Lesson Explainer: Inverse Trigonometric Functions | Nagwa Lesson Explainer: Inverse Trigonometric Functions | Nagwa

Lesson Explainer: Inverse Trigonometric Functions Mathematics

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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 function 𝑓𝑋𝑌 maps an input 𝑥 belonging to the domain 𝑋 to an output 𝑦=𝑓(𝑥) belonging to the range 𝑌.

The domain 𝑋 of the function 𝑓 is the set of all possible inputs 𝑥 such that 𝑓(𝑥) 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 𝑌 can be written in the form 𝑓(𝑥) for some 𝑥 in the domain 𝑋.

Let 𝑓𝑋𝑌 be an invertible function. The inverse of 𝑓 is the function 𝑓𝑌𝑋 with the property 𝑓(𝑥)=𝑦𝑓(𝑦)=𝑥.

Simply stated, the inverse of a function “reverses” the original function. We note that a function 𝑓 must be invertible in order to have an inverse function 𝑓. If the function 𝑦=𝑓(𝑥) is invertible, then an input 𝑥 will map to a unique output 𝑦 and 𝑓 will map 𝑦 back to the original 𝑥. This applies to every element 𝑥 in the domain 𝑋 and every element 𝑦 in the range 𝑌.

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 𝑓 and the range of 𝑓 equals the domain of 𝑓. If the domain of 𝑓 does not equal the range of 𝑓, 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(𝑥) and 𝑓(𝑥), we compute the composite function 𝑔(𝑓(𝑥)) by replacing each instance of 𝑥 in 𝑔(𝑥) by 𝑓(𝑥). 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 𝑓𝑋𝑌 and 𝑔𝑌𝑋 be inverse functions. Then, applying 𝑓 to any 𝑥 in 𝑋, followed by 𝑔, gives us back 𝑥. Equally, if we apply 𝑔 to any 𝑦 in 𝑌, followed by 𝑓, we get back 𝑦. That is, 𝑔(𝑓(𝑥))=𝑔(𝑦)=𝑥 and 𝑓(𝑔(𝑦))=𝑓(𝑥)=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 𝑘 returns exactly one value of 𝜃. With appropriate restrictions on 𝜃, we define the three inverse trigonometric functions as follows.

Definition: Domain and Range of Inverse Trigonometric Functions

For 𝜋2𝜃𝜋2 and 1𝑘1, we have 𝜃=(𝑘)arcsin if and only if 𝑘=(𝜃)sin.

For 0𝜃𝜋 and 1𝑘1, we have 𝜃=(𝑘)arccos if and only if 𝑘=(𝜃)cos.

For 𝜋2<𝜃<𝜋2 and 𝑘, we have 𝜃=(𝑘)arctan if and only if 𝑘=(𝜃)tan.

Now, we recall that when an angle 𝜃, in standard position, passes through a coordinate point (𝑥,𝑦), we can evaluate all six trigonometric functions using 𝑥, 𝑦, and 𝑟=𝑥+𝑦 according to the following definitions.

Definition: Trigonometric Functions in terms of (𝑥, 𝑦)

sinwherecoswheretanwhere𝜃=𝑦𝑟,𝑟0𝜃=𝑥𝑟,𝑟0𝜃=𝑦𝑥,𝑥0

When possible, it is convenient to use coordinate points from the unit circle, where 𝑟=1. The standard angles, in radians, found on the unit circle are multiples of 𝜋6 and 𝜋4 between 0 and 2𝜋. We can convert between degrees and radians, as needed, using the fact that 180=𝜋radians.

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 “in reverse” from a given standard trigonometric value 𝑘 to identify the corresponding angle 𝜃, where the value 𝑘 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 arcsin32.

Answer

We begin by recalling that arcsine is the inverse of the trigonometric function sine. In particular, arcsin(𝑘) gives us the angle 𝜃 in standard position, for which sin(𝜃)=𝑘. The range of the function 𝜃=(𝑘)arcsin is 𝜋2𝜃𝜋2 in radians. This means our angle 𝜃 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 𝜃 is in the first quadrant.

Now, we recall the definitions of trigonometric functions, expressed in terms of 𝑥- and 𝑦-coordinates from the unit circle: sincostanwhere𝜃=𝑦,𝜃=𝑥,𝜃=𝑦𝑥,𝑥0.

To evaluate arcsin32, we need to find the standard angle for which the sine of the angle is equal to 32.

Since sin𝜃=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 12,32, at 𝜃=𝜋3radians.

Since sin𝜋3=32, with 𝜋2𝜋3𝜋2, we conclude that arcsinradians32=𝜋3.

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)(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(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(𝑘) gives us the angle 𝜃, in standard position, if and only if tan(𝜃)=𝑘. The range of the function 𝜃=(𝑘)arctan is 𝜋2<𝜃<𝜋2.

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

In terms of 𝑥- and 𝑦-coordinates from the unit circle, we define tangent as tan𝜃=𝑦𝑥. So, we are looking for the angle that intersects the unit circle at a point (𝑥,𝑦), where 𝑦𝑥=1.

By multiplying each side of the above equation by 𝑥, we have (𝑥)𝑦𝑥=(𝑥)1.

Then, we simplify, resulting in the equation 𝑦=𝑥.

This means we are looking for the point where the 𝑥- 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 𝜋4 and 5𝜋4.

Since the range of 𝜃=(𝑘)arctan is 𝜋2<𝜃<𝜋2, we conclude that arctan(1) cannot be 5𝜋4. Thus, arctan(1)=𝜋4 because this angle is in the proper range and the 𝑥- and 𝑦-coordinates are equal.

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

Since sin𝜃=𝑦 on the unit circle, we are looking for the angle that intersects the unit circle at a 𝑦-coordinate of 1. The only place where we have this 𝑦-coordinate on the unit circle is at the quadrantal angle 3𝜋2.

We see from the unit circle that sin3𝜋2=1; however, arcsine has a range of 𝜋2𝜃𝜋2, so we need to find an angle coterminal with 3𝜋2 in that interval. To find a coterminal angle in radians, we add or subtract multiples of 2𝜋. Since 3𝜋2 is above the range of arcsine, we must subtract 2𝜋. Therefore, writing both terms over a common denominator of 2, we get 3𝜋22𝜋=3𝜋24𝜋2=𝜋2.

Therefore, arcsin(1)=𝜋2.

Finally, we evaluate arctanarcsin(1)(1) by substituting the values obtained above: arctanarcsin(1)(1)=𝜋4𝜋2.

Subtracting a negative is equivalent to adding a positive. Therefore, by using a common denominator of 4, we have 𝜋4𝜋2=𝜋4+𝜋2=𝜋4+2𝜋4=3𝜋4.

In conclusion, we have shown that arctanarcsin(1)(1)=3𝜋4.

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(𝑥) and 𝑓(𝑥), we compute the composite function 𝑔(𝑓(𝑥)) by replacing each instance of 𝑥 in 𝑔(𝑥) by 𝑓(𝑥). 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(𝑥)=(𝑥)sin and 𝑓(𝑥)=(𝑥)arccos; we have to evaluate 𝑔(𝑓(𝑥)) at 𝑥=0. We begin with evaluating the innermost function, 𝑓(𝑥), at 𝑥=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(𝜃)=𝑥. According to the definition of arccosine, 𝜃=(𝑘)arccos if and only if 𝑘=(𝜃)cos, where 0𝜃𝜋. Within this range on the unit circle, we see that 𝑥=0 at 𝜃=𝜋2. Therefore, cos𝜋2=0 and 𝑓(0)=(0)=𝜋2arccos.

To evaluate 𝑔(𝑓(0)), we substitute in the value of 𝑓(0): 𝑔(𝑓(0))=𝑔𝜋2.

Since 𝑔(𝑥)=(𝑥)sin, then 𝑔𝜋2=𝜋2.sin

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

Therefore, we use the 𝑦-coordinate of the point (0,1) to evaluate sin𝜋2: sin𝜋2=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 arccossin2𝜋6 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(𝑥) and 𝑓(𝑥), we compute the composite function 𝑔(𝑓(𝑥)) by replacing each instance of 𝑥 in 𝑔(𝑥) by 𝑓(𝑥).

In this case, we let 𝑔(𝑥)=(𝑥)arccos and 𝑓(𝑥)=2(𝑥)sin. To evaluate the composite function 𝑔(𝑓(𝑥)) at 𝑥=𝜋6, we first need to evaluate 𝑓𝜋6=2𝜋6sin.

We recall that sine is defined as sin𝜃=𝑦 on the unit circle. So, we are looking for the 𝑦-coordinate of the point where the angle 𝜋6 intersects the unit circle. The unit circle does not display negative angle measures, so we must add 2𝜋 radians to find the smallest positive angle coterminal with 𝜋6. We use a common denominator of 6 to find the coterminal angle: 𝜋6+2𝜋=𝜋6+12𝜋6=11𝜋6.

We see that 11𝜋6 is in the fourth quadrant of the unit circle. At 11𝜋6, we find the 𝑦-coordinate 12. Therefore, sinsin11𝜋6=𝜋6=12.

Now, we evaluate 𝑓(𝑥) at 𝑥=𝜋6. Considering 𝑓(𝑥)=2(𝑥)sin and sin𝜋6=12, we have 𝑓𝜋6=212=22.

Therefore, 𝑔𝑓𝜋6=𝑔22.

Since 𝑔(𝑥)=(𝑥)arccos, it follows that 𝑔22=22.arccos

Finally, to evaluate arccos22, we refer to the unit circle. We recall that arccosine is the inverse of cosine and that arccos22 gives us the angle 𝜃 in standard position, for which cos(𝜃)=22. The definition of cosine expressed in terms of coordinates from the unit circle is cos𝜃=𝑥. This means we are looking for the standard angle with an 𝑥-coordinate of 22.

Let us consider what else we know about the angle we want to find. We recall that the range of the function arccosine is 0𝜃𝜋. Thus, our angle 𝜃 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, 22. So, we can be sure that 𝜃 is in the second quadrant.

In the second quadrant, we find the 𝑥-coordinate 22 at 𝜃=3𝜋4. This means that arccos22=3𝜋4.

In conclusion, we have shown that arccossin2𝜋6=3𝜋4.

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(𝑥)=𝜃, where 0<𝑥<1 and 𝜃 is in radians. Find the first two positive values of 𝛼, in terms of 𝜃, such that cos(𝛼)=𝑥.

Answer

To begin, we recall the meaning of arccosine. Arccosine is the inverse of cosine. If arccos(𝑥)=𝜃, then it must also be true that cos(𝜃)=𝑥.

By definition, arccos(𝑥)=𝜃 has range 0𝜃𝜋. This indicates 𝜃 is in the first or second quadrant. However, we are told that 0<𝑥<1, and according to the CAST diagram, only the first or fourth quadrant has positive cosine values. Therefore, 𝜃 is in the first quadrant; so, the first positive value of 𝛼 is 𝜃.

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

To find the next positive angle where cosine has the same value, we subtract 𝜃 from 2𝜋. This means that coscos(2𝜋𝜃)=(𝜃).

For example, both cos𝜋6 and cos11𝜋6 are equal to 32 because coscos2𝜋𝜋6=11𝜋6. Similarly, both cos𝜋3 and cos5𝜋3 are equal to 12 because coscos2𝜋𝜋3=5𝜋3.

In conclusion, we have shown that the required first two positive values of 𝛼 are 𝜃 and 2𝜋𝜃.

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 𝜋2𝜃𝜋2 and 1𝑘1, 𝜃=(𝑘)𝑘=(𝜃)arcsinsin.
    • The inverse function of cosine is called arccosine.
      • For 0𝜃𝜋 and 1𝑘1, 𝜃=(𝑘)𝑘=(𝜃)arccoscos.
    • The inverse function of tangent is called arctangent.
      • For 𝜋2<𝜃<𝜋2 and 𝑘, 𝜃=(𝑘)𝑘=(𝜃)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 𝜋6 and 𝜋4 between 0 and 2𝜋. Each angle intersects the unit circle at a coordinate point (𝑥,𝑦), which helps us determine trigonometric functions at those angles.

  • Trigonometric functions can be defined in terms of the 𝑥- and 𝑦-coordinates shown on the unit circle, as follows: sincostanwhere𝜃=𝑦,𝜃=𝑥,𝜃=𝑦𝑥,𝑥0. This means that we look at the 𝑦-coordinates to evaluate arcsine, the 𝑥-coordinates to evaluate arccosine, and the quotient of corresponding 𝑦- and 𝑥-coordinates to evaluate arctangent.

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