latex]. In other words, what angle x would satisfy [latex]\sin(x)=\frac{1}{2}[/latex]? There are multiple values that would satisfy this relationship, such as [latex]\frac{\pi}{6}[/latex] and [latex]\frac{5\pi}{6}[/latex], but we know we need the angle in the interval [latex]\left[−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right][/latex], so the answer will be [latex]\sin^{−1}(\frac{1}{2})=\frac{\pi}{6}[/latex]. Remember that the inverse is a function, so for each input, we will get exactly one output.

b. To evaluate [latex]\sin^{−1}\left(−\frac{\sqrt{2}}{2}\right)[/latex], we know that [latex]\frac{5\pi}{4}[/latex] and [latex]\frac{7\pi}{4}[/latex] both have a sine value of [latex]−\frac{\sqrt{2}}{2}[/latex], but neither is in the interval [latex]\left[−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right][/latex]. For that, we need the negative angle coterminal with [latex]\frac{7\pi}{4}:\sin^{−1}\left(−\frac{\sqrt{2}}{2}\right)=−\frac{\pi}{4}[/latex].

c. To evaluate [latex]\cos^{−1}\left(−\frac{\sqrt{3}}{2}\right)[/latex], we are looking for an angle in the interval [0,π] with a cosine value of [latex]−\frac{\sqrt{3}}{2}[/latex]. The angle that satisfies this is [latex]\cos^{−1}\left(−\frac{\sqrt{3}}{2}\right)=\frac{5\pi}{6}[/latex].

d. Evaluating [latex]\tan^{−1}(1)[/latex], we are looking for an angle in the interval [latex](−\frac{\pi}{2}\text{, }\frac{\pi}{2})[/latex] with a tangent value of 1. The correct angle is [latex]\tan^{−1}(1)=\frac{\pi}{4}[/latex].

• [latex]\sin^{−1}(−1)[/latex]
• [latex]\tan^{−1}(−1)[/latex]
• [latex]\cos^{−1}(−1)[/latex]
• [latex]\cos^{−1}(\frac{1}{2})[/latex]

1. [latex]−\frac{\pi}{2}[/latex];

2. [latex]−\frac{\pi}{4}[/latex]

3. [latex]\pi[/latex]

4. [latex]\frac{\pi}{3}[/latex]

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, 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 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 [latex]\sin^{−1}(0.97)[/latex] using a calculator.

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, [latex]\sin^{−1}(0.97)\approx1.3252[/latex]. In degree mode, [latex]\sin^{−1}(0.97)\approx75.93^{\circ}[/latex]. Note that in calculus and beyond we will use radians in almost all cases.

Evaluate [latex]\cos^{−1}(−0.4)[/latex] using a calculator.

1.9823 or 113.578°

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

• If one given side is the hypotenuse of length h and the side of length a adjacent to the desired angle is given, use the equation [latex]\theta=\cos^{−1}\left(\frac{a}{h}\right)[/latex].
• If one given side is the hypotenuse of length h and the side of length p opposite to the desired angle is given, use the equation [latex]\theta=\sin^{−1}\left(\frac{p}{h}\right)[/latex].
• If the two legs (the sides adjacent to the right angle) are given, then use the equation [latex]\theta=\tan^{−1}\left(\frac{p}{a}\right)[/latex].

Example 4: Applying the Inverse Cosine to a Right Triangle

Solve the triangle in Figure 8 for the angle θ.

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

[latex]\begin{align}&\cos\theta=\frac{9}{12}\\ &\theta=\cos^{−1}\left(\frac{9}{12}\right) && \text{Apply definition of the inverse.} \\ &\theta\approx0.7227\text{ or about }41.4096^{\circ} && \text{Evaluate.} \end{align}[/latex]

Solve the triangle in Figure 9 for the angle θ.

[latex]\sin^{−1}(0.6)=36.87^{\circ}=0.6435[/latex] radians

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 ) and g ( x ) be two different trigonometric functions belonging to the set {sin( x ), cos( x ), tan( x )} and let [latex]f^{−1}(y)[/latex] and [latex]g^{−1}(y)[/latex] be their inverses.

Evaluating Compositions of the Form [latex]f\left(f^{−1}(y)\right)[/latex] and [latex]f^{−1}(f(x))[/latex]

For any trigonometric function, [latex]f(f^{−1}(y))=y[/latex] for all 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 was defined to be identical to the domain of [latex]f^{−1}[/latex]. However, we have to be a little more careful with expressions of the form [latex]f^{−1}(f(x))[/latex].

A General Note: Compositions of a trigonometric function and its inverse

[latex]\begin{align} &\sin(\sin^{−1}x)=x\text{ for }−1\leq x\leq1\\ &\cos(\cos^{−1}x)=x\text{ for }−\infty\leq x\leq1 \\ &\tan(\tan^{−1}x)=x\text{ for }−\infty\text{ < }x\text{ < }\infty \end{align}[/latex]

[latex]\begin{align} \hfill &\sin^{−1}(\sin x)=x\text{ only for }−\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \hfill \\ &\cos^{−1}(\cos x)=x\text{ only for }0\leq x\leq\pi \hfill \\ &\tan^{−1}(\tan x)=x\text{ only for }−\frac{\pi}{2}\text{ < }x\text{ < }\frac{\pi}{2} \end{align}[/latex]

Is it correct that [latex]\sin^{−1}(\sin x)=x[/latex]?

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

Given an expression of the form [latex]f^{−1}(f(\theta))[/latex] where [latex]f(\theta)=\sin\theta\text{, }\cos\theta\text{, or }\tan\theta[/latex], evaluate.

• If θ is in the restricted domain of f , then [latex]f^{−1}(f(\theta))=\theta[/latex].
• If not, then find an angle ϕ within the restricted domain of f such that [latex]f(\phi)=f(\theta)[/latex]. Then [latex]f^{−1}(f(\theta))=\phi[/latex].

Example 5: Using Inverse Trigonometric Functions

Evaluate the following:

• [latex]\sin^{−1}(\sin(\frac{\pi}{3}))[/latex]
• [latex]\sin^{−1}(\sin(\frac{2\pi}{3}))[/latex]
• [latex]\cos^{−1}(\cos(\frac{2\pi}{3}))[/latex]
• [latex]\cos^{−1}(\cos(−\frac{\pi}{3}))[/latex]
• [latex]\frac{\pi}{3}[/latex] is in [latex]\left[−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right][/latex], so [latex]\sin^{−1}(\sin(\frac{\pi}{3}))=\frac{\pi}{3}[/latex].
• [latex]\frac{2\pi}{3}[/latex] is not in [latex]\left[−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right][/latex], but [latex]\sin\left(\frac{2\pi}{3}\right)=\sin\left(\frac{\pi}{3}\right)[/latex], so [latex]\sin^{−1}\left(\sin\left(\frac{2\pi}{3}\right)\right)=\frac{\pi}{3}[/latex].
• [latex]\frac{2\pi}{3}[/latex] is in [0,π], so [latex]\cos^{−1}\left(\cos\left(\frac{2\pi}{3}\right)\right)=\frac{2\pi}{3}[/latex].
• [latex]−\frac{\pi}{3}[/latex] is not in [0,π], but [latex]\cos(−\frac{\pi}{3})=\cos\left(\frac{\pi}{3}\right)[/latex] because cosine is an even function.

Evaluate [latex]\tan^{−1}\left(\tan\left(\frac{\pi}{8}\right)\right)[/latex] and [latex]\tan^{−1}\left(\tan\left(\frac{11\pi}{9}\right)\right)[/latex].

[latex]\frac{\pi}{8}\text{; }\frac{2\pi}{9}[/latex]

Evaluating Compositions of the Form [latex]f^{−1}(g(x))[/latex]

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 [latex]f^{−1}(g(x))[/latex]. For special values of 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, making the other [latex]\frac{\pi}{2}−\theta[/latex]. Consider the sine and cosine of each angle of the right triangle in Figure 10.

Figure 10. Right triangle illustrating the cofunction relationships

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

How To: Given functions of the form [latex]\sin^{−1}(\cos x)\text{ and }\cos^{−1}(\sin x)[/latex], evaluate them.

• If x  is in [0,π], then [latex]\sin^{−1}(\cos x)=\frac{\pi}{2}−x[/latex].
• If x  is not in [0,π], then find another angle y  in [0,π] such that [latex]\cos y=\cos x[/latex]. [latex]\sin^{−1}(\cos x)=\frac{\pi}{2}−y[/latex]
• If x  is in [latex]\left[−\frac{\pi}{2},\frac{\pi}{2}\right][/latex], then [latex]\cos^{−1}(\sin x)=\frac{\pi}{2}−x[/latex].
• If x  is not in [latex]\left[−\frac{\pi}{2},\frac{\pi}{2}\right][/latex], then find another angle y  in [latex]\left[−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right][/latex] such that [latex]\sin y=\sin x[/latex]. [latex]\cos^{−1}(\sin x)=\frac{\pi}{2}−y[/latex]

Example 6: Evaluating the Composition of an Inverse Sine with a Cosine

Evaluate [latex]\sin^{−1}(\cos(\frac{13\pi}{6}))[/latex]

• by direct evaluation.
• by the method described previously.

Now, we can evaluate the inverse function as we did earlier.

• We have [latex]x=\frac{13\pi}{6}[/latex], [latex]y=\frac{\pi}{6}[/latex], and [latex]\begin{align}\sin^{−1}\left(\cos\left(\frac{13\pi}{6}\right)\right)=\frac{\pi}{2}−\frac{\pi}{6} =\frac{\pi}{3} \end{align}[/latex]

Evaluate [latex]\cos^{−1}(\sin(−\frac{11\pi}{4}))[/latex].

[latex]\frac{3\pi}{4}[/latex]

Evaluating Compositions of the Form [latex]f(g^{−1}(x))[/latex]

To evaluate compositions of the form [latex]f(g^{−1}(x))[/latex], where f and g are any two of the functions sine, cosine, or tangent and x is any input in the domain of [latex]g−1[/latex], we have exact formulas, such as [latex]\sin\left({\cos}^{−1}x\right)=\sqrt{1−{x}^{2}}[/latex]. 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, [latex]\sin^{2}x+cos^{2}x=1[/latex], 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 [latex]\sin\left(\cos^{−1}\left(\frac{4}{5}\right)\right)[/latex].

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

[latex]\begin{align} &\sin^{2}\theta+\cos^{2}\theta=1 && \text{Use our known value for cosine.} \\ &\sin^{2}\theta+\left(\frac{4}{5}\right)^{2}=1 && \text{Solve for sine.} \\ &\sin^{2}\theta=1−\frac{16}{25} \\ &\sin\theta=\pm\sqrt{\frac{9}{25}}=\pm\frac{3}{5} \end{align}[/latex]

Since [latex]\theta=\cos^{−1}(\frac{4}{5})[/latex] is in quadrant I, [latex]\sin{\theta}[/latex] must be positive, so the solution is [latex]\frac{3}{5}[/latex]. See Figure 11.

Figure 11. Right triangle illustrating that if [latex]\cos\theta=\frac{4}{5}[/latex], then [latex]\sin\theta=\frac{3}{5}[/latex]

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

Evaluate [latex]\cos(\tan^{−1}(\frac{5}{12}))[/latex].

[latex]\frac{12}{13}[/latex]

Example 8: Evaluating the Composition of a Sine with an Inverse Tangent

Find an exact value for [latex]\sin\left(\tan^{−1}\left(\frac{7}{4}\right)\right)[/latex].

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 [latex]\tan\theta=\frac{7}{4}[/latex]. 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.

[latex]\begin{gathered}4^{2}+7^{2}=\text{hypotenuse}^{2} \\ \text{hypotenuse}=\sqrt{65} \end{gathered}[/latex]

Evaluate [latex]\cos(\sin^{−1}(\frac{7}{9}))[/latex].

[latex]\frac{4\sqrt{2}}{9}[/latex]

Example 9: Finding the Cosine of the Inverse Sine of an Algebraic Expression

Find a simplified expression for [latex]\cos\left(\sin^{−1}\left(\frac{x}{3}\right)\right)[/latex] for [latex]−3\leq x\leq3[/latex].

We know there is an angle θ such that [latex]\sin\theta=\frac{x}{3}\\[/latex]

[latex]\begin{align}&\sin^{2}\theta+\cos^{2}\theta=1 7&& \text{Use the Pythagorean Theorem.} \\ &\left(\frac{x}{3}\right)^{2}+\cos^{2}+\cos^2\theta=1 && \text{Solve for cosine.} \\ &\cos^{2}\theta=1−\frac{x^{2}}{9} \\ &\cos\theta=\pm\sqrt{\frac{9−x^{2}}{9}}=\pm\frac{\sqrt{9−x^{2}}}{3} \end{align}[/latex]

Because we know that the inverse sine must give an angle on the interval [latex]\left[−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right][/latex], we can deduce that the cosine of that angle must be positive.

[latex]\cos\left(\sin^{−1}\left(\frac{x}{3}\right)\right)=\frac{\sqrt{9−x^{2}}}{3}[/latex]

Find a simplified expression for [latex]\sin\left(\tan^{−1}\left(4x\right)\right)\\[/latex] for [latex]−\frac{1}{4}\leq x \leq\frac{1}{4}[/latex].

[latex]\frac{4x}{\sqrt{16x^{2}+1}}[/latex]

Solving Linear Trigonometric Equations in Sine and Cosine

Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is [latex]2\pi [/latex]. In other words, every [latex]2\pi [/latex] units, the y- values repeat. If we need to find all possible solutions, then we must add [latex]2\pi k[/latex], where [latex]k[/latex] is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is [latex]2\pi :[/latex]

There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.

Example 10: Solving a Linear Trigonometric Equation Involving the Cosine Function

Find all possible exact solutions for the equation [latex]\cos \theta =\frac{1}{2}[/latex].

From the unit circle , we know that

[latex]\begin{gathered}\cos \theta =\frac{1}{2} \\ \theta =\frac{\pi }{3},\frac{5\pi }{3} \end{gathered}[/latex]

These are the solutions in the interval [latex]\left[0,2\pi \right][/latex]. All possible solutions are given by

[latex]\theta =\frac{\pi }{3}\pm 2k\pi \text{ and }\theta =\frac{5\pi }{3}\pm 2k\pi [/latex]

where [latex]k[/latex] is an integer.

Example 11: Solving a Linear Equation Involving the Sine Function

Find all possible exact solutions for the equation [latex]\sin t=\frac{1}{2}[/latex].

Solving for all possible values of t means that solutions include angles beyond the period of [latex]2\pi [/latex]. From the unit circle, we can see that the solutions are [latex]t=\frac{\pi }{6}[/latex] and [latex]t=\frac{5\pi }{6}[/latex]. But the problem is asking for all possible values that solve the equation. Therefore, the answer is

[latex]t=\frac{\pi }{6}\pm 2\pi k\text{ and }t=\frac{5\pi }{6}\pm 2\pi k[/latex]

How To: Given a trigonometric equation, solve using algebra.

• Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.
• Substitute the trigonometric expression with a single variable, such as [latex]x[/latex] or [latex]u[/latex].
• Solve the equation the same way an algebraic equation would be solved.
• Substitute the trigonometric expression back in for the variable in the resulting expressions.
• Solve for the angle.

Example 12: Solve the Trigonometric Equation in Linear Form

Solve the equation exactly: [latex]2\cos \theta -3=-5,0\le \theta <2\pi [/latex].

[latex]\begin{gathered}2\cos \theta -3=-5 \\ \cos \theta =-2 \\ \cos \theta =-1 \\ \theta =\pi \end{gathered}[/latex]

Solve exactly the following linear equation on the interval [latex]\left[0,2\pi \right):2\sin x+1=0[/latex].

[latex]x=\frac{7\pi }{6},\frac{11\pi }{6}[/latex]

Solve Trigonometric Equations Using a Calculator

Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.

Example 13: Using a Calculator to Solve a Trigonometric Equation Involving Sine

Use a calculator to solve the equation [latex]\sin \theta =0.8[/latex], where [latex]\theta [/latex] is in radians.

Make sure mode is set to radians. To find [latex]\theta [/latex], use the inverse sine function. On most calculators, you will need to push the 2 ND button and then the SIN button to bring up the [latex]{\sin }^{-1}[/latex] function. What is shown on the screen is [latex]{\sin}^{-1}[/latex](. The calculator is ready for the input within the parentheses. For this problem, we enter [latex]{\sin }^{-1}\left(0.8\right)[/latex], and press ENTER. Thus, to four decimals places,

[latex]{\sin }^{-1}\left(0.8\right)\approx 0.9273[/latex]

This is the solution in quadrant I. There is also a solution in quadrant II. To find this we subtract [latex]/pi - 0.9273 \approx 2.2143 [/latex]

The general solution is

[latex]\theta \approx 0.9273\pm 2\pi k \text{ and } \theta \approx 2.2143 \pm 2\pi k[/latex]

The angle measurement in degrees is

[latex]\begin{align} \theta &\approx {53.1}^{\circ } \\ \theta &\approx {180}^{\circ }-{53.1}^{\circ } \\ &\approx {126.9}^{\circ } \end{align}[/latex]

Analysis of the Solution

Note that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using [latex]\pi -\theta [/latex].

Example 14: Using a Calculator to Solve a Trigonometric Equation Involving Secant

Use a calculator to solve the equation [latex]\sec \theta =-4[/latex], giving your answer in radians.

We can begin with some algebra.

[latex]\begin{gathered}\sec \theta =-4\\ \frac{1}{\cos \theta }=-4\\ \cos \theta =-\frac{1}{4}\end{gathered}[/latex]

Check that the MODE is in radians. Now use the inverse cosine function.

[latex]\begin{gathered}{\cos }^{-1}\left(-\frac{1}{4}\right)\approx 1.8235 \\ \theta \approx 1.8235+2\pi k \end{gathered}[/latex]

Since [latex]\frac{\pi }{2}\approx 1.57[/latex] and [latex]\pi \approx 3.14[/latex], 1.8235 is between these two numbers, thus [latex]\theta \approx \text{1}\text{.8235}[/latex] is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine.

So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is [latex]\theta \text{ }\text{ }\text{‘}\approx \pi -\text{1}\text{.8235}\approx \text{1}\text{.3181}\text{.}[/latex] The other solution in quadrant III is [latex]\theta \text{ }\text{ }\text{‘}\approx \pi +\text{1}\text{.3181}\approx \text{4}\text{.4597}\text{.}[/latex]

The solutions are [latex]\theta \approx 1.8235\pm 2\pi k[/latex] and [latex]\theta \approx 4.4597\pm 2\pi k[/latex].

Solve [latex]\cos \theta =-0.2[/latex].

[latex]\theta \approx 1.7722\pm 2\pi k[/latex] and [latex]\theta \approx 4.5110\pm 2\pi k[/latex]

Solving Equations Involving a Single Trigonometric Function

When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle. We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is [latex]\pi [/latex], not [latex]2\pi [/latex]. Further, the domain of tangent is all real numbers with the exception of odd integer multiples of [latex]\frac{\pi }{2}[/latex], unless, of course, a problem places its own restrictions on the domain.

Example 15: Solving a Problem Involving a Single Trigonometric Function

Solve the problem exactly: [latex]2{\sin }^{2}\theta -1=0,0\le \theta <2\pi [/latex].

As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate [latex]\sin \theta [/latex]. Then we will find the angles.

[latex]\begin{gathered}2{\sin }^{2}\theta -1=0 \\ 2{\sin }^{2}\theta =1 \\ {\sin }^{2}\theta =\frac{1}{2} \\ \sqrt{{\sin }^{2}\theta }=\pm \sqrt{\frac{1}{2}} \\ \sin \theta =\pm \frac{1}{\sqrt{2}}=\pm \frac{\sqrt{2}}{2} \\ \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4} \end{gathered}[/latex]

Example 16: Solving a Trigonometric Equation Involving Cosecant

Solve the following equation exactly: [latex]\csc \theta =-2,0\le \theta <4\pi [/latex].

We want all values of [latex]\theta [/latex] for which [latex]\csc \theta =-2[/latex] over the interval [latex]0\le \theta <4\pi [/latex].

[latex]\begin{gathered}\csc \theta =-2 \\ \frac{1}{\sin \theta }=-2 \\ \sin \theta =-\frac{1}{2} \\ \theta =\frac{7\pi }{6},\frac{11\pi }{6},\frac{19\pi }{6},\frac{23\pi }{6} \end{gathered}[/latex]

As [latex]\sin \theta =-\frac{1}{2}[/latex], notice that all four solutions are in the third and fourth quadrants.

Example 17: Solving an Equation Involving Tangent

Solve the equation exactly: [latex]\tan \left(\theta -\frac{\pi }{2}\right)=1,0\le \theta <2\pi [/latex].

Recall that the tangent function has a period of [latex]\pi [/latex]. On the interval [latex]\left[0,\pi \right)[/latex], and at the angle of [latex]\frac{\pi }{4}[/latex], the tangent has a value of 1. However, the angle we want is [latex]\left(\theta -\frac{\pi }{2}\right)[/latex]. Thus, if [latex]\tan \left(\frac{\pi }{4}\right)=1[/latex], then

[latex]\begin{gathered}\theta -\frac{\pi }{2}=\frac{\pi }{4}\\ \theta =\frac{3\pi }{4}\pm k\pi \end{gathered}[/latex]

Over the interval [latex]\left[0,2\pi \right)[/latex], we have two solutions:

[latex]\theta =\frac{3\pi }{4}\text{ and }\theta =\frac{3\pi }{4}+\pi =\frac{7\pi }{4}[/latex]

Find all solutions for [latex]\tan x=\sqrt{3}[/latex].

[latex]\frac{\pi }{3}\pm \pi k[/latex]

Example 18: Identify all Solutions to the Equation Involving Tangent

Identify all exact solutions to the equation [latex]2\left(\tan x+3\right)=5+\tan x,0\le x<2\pi [/latex].

We can solve this equation using only algebra. Isolate the expression [latex]\tan x[/latex] on the left side of the equals sign.

[latex]\begin{gathered} 2\left(\tan x\right)+2\left(3\right) =5+\tan x \\ 2\tan x+6 =5+\tan x \\ 2\tan x-\tan x =5 - 6 \\ \tan x =-1\end{gathered}[/latex]

There are two angles on the unit circle that have a tangent value of [latex]-1:\theta =\frac{3\pi }{4}[/latex] and [latex]\theta =\frac{7\pi }{4}[/latex].

Solving Trigonometric Equations in Quadratic Form

Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as [latex]x[/latex] or [latex]u[/latex]. If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.

Example 19: Solving a Trigonometric Equation in Quadratic Form

Solve the equation exactly: [latex]{\cos }^{2}\theta +3\cos \theta -1=0,0\le \theta <2\pi[/latex].

We begin by using substitution and replacing cos [latex]\theta[/latex] with [latex]x[/latex]. It is not necessary to use substitution, but it may make the problem easier to solve visually. Let [latex]\cos \theta =x[/latex]. We have

[latex]{x}^{2}+3x - 1=0[/latex]

The equation cannot be factored, so we will use the quadratic formula [latex]x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex].

[latex]\begin{align} x&=\frac{-3\pm \sqrt{{\left(-3\right)}^{2}-4\left(1\right)\left(-1\right)}}{2}&=\frac{-3\pm \sqrt{13}}{2} \end{align}[/latex]

Replace [latex]x[/latex] with [latex]\cos \theta[/latex], and solve. Thus,

[latex]\begin{gathered} \cos \theta =\frac{-3\pm \sqrt{13}}{2}\theta ={\cos }^{-1}\left(\frac{-3+\sqrt{13}}{2}\right)\end{gathered}[/latex]

Note that only the + sign is used. This is because we get an error when we solve [latex]\theta ={\cos }^{-1}\left(\frac{-3-\sqrt{13}}{2}\right)[/latex] on a calculator, since the domain of the inverse cosine function is [latex]\left[-1,1\right][/latex]. However, there is a second solution:

[latex]\begin{align}\theta &={\cos }^{-1}\left(\frac{-3+\sqrt{13}}{2}\right) \\ &\approx 1.26 \end{align}[/latex]

This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is

[latex]\begin{align}\theta &=2\pi -{\cos }^{-1}\left(\frac{-3+\sqrt{13}}{2}\right) \\ &\approx 5.02 \end{align}[/latex]

Example 20: Solving a Trigonometric Equation in Quadratic Form by Factoring

Solve the equation exactly: [latex]2{\sin }^{2}\theta -5\sin \theta +3=0,0\le \theta \le 2\pi[/latex].

Using grouping, this quadratic can be factored. Either make the real substitution, [latex]\sin \theta =u[/latex], or imagine it, as we factor:

[latex]\begin{gathered}2{\sin }^{2}\theta -5\sin \theta +3=0 \\ \left(2\sin \theta -3\right)\left(\sin \theta -1\right)=0 \end{gathered}[/latex]

Now set each factor equal to zero.

[latex]\begin{gathered}2\sin \theta -3=0 \\ 2\sin \theta =3 \\ \sin \theta =\frac{3}{2} \\ \text{ } \\ \sin \theta -1=0 \\ \sin \theta =1 \end{gathered}[/latex]

Next solve for [latex]\theta :\sin \theta \ne \frac{3}{2}[/latex], as the range of the sine function is [latex]\left[-1,1\right][/latex]. However, [latex]\sin \theta =1[/latex], giving the solution [latex]\theta =\frac{\pi }{2}[/latex].

Make sure to check all solutions on the given domain as some factors have no solution.

Solve [latex]{\sin }^{2}\theta =2\cos \theta +2,0\le \theta \le 2\pi[/latex]. [Hint: Make a substitution to express the equation only in terms of cosine.]

[latex]\cos \theta =-1,\theta =\pi [/latex]

Example 21: Solving a Trigonometric Equation Using Algebra

Solve exactly:

[latex]2{\sin }^{2}\theta +\sin \theta =0;0\le \theta <2\pi[/latex]

This problem should appear familiar as it is similar to a quadratic. Let [latex]\sin \theta =x[/latex]. The equation becomes [latex]2{x}^{2}+x=0[/latex]. We begin by factoring:

[latex]\begin{gathered}2{x}^{2}+x=0\\ x\left(2x+1\right)=0\end{gathered}[/latex]

Set each factor equal to zero.

[latex]\begin{gathered}x=0 \\ 2x+1=0 \\ x=-\frac{1}{2} \end{gathered}[/latex]

Then, substitute back into the equation the original expression [latex]\sin \theta[/latex] for [latex]x[/latex]. Thus,

[latex]\begin{gathered}\sin \theta =0 \\ \theta =0,\pi \\ \text{ } \\ \sin \theta =-\frac{1}{2} \\ \theta =\frac{7\pi }{6},\frac{11\pi }{6} \end{gathered}[/latex]

The solutions within the domain [latex]0\le \theta <2\pi[/latex] are [latex]\theta =0,\pi ,\frac{7\pi }{6},\frac{11\pi }{6}[/latex].

If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.

[latex]\begin{gathered}2{\sin }^{2}\theta +\sin \theta =0 \\ \sin \theta \left(2\sin \theta +1\right)=0 \\ \sin \theta =0 \\ \theta =0,\pi \\ \text{ } \\ 2\sin \theta +1=0 \\ 2\sin \theta =-1 \\ \sin \theta =-\frac{1}{2} \\ \theta =\frac{7\pi }{6},\frac{11\pi }{6} \end{gathered}[/latex]

We can see the solutions on the graph in Figure 3. On the interval [latex]0\le \theta <2\pi[/latex], the graph crosses the x- axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.

We can verify the solutions on the unit circle in Sum and Difference Identities as well.

Example 22: Solving a Trigonometric Equation Quadratic in Form

Solve the equation quadratic in form exactly: [latex]2{\sin }^{2}\theta -3\sin \theta +1=0,0\le \theta <2\pi[/latex].

We can factor using grouping. Solution values of [latex]\theta[/latex] can be found on the unit circle:

[latex]\begin{gathered}\left(2\sin \theta -1\right)\left(\sin \theta -1\right)=0 \\ 2\sin \theta -1=0 \\ \sin \theta =\frac{1}{2} \\ \theta =\frac{\pi }{6},\frac{5\pi }{6} \\ \text{ } \\ \sin \theta =1 \\ \theta =\frac{\pi }{2} \end{gathered}[/latex]

Solve the quadratic equation [latex]2{\cos }^{2}\theta +\cos \theta =0[/latex].

[latex]\frac{\pi }{2},\frac{2\pi }{3},\frac{4\pi }{3},\frac{3\pi }{2}[/latex]

Key Concepts

• An inverse function is one that “undoes” another function. The domain of an inverse function is the range of the original function and the range of an inverse function is the domain of the original function.
• Because the trigonometric functions are not one-to-one on their natural domains, inverse trigonometric functions are defined for restricted domains.
• For any trigonometric function [latex]f(x)[/latex], if [latex]x=f^{−1}(y)[/latex], then [latex]f(x)=y[/latex]. However, [latex]f(x)=y[/latex] only implies [latex]x=f^{−1}(y)[/latex] if x is in the restricted domain of f .
• Special angles are the outputs of inverse trigonometric functions for special input values; for example, [latex]\frac{\pi}{4}=\tan^{−1}( 1 )\text{ and }\frac{\pi}{6}=\sin^{−1}(\frac{1}{2})[/latex].
• A calculator will return an angle within the restricted domain of the original trigonometric function.
• Inverse functions allow us to find an angle when given two sides of a right triangle.
• In function composition, if the inside function is an inverse trigonometric function, then there are exact expressions; for example, [latex]\sin\left(\cos^{−1}\left(x\right)\right)=\sqrt{1−x^{2}}[/latex].
• If the inside function is a trigonometric function, then the only possible combinations are [latex]\sin^{−1}\left(\cos x\right)=\frac{\pi}{2}−x[/latex] if [latex]0\leq x\leq\pi[/latex] and [latex]\cos^{−1}\left(\sin x\right)=\frac{\pi}{2}−x[/latex] if [latex]−\frac{\pi}{2}\leq x \leq\frac{\pi}{2}[/latex].
• When evaluating the composition of a trigonometric function with an inverse trigonometric function, draw a reference triangle to assist in determining the ratio of sides that represents the output of the trigonometric function.
• When evaluating the composition of a trigonometric function with an inverse trigonometric function, you may use trig identities to assist in determining the ratio of sides.
• When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution.
• Equations involving a single trigonometric function can be solved or verified using the unit circle.
• We can also solve trigonometric equations using a graphing calculator.
• Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc.
• Precalculus. Authored by : OpenStax College. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:1/Preface . License : CC BY: Attribution

Inverse Trigonometric Functions

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• Inverse Trigonometric Functions

Inverse Trigonometric Functions: Problems with Solutions

• Problems on Inverse Trigonometric Function

We will solve different types of problems on inverse trigonometric function.

1.  Find the values of sin (cos\(^{-1}\) 3/5)

Solution:  

Let, cos\(^{-1}\) 3/5 = θ 

Therefore, cos θ = 3/5

Therefore, sin θ = √(1 - cos\(^{2}\)  θ) = √(1 - 9/25) = √(16/25) = 4/5 .

Therefore, sin (cos\(^{-1}\) 3/5) = sin θ = 4/5.

2.  Find the values of tan\(^{-1}\) sin (- π/2)

tan\(^{-1}\) sin (- π/2)

= tan\(^{-1}\) (- sin π/2)

= tan\(^{-1}\) (- 1), [Since - sin π/2 = -1]

= tan\(^{-1}\)(- tan π/4), [Since tan π/4 = 1]

= tan\(^{-1}\) tan (-π/4)

= - π/4. 

Therefore, tan\(^{-1}\) sin (- π/2) = - π/4

3. Evaluate: sin\(^{-1}\) (sin 10)

We know that sin\(^{-1}\) (sin θ) = θ, if - \(\frac{π}{2}\) ≤ θ ≤ \(\frac{π}{2}\).

Here, θ = 10 radians which does not lie between - \(\frac{π}{2}\) and \(\frac{π}{2}\). But 3π - θ i.e., 3π - 10 lies between - \(\frac{π}{2}\) and \(\frac{π}{2}\) and sin (3π - 10) = sin 10.

Now, sin\(^{-1}\) (sin 10)

= sin^-1 (sin (3π - 10)

Therefore, sin\(^{-1}\) (sin 10) = 3π - 10.

4. Find the values of cos (tan\(^{-1}\) ¾)

Let, tan\(^{-1}\) ¾ = θ 

Therefore, tan θ = ¾

We know that sec\(^{2}\) θ - tan\(^{2}\) θ = 1

⇒ sec θ = √(1 + tan\(^{2}\) θ)

⇒ sec θ = √(1 + (3/4)\(^{2}\))

⇒ sec θ = √(1 + 9/16)

⇒ sec θ = √(25/16)

⇒ sec θ = 5/4

Therefore, cos θ = 4/5

⇒ θ = cos\(^{-1}\) 4/5

Now, cos (tan\(^{-1}\) ¾) = cos (cos\(^{-1}\) 4/5) = 4/5

Therefore, cos (tan\(^{-1}\) ¾) = 4/5

5. Find the values of sec csc\(^{-1}\) (2/√3)

sec csc\(^{-1}\) (2/√3)

= sec csc\(^{-1}\) (csc π/3)

= sec (csc\(^{-1}\)csc π/3)

Therefore, sec csc\(^{-1}\) (2/√3) = 2

●   Inverse Trigonometric Functions

• General and Principal Values of sin\(^{-1}\) x
• General and Principal Values of cos\(^{-1}\) x
• General and Principal Values of tan\(^{-1}\) x
• General and Principal Values of csc\(^{-1}\) x
• General and Principal Values of sec\(^{-1}\) x
• General and Principal Values of cot\(^{-1}\) x
• Principal Values of Inverse Trigonometric Functions
• General Values of Inverse Trigonometric Functions
• arcsin(x) + arccos(x) = \(\frac{π}{2}\)
• arctan(x) + arccot(x) = \(\frac{π}{2}\)
• arctan(x) + arctan(y) = arctan(\(\frac{x + y}{1 - xy}\))
• arctan(x) - arctan(y) = arctan(\(\frac{x - y}{1 + xy}\))
• arctan(x) + arctan(y) + arctan(z)= arctan\(\frac{x + y + z – xyz}{1 – xy – yz – zx}\)
• arccot(x) + arccot(y) = arccot(\(\frac{xy - 1}{y + x}\))
• arccot(x) - arccot(y) = arccot(\(\frac{xy + 1}{y - x}\))
• arcsin(x) + arcsin(y) = arcsin(x \(\sqrt{1 - y^{2}}\) + y\(\sqrt{1 - x^{2}}\))
• arcsin (x) - arcsin(y) = arcsin (x \(\sqrt{1 - y^{2}}\) - y\(\sqrt{1 - x^{2}}\))
• arccos (x) + arccos(y) = arccos(xy - \(\sqrt{1 - x^{2}}\)\(\sqrt{1 - y^{2}}\))
• arccos(x) - arccos(y) = arccos(xy + \(\sqrt{1 - x^{2}}\)\(\sqrt{1 - y^{2}}\))
• 2 arcsin(x) = arcsin(2x\(\sqrt{1 - x^{2}}\)) 
• 2 arccos(x) = arccos(2x\(^{2}\) - 1)
• 2 arctan(x) = arctan(\(\frac{2x}{1 - x^{2}}\)) = arcsin(\(\frac{2x}{1 + x^{2}}\)) = arccos(\(\frac{1 - x^{2}}{1 + x^{2}}\))
• 3 arcsin(x) = arcsin(3x - 4x\(^{3}\))
• 3 arccos(x) = arccos(4x\(^{3}\) - 3x)
• 3 arctan(x) = arctan(\(\frac{3x - x^{3}}{1 - 3 x^{2}}\))
• Inverse Trigonometric Function Formula

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Calcworkshop

How to Solve Inverse Trig Functions 19 Awesome Examples!

// Last Updated: January 21, 2020 - Watch Video //

Have you ever put on an article of clothing, only to find out that it wasn’t the right size?

Well, guess what, this same dilemma can be seen in mathematics…. albeit, not with clothing but with Inverse Trigonometric Functions !

Remember how the domain of our basic Trigonometric Functions (i.e., Sine and Cosine) are all real numbers ?

And remember, how these trigonometric functions are periodic, and make a fun “wave” pattern?

Well, what we quickly realize is that these functions don’t pass the Horizontal Line Test, which means they don’t have inverses!

Graph of the Arcsine Function

Gasp! What are we to do?

Well, we’re going to become mathematical tailors and hem our functions just like we hem an article of clothing!

Ok, so we are not going to use needle and thread, but we are going to shorten the domain to make it fit just right!

Together, we are going to see how to restrict the domain of all six of our Trigonometric Functions, which are sometimes referred to as Arcfunctions, and then see how to apply this new domain, or our principle range, as Khan Academy accurately states, to our Unit Circle.

As the Math Page nicely points out, the reason why Inverse Trig Functions are commonly referred to as arcfunctions is because we are looking for the arc (i.e., the angle in radians) whose sine, cosine or tangent is the given value.

In other words, we’re going to do the exact same thing we did when we learned the Unit Circle, just in reverse!

Then, we will learn how to find the exact value of an Inverse Trig Function without using a calculator by using the Unit Circle, Reference Triangles, and our Trigonometric Identities .

Inverse Trig Functions – Video

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Inverse trigonometric function

An Inverse trigonometric function is an inverse function of trigonometric functions, such as sin and cosine.

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• Math Article

Inverse Trigonometric Functions

Inverse trigonometric functions are defined as the inverse functions of the basic trigonometric functions, which are sine, cosine, tangent, cotangent, secant and cosecant functions. They are also termed arcus functions, antitrigonometric functions or cyclometric functions. These inverse functions in trigonometry are used to get the angle with any of the trigonometry ratios . The inverse trigonometry functions have major applications in the field of engineering, physics, geometry and navigation.

Table of Contents

• Derivatives

What Are Inverse Trigonometric Functions?

Inverse trigonometric functions are also called “ arc functions ” since, for a given value of trigonometric functions, they produce the length of arc needed to obtain that particular value. The inverse trigonometric functions perform the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant and cotangent. We know that trigonometric functions are applicable, especially to the right-angle triangle. These six important functions are used to find the angle measure in the right-angle triangle when two sides of the triangle measures are known.

The basic inverse trigonometric formulas are as follows:

Inverse Trigonometric  Functions Graphs

There are particularly six inverse trig functions for each trigonometry ratio . The inverse of six important trigonometric functions are

• Arccotangent
• Arccosecant

Let us discuss all six important types of inverse trigonometric functions, along with its definition, formulas, graphs, properties and solved examples.

Arcsine Function

The arcsine function is an inverse of the sine function denoted by sin -1 x . It is represented in the graph as shown below.

Arccosine Function

The arccosine function is the inverse of the cosine function denoted by cos -1 x . It is represented in the graph as shown below.

Therefore, the inverse of the cos function can be expressed as y = cos -1 x (arccosine x )

Domain and range of arcsine function are as follows:

Arctangent Function

The arctangent function is the inverse of the tangent function denoted by tan -1 x . It is represented in the graph as shown below.

Therefore, the inverse of the tangent function can be expressed as y = tan -1 x (arctangent x )

Domain and range of arctangent are as follows:

Arccotangent (Arccot) Function

The arccotangent function is the inverse of the cotangent function denoted by cot -1 x .

Therefore, the inverse of the cotangent function can be expressed as y = cot -1 x (arccotangent x )

Domain and range of arccotangent are as follows:

Arcsecant Function

What is the arcsecant (arcsec) function? The arcsecant function is the inverse of the secant function denoted by sec -1 x . It is represented in the graph as shown below.

Therefore, the inverse of the secant function can be expressed as y = sec -1 x (arcsecant x )

Domain and range of arcsecant are as follows:

Arccosecant Function

What is the arccosecant (arccsc x ) function? The arccosecant function is the inverse of the cosecant function denoted by cosec -1 x . It is represented in the graph as shown below.

Therefore, the inverse of the cosecant function can be expressed as y = cosec -1 x (arccosecant x )

Domain and range of arccosecant are:

Inverse Trigonometric Functions Table

Let us look at all the inverse trigonometric functions with their notation, definition, domain and range.

Inverse Trigonometric Functions Derivatives

The derivatives of inverse trigonometric functions are first-order derivatives. Let us check out the derivatives of all six inverse functions here.

Learn in detail the derivation of these functions here:  Derivative Inverse Trigonometric Functions

Inverse Trigonometric Functions Properties

The inverse trigonometric functions are also known as arc functions. Inverse trigonometric functions are defined in a certain interval (under restricted domains). Read more on inverse trigonometric properties here .

Trigonometry Basics

Trigonometry basics include the basic trigonometry and trigonometric ratios, such as sin x, cos x, tan x, cosec x, sec x and cot x.

Video Lessons on Inverse Trigonometric Functions

Multiple Angles

Inverse Trigonometric Functions Problems

Example 1: Find the value of x for sin(x) = 2.

Solution:  Given: sin x = 2

x =sin -1 (2), which is not possible.

Hence, there is no value of x for which sin x = 2, so the domain of sin -1 x is -1 to 1 for the values of x.

Example 2: Find the value of sin -1 (sin (π/6)).

sin -1 (sin (π/6) = π/6 (Using identity sin -1 (sin (x) ) = x)

Example 3: Find sin (cos -1 3/5).

Suppose that, cos -1 3/5 = x

So, cos x = 3/5

This implies, sin x = sin (cos -1 3/5) = 4/5

Example 4: 

Practice Problems

Problem 1: Solve: tan(arcsin 12/13)

Problem 2: Find the value of x, cos(arccos 1) = cos x

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How to Solve Inverse Trigonometric Functions?

Inverse trigonometric functions are defined as the inverse functions of the basic trigonometric functions. Learn how to solve inverse trigonometric functions by the following step-by-step guide.

Inverse trigonometric functions are also known as anti-trigonometric functions, arcus functions, or cyclometric functions. Inverse trigonometric functions are the inverse functions of the basic trigonometric functions sine, cosine, tangent, cotangent, secant, and cosecant functions.

Related Topics

• How to Evaluate Trigonometric Function
• How to Solve Angles and Angle Measure
• How to Solve Coterminal Angles and Reference Angles

A step-by-step guide to inverse trigonometric functions

The basic inverse trigonometric formulas are as follows:

• \(\color{blue}{sin^{-1}(-x)=-sin^{-1}x, x∈[-1,1]}\)
• \(\color{blue}{tan^{-1}(-x)=-tan^{-1}x, x∈ R}\)
• \(\color{blue}{cosec^{-1}(-x)=-cosec^{-1}x, x∈ R-(-1,1)}\)
• \(\color{blue}{cos^{-1}(-x)=π-cos^{-1}x, x∈[-1,1]}\)
• \(\color{blue}{sec^{-1}(-x)=π-sec^{-1}x, x∈ R -(-1,1)}\)
• \(\color{blue}{cot^{-1}(-x)=π-cot^{-1}x, x∈ R}\)

Inverse trigonometric function formulas for reciprocal functions

The inverse trigonometric function for reciprocal values x converts the given inverse trigonometric function to its reciprocal function. This follows from the trigonometric functions where sin and cosecant are reciprocal to each other, tangent and cotangent are reciprocal to each other, and cos and secant are reciprocal to each other.

The inverse triangular formula of inverse sine, inverse cosine, and inverse tangent can also be expressed as follows.

• \(\color{blue}{sin^{-1}(x)=cosec^{-1}\frac{1}{x}, x∈ R -(-1,1)}\)
• \(\color{blue}{cos^{-1}(x)=sec^{-1}\frac{1}{x}, x∈ R -(-1,1)}\)
• \(\color{blue}{tan^{-1}(x)=cot^{-1}\frac{1}{x}, x >0}\)
• \(\color{blue}{tan^{-1}(x)=-π+cot^{-1}x, x<0}\)

Inverse trigonometric function formulas for complementary functions

The complementary functions, sine-cosine, tangent-cotangent, and secant-cosecant can be interpreted as:

• \(\color{blue}{sin^{-1}(x)+cos^{-1}x=\frac{\pi}{2}, x ∈ [-1,1]}\)
• \(\color{blue}{tan^{-1}(x)+cot^{-1}x=\frac{\pi}{2}, x ∈ R}\)
• \(\color{blue}{sec^{-1}(x)+cosec^{-1}x=\frac{\pi}{2}, x ∈ R -[-1,1]}\)

Sum and difference of inverse trigonometric function formulas

The sum and difference of two inverse trigonometric functions can be combined to form an inverse function, according to the following set of formulas:

• \(\color{blue}{sin^{-1}x+sin^{-1}y=sin^{-1}(x.\sqrt{\left(1-y^2\right)}+y\sqrt{\left(1-x^2\right)})}\)
• \(\color{blue}{sin^{-1}x-sin^{-1}y=sin^{-1}(x.\sqrt{\left(1-y^2\right)}-y\sqrt{\left(1-x^2\right)})}\)
• \(\color{blue}{cos^{-1}x+cos^{-1}y=cos^{-1}\left(xy-\sqrt{\left(1-x^2\right)}.\sqrt{\left(1-y^2\right)}\right)}\)
• \(\color{blue}{cos^{-1}x-cos^{-1}y=cos^{-1}\left(xy+\sqrt{\left(1-x^2\right)}.\sqrt{\left(1-y^2\right)}\right)}\)
• \(\color{blue}{tan^{-1}x+tan^{-1}y=tan^{-1}\frac{(x+y)}{(1-xy)}, \:if\: xy<1}\)
• \(\color{blue}{tan^{-1}x+tan^{-1}y=tan^{-1}\frac{(x-y)}{(1+xy)}, \:if\: xy> -1}\)

Double of inverse trigonometric function formulas

Twice an inverse trigonometric function can be solved to form a single trigonometric function according to the following set of formulas:

• \(\color{blue}{2sin^{-1}x=sin^{-1}\left(2x.\sqrt{\left(1-x^2\right)}\right)}\)
• \(\color{blue}{2cos^{-1}x=cos^{-1}(2x^2-1)}\)
• \(\color{blue}{2tan^-1x=tan^{-1}(\frac{2x}{1}-x^2)}\)

Inverse Trigonometric Functions – Example 1:

Find the value of \(sin^{-1}(-1)\).

Use this formula: \(\color{blue}{sin^{-1}(-x)=-sin^{-1}x, x∈[-1,1]}\)

\(sin^{-1}(-1)=-sin^{-1}(1)\)

\(-sin^{-1} (1)=\frac{\pi }{2}\)

Exercises for Inverse Trigonometric Functions

Evaluate each of the following..

• \(\color{blue}{tan^{-1}(\sqrt{3})-cot^{-1}(-\sqrt{3})}\)
• \(\color{blue}{sin(cot^{-1}x)}\)
• \(\color{blue}{tan^{-1}(1)+cos^{-1}(-\frac{1}{2})+sin^{-1}(-\frac{1}{2})}\)
• \(\color{blue}{cos^{-1}(cos(-\frac{\pi}{3}))}\)
• \(\color{blue}{sin\left(cos^{-1}\frac{3}{5}\right)}\)
• \(\color{blue}{-\frac{\pi}{2}}\)
• \(\color{blue}{\frac{\sqrt{1+x^2}}{1+x^2}}\)
• \(\color{blue}{\frac{3\pi}{4}}\)
• \(\color{blue}{\frac{\pi}{3}}\)
• \(\color{blue}{\frac{4}{5}}\)

by: Effortless Math Team about 2 years ago (category: Articles )

Effortless Math Team

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Trig Word Problems #2

Now that you understand inverse trig functions, this opens up a whole new set of problems you can solve. In the previous set of problems, you were given one side length and one angle. From this you could determine other information about the triangle. However, in real world situations you won't always know these specific pieces of information; instead you might know two side lengths. Here are some examples: Sample #1 A 10 foot pole casts a 30 foot shadow. What is the angle of inclination of the sun?  

Solution Using the image above, tan -1  (x/y) = X tan -1  (10/30) = 18.43 degrees Sample #2 A man walks in a northeasterly direction for 30 miles, and he ends up 5 miles east of his starting point. In what direction was he walking? Solution Using the image above, cos -1  (y/z) = X cos -1  (5/30) = 80.4 degrees North of East

Sample #3 After walking for awhile, Jack ends up twice as far east of his starting point as he is north of his starting point. In what direction was he walking? Solution His east distance is 2n, and his north distance is n.  tan -1  (n/2n) = tan -1  (1/2) = 26.57 degrees  

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