Lesson Video: Inverse Trigonometric Functions Mathematics • 10th Grade

In this video, 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.

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Video Transcript

In this video, 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. Letβs begin by recalling some key definitions and notations related to inverse functions in general. Then, we will get into specifics of how to evaluate inverse trigonometric functions.

A function π maps an input π₯ belonging to the set π, called the domain, to an output π¦ equal to π of π₯ belonging to the set π, called the range. The domain π of a function π is the set of all possible inputs π₯ such that π of π₯ is defined. The range π of a function π is the set of all outputs we can get from applying π to the elements of π.

A function is invertible if it is a one-to-one and onto function. One-to-one or injective means that every input has one unique output, and onto or surjective means every element of the range can be written in the form π of π₯ for some π₯ in the domain. If we let π, mapping elements from π to π, be an invertible function, then the inverse of π is the function π inverse, which maps elements from π to π, with the property π of π₯ equals π¦ if and only if π inverse of π¦ equals π₯.

Simply stated, the inverse of a function reverses the original function. If π is invertible, then π maps an input π₯ to a unique output π¦ and π inverse maps π¦ back to the original π₯. As a result, the domain and range of the inverse function are essentially swapped around, compared to the original function. If the domain of π does not equal the range of π inverse, we may be able to restrict the original domain to ensure they do match. Letβs now recall the definition of a composite function.

Given two functions π of π₯ and π of π₯, we compute the composite function π of π of π₯ by replacing each instance of π₯ in π of π₯ by π of π₯. For example, if π of π₯ equals two π₯ plus one and π of π₯ equals π₯ squared, then the composite function π of π of π₯ equals two π₯ squared plus one. If the functions are inverses, their composition will have a very predictable result, according to the following rule. Let π and π be inverse functions. Then, applying π to any element π₯ in the set π followed by π gives back the original element π₯. Equally, if we apply π to any π¦ in set π followed by π, we get back π¦.

Now that we have reviewed the foundational ideas about inverse functions, we will go over the special names for inverse trigonometric functions. First, we have arcsine, which is the inverse of sine, arccosine is the inverse of cosine, and arctangent is the inverse of tangent. 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 be sure each value of π returns exactly one value of π.

With appropriate restrictions on π, we define the three inverse trigonometric functions as follows. For sine and tangent, we restrict the domain to the first and fourth quadrants of the unit circle. This is the interval from negative π over two to positive π over two radians. For cosine, we restrict the domain to the first and second quadrants. This is the interval from zero to π radians. Now, we recall that the range of sine and cosine are already restricted to values greater than or equal to negative one and less than or equal to positive one. And the range of tangent is the set of all real numbers. Next, we will review the trigonometric function definitions in terms of coordinate points.

We recall that when the angle π in standard position passes through a coordinate point π₯, π¦, we can evaluate all six trigonometric functions using π₯, π¦, and π equal to the square root of π₯ squared plus π¦ squared according to the following definitions. sin of π equals π¦ over π, where π is nonzero. cos of π equals π₯ over π, where π is also nonzero. And tan of π equals π¦ over π₯, where π₯ is nonzero. When possible, it is convenient to use coordinate points from the unit circle, where π equals one.

Standard angles in radians found on the unit circle are multiples of π over six and π over four between zero and two π radians. The first diagram of the unit circle shows increments of π over six. The second diagram shows increments of π over four. We have included the degree conversions on the diagrams. But we can convert between degrees and radians as needed using the fact that 180 degrees equals π radians.

With the unit circle, we can use simplified coordinate definitions for sine and cosine. To evaluate sin of π, we would look at the π¦-coordinate of the point where π intersects the unit circle. To evaluate cos of π, we look at the corresponding π₯-coordinate. Now that we have reviewed how to evaluate standard trigonometric functions around the unit circle, we will evaluate our first inverse trigonometric function.

Evaluate the expression arcsin of the square root of three over two.

We begin by recalling that arcsine is the inverse of the trigonometric function sine. In particular, arcsin of π gives us the angle π in standard position, for which sin of π equals π, where the range of arcsin is π greater than or equal to negative π over two and less than or equal to positive π over two radians. This means our angle π 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, we can be sure that π will be found in the first quadrant.

Now, recall the definitions of trigonometric functions expressed in terms of π₯- and π¦-coordinates from the unit circle as follows. In particular, we will be using the definition of sine evaluated at π, which is π¦. To evaluate arcsin of the square root of three over two, we need to find the standard angle in the first quadrant for which the sine of the angle is equal to the square root of three over two. Since sin of π equals π¦, we are looking for the angle that intersects the unit circle at a π¦-coordinate of the square root of three over two. So, we look at the first quadrant of the unit circle, which is shown here in increments of π over six and π over four.

We spot the point with a π¦-value of the square root of three over two. The angle that intersects this point on the unit circle is π over three radians or 60 degrees. Since sin of π over three equals the square root of three over two and π over three is in the proper range for arcsine, we conclude that arcsin of the square root of three over two is equal to π over three radians.

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.

Evaluate the expression arctan of one minus arcsin of negative one.

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 of one and arcsin of negative one separately, then take their difference. We begin by recalling that arctan of π gives us the angle π in standard position if and only if tan of π equals π. The range of arctan is π greater than or equal to negative π over two and less than or equal to positive π over two radians. We need to find the standard angle for which tangent of the angle is equal to one. In terms of π₯- and π¦-coordinates from the unit circle, we define tangent as π¦ over π₯.

So, we are looking for the angle that intersects the unit circle at a point where π¦ divided by π₯ equals one. By multiplying each side of this equation by π₯, we have π¦ equals π₯. This means we are looking for the point where the π₯- and π¦-coordinates are equal. This only happens at two locations on the unit circle, π over four and five π over four radians. Since five π over four is not in the range of arctangent, we conclude that arctan of one cannot be five π over four. However, π over four is in the proper range, and the π₯- and π¦-coordinates are equal. So, we have found that tan of π over four equals one, which means that arctan of one equals π over four.

Next, we need to evaluate arcsin of negative one. We recall that arcsin of π gives us the angle π in standard position if and only if sin of π equals π. And we recall that the range of arcsine is the same as the range of arctangent. Since sin of π equals π¦ on the unit circle, we are looking for the angle that intersects the unit circle at a π¦-coordinate of negative one. There is only one place where we have this π¦-coordinate on the unit circle. And that is at the quadrantal angle three π over two. We see from the unit circle that sin of three π over two equals negative one. So, we would expect that arcsin of negative one would equal three π over two.

Unfortunately, three π over two is not in the proper range for the arcsine function. We have the correct location on the unit circle, but we need to find a coterminal angle with three π over two in the given interval. To find a coterminal angle in radians, we add or subtract multiples of two π. Since three π over two is above the range of arcsine, we must subtract two π. Therefore, after writing both terms over a common denominator of two and finding their difference, we get negative π over two. Therefore, arcsin of negative one equals negative π over two because negative π over two is in the proper range and sin of negative π over two equals negative one.

Finally, we evaluate arctan of one minus arcsin of negative one by substituting the values obtained. Subtracting a negative is equivalent to adding a positive. Therefore, by using a common denominator of four, we have three π over four.

In conclusion, we have shown that arctan of one minus arcsin of negative one is equal to three π over four radians.

In our last example, we will consider how to evaluate a composition of standard and inverse trigonometric functions.

Find the exact value of arccos of the square root of two times sin of negative π over six without using a calculator.

To evaluate this expression, we must first recognize it as a composition of functions arccosine, the inverse of cosine, and sine. Letβs recall the meaning of a composition of two functions given two functions π of π₯ and π of π₯. The composition of π of π of π₯ is computed by replacing each instance of π₯ in π of π₯ with π of π₯. In this case, we let π of π₯ be arccos of π₯ and π of π₯ be the square root of two times sin of π₯. To evaluate π of π of π₯ at π₯ equals negative π over six, we first evaluate π of negative π over six. We recall that sine is defined as π¦ on the unit circle. So, we are looking for the π¦-coordinate of the point where the angle negative π over six intersects the unit circle.

The unit circle does not display negative angle measures, so we must add two π radians to find the smallest positive angle coterminal with negative π over six. We can use a common denominator of six to find the coterminal angle. So, we have negative π over six plus 12π over six, which equals 11π over six. We see that 11π over six is in the fourth quadrant of the unit circle. At 11π over six, we find the π¦-coordinate negative one-half. Therefore, we have shown that sin of negative π over six equals sin of 11π over six, so they both equal negative one-half.

Now, we can evaluate π of π₯ at π₯ equals negative π over six. Using substitution, we find that π of negative π over six equals the square root of two times negative one-half. So, π of negative π over six equals the negative square root of two over two.

Letβs clear some space for our next step. Thus far, we have π of negative π over six equals the negative square root of two over two. Now, to evaluate π of π of negative π over six, we must evaluate π of negative square root of two over two. Since π of π₯ equals arccos of π₯, it follows that π of negative square root of two over two is equal to arccos of negative square root of two over two. Finally, to evaluate arccos of negative square root of two over two, we once again refer to the unit circle. We recall that arccosine is the inverse of cosine and that arccos of π gives us the angle π in standard position, for which cos of π equals π. In this case, π is the square root of two over two.

The definition of cosine expressed in terms of coordinates from the unit circle is π₯. This means we are looking for the standard angle with an π₯-coordinate of negative square root of two over two. Letβs consider what else we know about the angle we want to find. We recall that the range of the function arccosine is limited to the first or second quadrants, specifically from zero to π radians. So we can ignore anything below the π₯-axis.

According to the CAST diagram, we know all trigonometric functions are positive in the first quadrant. We wish to evaluate the inverse cosine at a negative value, so we can be sure that π is in the second quadrant, where only sine is positive. In the second quadrant, we find the π₯-coordinate negative square root of two over two at π equals three π over four. This means that arccos of the negative square root of two over two equals three π over four.

In conclusion, we have shown that arccos of the square root of two times sin of negative π over six is equal to three π over four radians.

Letβs finish by recapping some important points from this video. The three inverse trigonometric functions arcsine, arccosine, and arctangent are defined in terms of the standard trigonometric functions. The inverse function of sine is called arcsine, the inverse function of cosine is called arccosine, and the inverse function of tangent is called arctangent.

Trigonometric functions are only invertible when we restrict their domain like so. Then, they can be defined as follows in terms of π and π. The unit circle is a circle with a radius of one whose center lies at the origin of a coordinate plane. The standard angles in radians found on the unit circle are multiples of π over six and π over four between zero and two π radians. Each angle intersects the unit circle at a coordinate point π₯, π¦, which helps us determine the trigonometric functions at those angles.

These two diagrams display the unit circle divided into increments of π over six and π over four. They are a very helpful reference whenever evaluating standard or inverse trigonometric functions. Trigonometric functions can be defined in terms of the π₯- and π¦-coordinates shown on the unit circle as follows. To evaluate the sin of π, we look at the π¦-coordinate. To evaluate the cos of π, we look at the π₯-coordinate. To evaluate the tan of π, we take the quotient of the corresponding π¦- and π₯-coordinates. We note that the tangent function is undefined when the π₯-coordinate equals zero.