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Lesson Explainer: Inverse Trigonometric Functions Mathematics • 10th Grade

In this explainer, we will learn how to calculate exact values of trigonometric inverses and evaluate compositions of trigonometric and inverse trigonometric functions at standard angles in radians.

Let us begin by recalling some key definitions and notations related to inverse functions in general. Then, we will get into the specifics of how to evaluate inverse trigonometric functions.

Definition: Invertible Function and Related Concepts

A functionΒ π‘“βˆΆπ‘‹β†’π‘Œ maps an input π‘₯ belonging to the domain 𝑋 to an output 𝑦=𝑓(π‘₯) belonging to the range π‘Œ.

The domain 𝑋 of the function 𝑓 is the set of all possible inputs π‘₯ such that 𝑓(π‘₯) is defined.

The range of 𝑓(𝑋) is the set of all outputs we can get from applying 𝑓 to elements of 𝑋.

A function is invertible if it is a one-to-one and onto function; that is, every input has one unique output and every element of the range π‘Œ can be written in the form 𝑓(π‘₯) for some π‘₯ in the domain 𝑋.

Let π‘“βˆΆπ‘‹β†’π‘Œ be an invertible function. The inverse of 𝑓 is the function π‘“βˆΆπ‘Œβ†’π‘‹οŠ±οŠ§ with the property 𝑓(π‘₯)=π‘¦βŸΊπ‘“(𝑦)=π‘₯.

Simply stated, the inverse of a function β€œreverses” the original function. We note that a function 𝑓 must be invertible in order to have an inverse function π‘“οŠ±οŠ§. If the function 𝑦=𝑓(π‘₯) is invertible, then an input π‘₯ will map to a unique output 𝑦 and π‘“οŠ±οŠ§ will map 𝑦 back to the original π‘₯. This applies to every element π‘₯ in the domain 𝑋 and every element 𝑦 in the range π‘Œ.

As a result, the domain and range of the inverse function are essentially swapped around, compared to the original function. The range of 𝑓 equals the domain of π‘“οŠ±οŠ§ and the range of π‘“οŠ±οŠ§ equals the domain of 𝑓. If the domain of 𝑓 does not equal the range of π‘“οŠ±οŠ§, we may be able to restrict the original domain to ensure they match.

Let us recall the definition of a composite function. Given two functions 𝑔(π‘₯) and 𝑓(π‘₯), we compute the composite function 𝑔(𝑓(π‘₯)) by replacing each instance of π‘₯ in 𝑔(π‘₯) by 𝑓(π‘₯). If two functions are inverses, their compositions will have very predictable results, according to the following rule. As we will demonstrate in this explainer, these results hold true for trigonometric functions and their inverses as well.

Rule: Composition of Inverse Functions

Let π‘“βˆΆπ‘‹β†’π‘Œ and π‘”βˆΆπ‘Œβ†’π‘‹ be inverse functions. Then, applying 𝑓 to any π‘₯ in 𝑋, followed by 𝑔, gives us back π‘₯. Equally, if we apply 𝑔 to any 𝑦 in π‘Œ, followed by 𝑓, we get back 𝑦. That is, 𝑔(𝑓(π‘₯))=𝑔(𝑦)=π‘₯ and 𝑓(𝑔(𝑦))=𝑓(π‘₯)=𝑦.

Now that we have reviewed the foundational ideas about inverse functions, we will go over the special names for inverse trigonometric functions.

Key Terms: Arcsine, Arccosine, and Arctangent

The inverse function of sine is called arcsine.

The inverse function of cosine is called arccosine.

The inverse function of tangent is called arctangent.

We recall that trigonometric functions are not one to one. This means that sine, cosine, and tangent are not invertible unless we limit their domains to ensure each value of π‘˜ returns exactly one value of πœƒ. With appropriate restrictions on πœƒ, we define the three inverse trigonometric functions as follows.

Definition: Domain and Range of Inverse Trigonometric Functions

For βˆ’πœ‹2β‰€πœƒβ‰€πœ‹2 and βˆ’1β‰€π‘˜β‰€1, we have πœƒ=(π‘˜)arcsin if and only if π‘˜=(πœƒ)sin.

For 0β‰€πœƒβ‰€πœ‹ and βˆ’1β‰€π‘˜β‰€1, we have πœƒ=(π‘˜)arccos if and only if π‘˜=(πœƒ)cos.

For βˆ’πœ‹2<πœƒ<πœ‹2 and π‘˜βˆˆβ„, we have πœƒ=(π‘˜)arctan if and only if π‘˜=(πœƒ)tan.

Now, we recall that when an angle πœƒ, in standard position, passes through a coordinate point (π‘₯,𝑦), we can evaluate all six trigonometric functions using π‘₯, 𝑦, and π‘Ÿ=√π‘₯+π‘¦οŠ¨οŠ¨ according to the following definitions.

Definition: Trigonometric Functions in terms of (π‘₯, 𝑦)

sinwherecoswheretanwhereπœƒ=π‘¦π‘Ÿ,π‘Ÿβ‰ 0πœƒ=π‘₯π‘Ÿ,π‘Ÿβ‰ 0πœƒ=𝑦π‘₯,π‘₯β‰ 0

When possible, it is convenient to use coordinate points from the unit circle, where π‘Ÿ=1. The standard angles, in radians, found on the unit circle are multiples of πœ‹6 and πœ‹4 between 0 and 2πœ‹. We can convert between degrees and radians, as needed, using the fact that 180=πœ‹βˆ˜radians.

Now that we have reviewed how to evaluate standard trigonometric functions around the unit circle, we will evaluate our first inverse trigonometric function. This will require thinking β€œin reverse” from a given standard trigonometric value π‘˜ to identify the corresponding angle πœƒ, where the value π‘˜ is found on the unit circle. One of the challenges will be to keep in mind our newly discovered domain and range restrictions for inverse trigonometric functions.

Example 1: Evaluating Inverse Trigonometric Functions in Radians

Evaluate the expression arcsinο€Ώβˆš32.

Answer

We begin by recalling that arcsine is the inverse of the trigonometric function sine. In particular, arcsin(π‘˜) gives us the angle πœƒ in standard position, for which sin(πœƒ)=π‘˜. The range of the function πœƒ=(π‘˜)arcsin is βˆ’πœ‹2β‰€πœƒβ‰€πœ‹2 in radians. This means our angle πœƒ must lie somewhere within the first or fourth quadrant on the unit circle. According to the familiar CAST diagram, we know that only cosine is positive in the fourth quadrant. Since we wish to evaluate the inverse of sine at a positive value, √32, we can be sure that πœƒ is in the first quadrant.

Now, we recall the definitions of trigonometric functions, expressed in terms of π‘₯- and 𝑦-coordinates from the unit circle: sincostanwhereπœƒ=𝑦,πœƒ=π‘₯,πœƒ=𝑦π‘₯,π‘₯β‰ 0.

To evaluate arcsinο€Ώβˆš32, we need to find the standard angle for which the sine of the angle is equal to √32.

Since sinπœƒ=𝑦, we are looking for the angle that intersects the unit circle at a 𝑦-coordinate of √32. We already know it should be in the first quadrant, so we look there. We find the point ο€Ώ12,√32, at πœƒ=πœ‹3radians.

Since sinο€»πœ‹3=√32, with βˆ’πœ‹2β‰€πœ‹3β‰€πœ‹2, we conclude that arcsinradiansο€Ώβˆš32=πœ‹3.

In some questions, we may be asked to evaluate the sum or difference of inverse trigonometric functions. We will look at an example of this type next.

Example 2: Evaluating the Difference of Inverse Trigonometric Functions in Radians

Evaluate the expression arctanarcsin(1)βˆ’(βˆ’1).

Answer

First, we recall that arctangent is the inverse of tangent and arcsine is the inverse of sine. To approach this problem, we will evaluate arctan(1) and arcsin(βˆ’1) separately, then take their difference. We will proceed using radians, but we can always change angle measures to degrees if desired.

We begin by recalling that arctan(π‘˜) gives us the angle πœƒ, in standard position, if and only if tan(πœƒ)=π‘˜. The range of the function πœƒ=(π‘˜)arctan is βˆ’πœ‹2<πœƒ<πœ‹2.

We need to find the standard angle for which the tangent of the angle is equal to 1.

In terms of π‘₯- and 𝑦-coordinates from the unit circle, we define tangent as tanπœƒ=𝑦π‘₯. So, we are looking for the angle that intersects the unit circle at a point (π‘₯,𝑦), where 𝑦π‘₯=1.

By multiplying each side of the above equation by π‘₯, we have (π‘₯)𝑦π‘₯=(π‘₯)1.

Then, we simplify, resulting in the equation 𝑦=π‘₯.

This means we are looking for the point where the π‘₯- and 𝑦-coordinates are equal.

This only happens at two locations on the unit circle, 45∘ and 225∘ as shown below. In radians, these angle measures are πœ‹4 and 5πœ‹4.

Since the range of πœƒ=(π‘˜)arctan is βˆ’πœ‹2<πœƒ<πœ‹2, we conclude that arctan(1) cannot be 5πœ‹4. Thus, arctan(1)=πœ‹4 because this angle is in the proper range and the π‘₯- and 𝑦-coordinates are equal.

Next, we need to evaluate arcsin(βˆ’1). In other words, we need to find the standard angle for which the sine of the angle is equal to βˆ’1.

Since sinπœƒ=𝑦 on the unit circle, we are looking for the angle that intersects the unit circle at a 𝑦-coordinate of βˆ’1. The only place where we have this 𝑦-coordinate on the unit circle is at the quadrantal angle 3πœ‹2.

We see from the unit circle that sinο€Ό3πœ‹2=βˆ’1; however, arcsine has a range of βˆ’πœ‹2β‰€πœƒβ‰€πœ‹2, so we need to find an angle coterminal with 3πœ‹2 in that interval. To find a coterminal angle in radians, we add or subtract multiples of 2πœ‹. Since 3πœ‹2 is above the range of arcsine, we must subtract 2πœ‹. Therefore, writing both terms over a common denominator of 2, we get 3πœ‹2βˆ’2πœ‹=3πœ‹2βˆ’4πœ‹2=βˆ’πœ‹2.

Therefore, arcsin(βˆ’1)=βˆ’πœ‹2.

Finally, we evaluate arctanarcsin(1)βˆ’(βˆ’1) by substituting the values obtained above: arctanarcsin(1)βˆ’(βˆ’1)=πœ‹4βˆ’ο€»βˆ’πœ‹2.

Subtracting a negative is equivalent to adding a positive. Therefore, by using a common denominator of 4, we have πœ‹4βˆ’ο€»βˆ’πœ‹2=πœ‹4+πœ‹2=πœ‹4+2πœ‹4=3πœ‹4.

In conclusion, we have shown that arctanarcsin(1)βˆ’(βˆ’1)=3πœ‹4.

Now let us consider how to evaluate a composition of standard and inverse trigonometric functions.

Example 3: Evaluating a Composite Trigonometric and Inverse Trigonometric Function

Find the exact value of sinarccos((0)) without using a calculator.

Answer

To answer this question, we must first recognize that the given sine function contains arccosine, which is the inverse cosine function. We call this a composite function. Let us recall the meaning of a composition of two functions. Given two functions 𝑔(π‘₯) and 𝑓(π‘₯), we compute the composite function 𝑔(𝑓(π‘₯)) by replacing each instance of π‘₯ in 𝑔(π‘₯) by 𝑓(π‘₯). Another way to think of this is that we substitute for the innermost values first and then work outward.

In this case, we let 𝑔(π‘₯)=(π‘₯)sin and 𝑓(π‘₯)=(π‘₯)arccos; we have to evaluate 𝑔(𝑓(π‘₯)) at π‘₯=0. We begin with evaluating the innermost function, 𝑓(π‘₯), at π‘₯=0. This means we need to find the standard angle for which cosine equals 0.

We recall that, in terms of coordinates from the unit circle, cos(πœƒ)=π‘₯. According to the definition of arccosine, πœƒ=(π‘˜)arccos if and only if π‘˜=(πœƒ)cos, where 0β‰€πœƒβ‰€πœ‹. Within this range on the unit circle, we see that π‘₯=0 at πœƒ=πœ‹2. Therefore, cosο€»πœ‹2=0 and 𝑓(0)=(0)=πœ‹2arccos.

To evaluate 𝑔(𝑓(0)), we substitute in the value of 𝑓(0): 𝑔(𝑓(0))=π‘”ο€»πœ‹2.

Since 𝑔(π‘₯)=(π‘₯)sin, then π‘”ο€»πœ‹2=ο€»πœ‹2.sin

We recall the definition of sine in terms of coordinates from the unit circle: sinπœƒ=𝑦.

Therefore, we use the 𝑦-coordinate of the point (0,1) to evaluate sinο€»πœ‹2: sinο€»πœ‹2=1.

In conclusion, we have shown that sinarccos((0))=1.

Let us try another example involving the composition of functions. We will use arccosine again, but this time, it is the outer function of the composition.

Example 4: Evaluating a Composite Trigonometric and Inverse Trigonometric Function

Find the exact value of arccossinο€»βˆš2ο€»βˆ’πœ‹6 without using a calculator.

Answer

To evaluate this expression, we must first recognize it as a composition of the functions arccosine (the inverse of cosine) and sine. Let us recall the meaning of a composition of two functions. Given two functions 𝑔(π‘₯) and 𝑓(π‘₯), we compute the composite function 𝑔(𝑓(π‘₯)) by replacing each instance of π‘₯ in 𝑔(π‘₯) by 𝑓(π‘₯).

In this case, we let 𝑔(π‘₯)=(π‘₯)arccos and 𝑓(π‘₯)=√2(π‘₯)sin. To evaluate the composite function 𝑔(𝑓(π‘₯)) at π‘₯=βˆ’πœ‹6, we first need to evaluate π‘“ο€»βˆ’πœ‹6=√2ο€»βˆ’πœ‹6sin.

We recall that sine is defined as sinπœƒ=𝑦 on the unit circle. So, we are looking for the 𝑦-coordinate of the point where the angle βˆ’πœ‹6 intersects the unit circle. The unit circle does not display negative angle measures, so we must add 2πœ‹ radians to find the smallest positive angle coterminal with βˆ’πœ‹6. We use a common denominator of 6 to find the coterminal angle: βˆ’πœ‹6+2πœ‹=βˆ’πœ‹6+12πœ‹6=11πœ‹6.

We see that 11πœ‹6 is in the fourth quadrant of the unit circle. At 11πœ‹6, we find the 𝑦-coordinate βˆ’12. Therefore, sinsinο€Ό11πœ‹6=ο€»βˆ’πœ‹6=βˆ’12.

Now, we evaluate 𝑓(π‘₯) at π‘₯=βˆ’πœ‹6. Considering 𝑓(π‘₯)=√2(π‘₯)sin and sinο€»βˆ’πœ‹6=βˆ’12, we have π‘“ο€»βˆ’πœ‹6=√2ο€Όβˆ’12=βˆ’βˆš22.

Therefore, π‘”ο€»π‘“ο€»βˆ’πœ‹6=π‘”ο€Ώβˆ’βˆš22.

Since 𝑔(π‘₯)=(π‘₯)arccos, it follows that π‘”ο€Ώβˆ’βˆš22=ο€Ώβˆ’βˆš22.arccos

Finally, to evaluate arccosο€Ώβˆ’βˆš22, we refer to the unit circle. We recall that arccosine is the inverse of cosine and that arccosο€Ώβˆ’βˆš22 gives us the angle πœƒ in standard position, for which cos(πœƒ)=βˆ’βˆš22. The definition of cosine expressed in terms of coordinates from the unit circle is cosπœƒ=π‘₯. This means we are looking for the standard angle with an π‘₯-coordinate of βˆ’βˆš22.

Let us consider what else we know about the angle we want to find. We recall that the range of the function arccosine is 0β‰€πœƒβ‰€πœ‹. Thus, our angle πœƒ is somewhere within the first or second quadrant on the unit circle. According to the CAST diagram, we know that all trigonometric functions are positive in the first quadrant. We wish to evaluate the inverse of cosine at a negative value, βˆ’βˆš22. So, we can be sure that πœƒ is in the second quadrant.

In the second quadrant, we find the π‘₯-coordinate βˆ’βˆš22 at πœƒ=3πœ‹4. This means that arccosο€Ώβˆ’βˆš22=3πœ‹4.

In conclusion, we have shown that arccossinο€»βˆš2ο€»βˆ’πœ‹6=3πœ‹4.

In our last example, we will explore the relationship between cosine and its inverse for various angle measures around the unit circle.

Example 5: Rewriting an Equation Using a Trigonometric Inverse

Let arccos(π‘₯)=πœƒ, where 0<π‘₯<1 and πœƒ is in radians. Find the first two positive values of 𝛼, in terms of πœƒ, such that cos(𝛼)=π‘₯.

Answer

To begin, we recall the meaning of arccosine. Arccosine is the inverse of cosine. If arccos(π‘₯)=πœƒ, then it must also be true that cos(πœƒ)=π‘₯.

By definition, arccos(π‘₯)=πœƒ has range 0β‰€πœƒβ‰€πœ‹. This indicates πœƒ is in the first or second quadrant. However, we are told that 0<π‘₯<1, and according to the CAST diagram, only the first or fourth quadrant has positive cosine values. Therefore, πœƒ is in the first quadrant; so, the first positive value of 𝛼 is πœƒ.

To find the next positive value of 𝛼, we need to find an angle that intersects the unit circle at the same π‘₯-value as πœƒ. As highlighted in the diagram below, the π‘₯-coordinates in the first quadrant correspond to the π‘₯-coordinates in the fourth quadrant.

To find the next positive angle where cosine has the same value, we subtract πœƒ from 2πœ‹. This means that coscos(2πœ‹βˆ’πœƒ)=(πœƒ).

For example, both cosο€»πœ‹6 and cosο€Ό11πœ‹6 are equal to √32 because coscosο€»2πœ‹βˆ’πœ‹6=ο€Ό11πœ‹6. Similarly, both cosο€»πœ‹3 and cosο€Ό5πœ‹3 are equal to 12 because coscosο€»2πœ‹βˆ’πœ‹3=ο€Ό5πœ‹3.

In conclusion, we have shown that the required first two positive values of 𝛼 are πœƒ and 2πœ‹βˆ’πœƒ.

Let us finish by recapping some important points from the explainer.

Key Points

  • The inverse trigonometric functions arcsine, arccosine, and arctangent are defined in terms of the standard trigonometric functions, as follows:
    • The inverse function of sine is called arcsine.
      • For βˆ’πœ‹2β‰€πœƒβ‰€πœ‹2 and βˆ’1β‰€π‘˜β‰€1, πœƒ=(π‘˜)β‡”π‘˜=(πœƒ)arcsinsin.
    • The inverse function of cosine is called arccosine.
      • For 0β‰€πœƒβ‰€πœ‹ and βˆ’1β‰€π‘˜β‰€1, πœƒ=(π‘˜)β‡”π‘˜=(πœƒ)arccoscos.
    • The inverse function of tangent is called arctangent.
      • For βˆ’πœ‹2<πœƒ<πœ‹2 and π‘˜βˆˆβ„, πœƒ=(π‘˜)β‡”π‘˜=(πœƒ)arctantan.
  • The unit circle is a circle with a radius of 1 whose center lies at the origin of a coordinate plane. The standard angles, in radians, found on the unit circle are multiples of πœ‹6 and πœ‹4 between 0 and 2πœ‹. Each angle intersects the unit circle at a coordinate point (π‘₯,𝑦), which helps us determine trigonometric functions at those angles.

  • Trigonometric functions can be defined in terms of the π‘₯- and 𝑦-coordinates shown on the unit circle, as follows: sincostanwhereπœƒ=𝑦,πœƒ=π‘₯,πœƒ=𝑦π‘₯,π‘₯β‰ 0. This means that we look at the 𝑦-coordinates to evaluate arcsine, the π‘₯-coordinates to evaluate arccosine, and the quotient of corresponding 𝑦- and π‘₯-coordinates to evaluate arctangent.

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