# 1.4: Inverse Functions and Trigonometric Equations

**At Grade**Created by: CK-12

## Inverse Trigonometric Functions

**Inverse Functions**

The definition of “one-to-one” bears reviewing here; a function is one-to-one if, in addition to having at most one \begin{align*}y-\end{align*}

Students may get confused about the “inverse reflection principle.” It doesn’t mean that the graph of \begin{align*}f(x)\end{align*}

The inverse reflection principle, in turn, should make it clear why a function has to pass the horizontal line test in order for its inverse to also be a function. When the graph of a relation is flipped about that diagonal line to create the graph of the inverse relation, any horizontal lines drawn through the graph of the original relation would become vertical lines drawn through the graph of the inverse relation—so if the original relation would not pass the horizontal line test, it follows that the inverse relation would not pass the vertical line test, and so would not be a function.

When students work through the examples in the text, they should make sure their calculators are in degree mode. With a calculator, it is much easier to find inverse trig functions in degrees than in radians, because a calculator in radian mode will generally only give decimal approximations for the answers instead of telling you precisely what the answers are in multiples of \begin{align*}\pi\end{align*}

**Points to Consider**

The inverse relations of the trig functions are one-to-one, but they are not functions. Under the restricted domains that will be discussed later, they are one-to-one functions.

**Using the “Inverse” Notation**

The graphs of \begin{align*}\sin(x)\end{align*}

**Points to Consider**

We can determine exact values for inverse trig functions when those functions correspond with the various special angles on the unit circle.

## Exact Values of Inverse Functions

*Ranges of Inverse Circular Functions*

**Domain and Range of the Circular Functions and their Inverses**

Remind students if necessary that the notation \begin{align*}|q|\end{align*}

*Exact Values of Special Inverse Circular Functions*

**Introduction**

A much easier way to do the problems in example 1 is to think of \begin{align*}\cot, \csc,\end{align*}

**Points to Consider**

The inverse composition rule has not previously been discussed in this book, although students may have encountered it in other courses. It states that if the functions \begin{align*}f(x)\end{align*}

For example, the inverse of \begin{align*}f(x) = 3x\end{align*}

So, does the rule apply to trig functions? Yes, if we use the appropriate domain restrictions. For example, the tangent of \begin{align*}45^\circ\end{align*} is \begin{align*}1\end{align*}, and the inverse tangent of \begin{align*}1\end{align*} is \begin{align*}45^\circ\end{align*}, so \begin{align*}\tan (\tan^{-1}(1)) = \tan (45^\circ) = 1\end{align*}, and \begin{align*}\tan^{-1}(\tan (45^\circ)) = \tan^{1} (1) = 45^\circ\end{align*}. But the rule doesn’t apply if we pick an angle measure outside of our restricted domain. The tangent of \begin{align*}225^\circ\end{align*} is also \begin{align*}1\end{align*}, but the inverse tangent of \begin{align*}1\end{align*} is still \begin{align*}45^\circ\end{align*}. (Of the many angles whose tangent is \begin{align*}1\end{align*}, \begin{align*}45^\circ\end{align*} is the only “official” inverse tangent of \begin{align*}1\end{align*}. Similarly, although \begin{align*}2\end{align*} and \begin{align*}-2\end{align*}, when squared, each yield \begin{align*}4\end{align*}, \begin{align*}2\end{align*} is the only “official” square root of \begin{align*}4\end{align*}.) So \begin{align*}\tan^{-1}(\tan(225^\circ))\end{align*} would equal \begin{align*}45^\circ\end{align*} instead of \begin{align*}225^\circ.\end{align*}

(This rule will be explored in more detail in the next section.)

## Properties of Inverse Circular Functions

*Derive Properties of Other Five Inverse Circular Functions in terms of Arctan*

**Composing Trigonometric Functions with Arctan**

The notation in this section may be a bit confusing; students may wonder why we would want to find the values of expressions like \begin{align*}\sin (\tan^{-1}(x))\end{align*} and \begin{align*}\csc (\tan^{-1}(x))\end{align*}, or may be unclear on what these expressions really mean. Basically, all we are doing here is the same thing we did in chapter 1, when we were given, say, the sine of an angle and had to find the cosine or the tangent of that same angle without knowing the measure of the angle itself. Based on our knowledge of right triangles and the Pythagorean Theorem, we know that if the tangent of an angle is \begin{align*}x\end{align*}, we can find all the other trig ratios in terms of \begin{align*}x\end{align*} just by drawing an appropriate triangle. If one leg of the triangle measures \begin{align*}x\;\mathrm{units}\end{align*} and the other measures \begin{align*}1\;\mathrm{unit}\end{align*} (making the tangent \begin{align*}\frac{x}{1}\end{align*}, or just \begin{align*}x\end{align*}), then we can find the length of the hypotenuse, and once we know the lengths of all three sides, all the trig functions are simply ratios of certain pairs of sides.

So, for example, part b of Example 1 simply asks “What is the sine of an angle whose tangent is \begin{align*}1\end{align*}?” and the answer can be found by drawing a triangle whose legs both measure \begin{align*}1\end{align*}, finding the length of the hypotenuse, and then finding the sine of the acute angle.

**Points to Consider**

It is indeed possible to graph these composite expressions, as they are simply algebraic functions. Analyzing them will show that their domains are unlimited but their ranges are limited, because the expression \begin{align*}(x^2 + 1)\end{align*} can only take on values greater than or equal to \begin{align*}1\end{align*}.

*Derive Inverse Cofunction Properties*

**Cofunction Identities**

Technically, the graph of \begin{align*}y = \sin \theta\end{align*} is not really the graph of \begin{align*}y = \cos \theta\end{align*} shifted \begin{align*}\frac{\pi}{2}\;\mathrm{units}\end{align*} to the right, but rather the graph of \begin{align*}y = \cos \theta\end{align*} flipped upside down and shifted \begin{align*}\frac{\pi}{2}\;\mathrm{units}\end{align*} to the left. Since the resulting graph is exactly the same, though, you needn’t stress this technicality, but it’s useful to keep in mind in case some sharp student spots it. (You might, in that case, challenge them to work out why those two different operations are equivalent.)

*Find Exact Values of Functions of Inverse Functions using Pythagorean Triples.*

The phrase “functions of inverse functions” may be confusing. In this case it simply means “trig functions applied to inverse trig functions,” or in other words “inverse trig functions plugged into trig functions”—for example, \begin{align*}\cos(\sin^{-1}(x))\end{align*} or \begin{align*}\tan(\cos^{-1}(x))\end{align*}—which is just the sort of thing we were working with earlier in the lesson.

## Applications of Inverse Circular Functions

*Revisiting \begin{align*}y = c + a cos b(x - d)\end{align*}*

**Transformations of \begin{align*}y = cos x\end{align*}**

You may want to walk through examples 1 and 2 in more detail, as these concepts haven’t been covered since chapter 2.

For example 1, first we need to pull out the amplitude, frequency, phase shift, and vertical shift from the equation: they are \begin{align*}3, 2, \frac{-\pi}{4}\end{align*}, and \begin{align*}5\end{align*} respectively. Then we need to think about how all of these values affect the basic cosine graph: it will be stretched vertically by a factor of \begin{align*}3\end{align*}, compressed horizontally by a factor of \begin{align*}2\end{align*}, shifted \begin{align*}\frac{\pi}{2}\;\mathrm{units}\end{align*} to the left, and shifted \begin{align*}5\;\mathrm{units}\end{align*} up. Only after determining all this can we actually draw the graph.

For example 2, we perform almost the same procedure in reverse. First we inspect the graph to discover the amplitude, frequency, phase shift, and vertical shift: they are \begin{align*}5, 4\end{align*} (because the period is \begin{align*}90^\circ\end{align*}), \begin{align*}-20^\circ,\end{align*} and \begin{align*}-3\end{align*}, respectively. Then we recall where those numbers fit into the equation, and then we can write the equation. (The equation given in the text should say \begin{align*}20^\circ\end{align*} where it says \begin{align*}70^\circ\end{align*}.)

**Points to Consider**

Given an equation for \begin{align*}y\end{align*} in terms of \begin{align*}x\end{align*}, it is generally possible to solve for \begin{align*}x\end{align*} in terms of \begin{align*}y\end{align*} (that is, to find the inverse of the equation); the result simply may not be a function. In this case, since we know we can find the inverse of the plain old cosine function, it seems reasonable that we can still find an inverse of the cosine function when it is stretched and shifted as it is here. In the next part of this lesson, we will find out how.

*Solving for Particular Values in Trigonometric Equations*

**Points to Consider**

Degrees and radians are simply two different ways of expressing the same angle measures, so anything that can be done when working in degrees can also be done when working in radians.

*Applications, Technological Tools*

**Examples**

In Example 1, \begin{align*}4.34\;\mathrm{seconds}\end{align*} is one time when the dolphin is at a height of \begin{align*}4\;\mathrm{feet}\end{align*}, but it isn’t the first time or even the only time. Try having students also determine the first time the dolphin reaches that height, and two other times when it does so again.

(They may think that since the graph has a period of \begin{align*}3\;\mathrm{seconds}\end{align*}, the dolphin reaches a height of \begin{align*}4\;\mathrm{feet}\end{align*} every \begin{align*}3\;\mathrm{seconds}\end{align*}. This is only partly true; the dolphin does reach that height at \begin{align*}4.34\;\mathrm{seconds}\end{align*} and every \begin{align*}3\;\mathrm{seconds}\end{align*} thereafter, but it also reaches that height at \begin{align*}3.66\;\mathrm{seconds}\end{align*} and every \begin{align*}3\;\mathrm{seconds}\end{align*} thereafter. Any height on the graph that is not a maximum or minimum will be reached not once, but twice per period: once on the way up and once on the way down. Tracing the graph should confirm this.)

## Trigonometric Equations

*Solving Trigonometric Equations Analytically*

**Introduction**

Before beginning this lesson, you will probably need to review all of the trig identities that were introduced in chapter 3: the Pythagorean identities, sum and difference identities, double- and half-angle identities, and sum and product identities. (If you haven’t covered that chapter, you’ll need to introduce those identities some other way, as they will be needed to solve the problems in this lesson—except for the double- and half-angle identities, which are covered in the next lesson.)

Students may try applying trigonometric identities to example 2. That isn’t a useful technique, though, because the only trigonometric expression in the equation is already as simple as it can get. All that remains is to perform basic algebra to reduce the left-hand side of the equation to \begin{align*}\tan x\end{align*}, and then to take the inverse tangent of both sides to get \begin{align*}x\end{align*} by itself. Except for the inverse tangent step, this problem has more to do with algebra than with trigonometry (proving that algebra skills are essential even as we study more advanced math!).

(Without a calculator, students may also forget the inverse tangent of \begin{align*}\sqrt{3}\end{align*}. Remind them if necessary to draw an appropriate right triangle to figure it out.)

**Points to Consider**

Any equation-solving method that works for non-trigonometric equations should also work for trigonometric ones, if applied appropriately. And it is certainly possible that a quadratic equation with trigonometric expressions in it might turn up—for example, \begin{align*}\sin^2 x + \sin x + 2 = 0\end{align*}. We would first solve this equation for \begin{align*}\sin x\end{align*}, using any standard technique for solving quadratics, and then we would take the arcsine of both sides to find \begin{align*}x\end{align*}.

*Solve Trig Equations (Factoring)*

**Introduction**

“Principal values,” in case students are confused by this, is simply another term for the values that are in the limited domain of a trig function, or in other words are in the range of the inverse function. In example 1, when \begin{align*}\sin x = \frac{1}{2}\end{align*}, the principal value of \begin{align*}x\end{align*} is \begin{align*}\frac{\pi}{6}\end{align*} or \begin{align*}30^\circ\end{align*}, because that is the one \begin{align*}x-\end{align*}value between \begin{align*}\frac{-\pi}{2}\end{align*} and \begin{align*}\frac{\pi}{2}\end{align*} whose sine is \begin{align*}\frac{1}{2}\end{align*}.

**Points to Consider**

The quadratic formula will work on trig equations; we just need to remember that we are solving for \begin{align*}\sin x\end{align*} rather than for \begin{align*}x\end{align*}, and still need to take the arcsine afterward.

## Trigonometric Equations with Multiple Angles

*Solve equations (with double angles)*

**Double Angle Identity for the Sine Function**

Example 1, in addition to the double angle identity, also employs algebraic techniques that students may not have encountered for a while, such as the zero product property. It might be necessary to go through the last few steps slowly.

**Double Angle Identity for the Cosine Function**

There is a typo in the first formula derived here: the final line of the derivation should have a minus sign in place of the second equals sign. Also, there should be a line break after the line in the next paragraph that reads \begin{align*}\cos 2a = 2 \cos^2 - 1\end{align*}; the next three lines are a separate derivation of a different formula. (All three formulas are summarized immediately below.)

To solve example 2, we must note that \begin{align*}4 \theta\end{align*} is twice \begin{align*}2\theta\end{align*}, so the double-angle formula applies if we plug in \begin{align*}2 \theta\end{align*} in place of the usual \begin{align*}\theta\end{align*}.

**Double Angle Identity for the Tangent Function**

You may want to point out that the solution to example 4 makes a great deal of sense if you think of angles as rotations around the unit circle. Remember that \begin{align*}\sin x\end{align*} is equal to the \begin{align*}y-\end{align*}coordinate of the point on the unit circle that corresponds to the given angle \begin{align*}x\end{align*}. (Using \begin{align*}x\end{align*} to stand for the measure of the angle can get confusing—remember that it does stand for the angle measure, in this case, and not for the corresponding \begin{align*}x-\end{align*}coordinate.) The question this problem asks, therefore, is “What angle has the same \begin{align*}y-\end{align*}coordinate as another angle that’s twice as big?” Now, two angles can only have the same \begin{align*}y-\end{align*}coordinate if they are the same angle, or if one of them is the reflection of the other across the \begin{align*}y-\end{align*}axis. And the only angle whose reflection across the \begin{align*}y-\end{align*}axis is also twice as big as itself is an angle of \begin{align*}\pi\end{align*} (or \begin{align*}- \pi\end{align*}) radians (a rotation of half a circle). \begin{align*}\pi\end{align*} is outside the range of permissible answers we were given, so \begin{align*}-\pi\end{align*} is the solution we want. \begin{align*}0\end{align*} is also a solution, because \begin{align*}0\end{align*} doubled is still \begin{align*}0\end{align*} and so an angle of measure \begin{align*}0\end{align*} has the same coordinates as an angle twice as big.

*Solving Trigonometric Equations Using Half Angle Formulas*

**Half-Angle Identity for the Sine Function**

Basically, what we are doing here is applying the double-angle formula more or less in reverse to find the half-angle formula. Because we have to take the square root of both sides, however, we end up with two possible answers, one positive and one negative. We can tell which one is correct in any particular case, though, based on what quadrant the angle \begin{align*}\frac{\alpha}{2}\end{align*} is in: if it is in the first two quadrants, its sine must be positive, and if it is in the third or fourth quadrant, its sine must be negative.

(Similar reasoning is used below for the cosine half-angle formula.)

Example 1 should read “Use the half angle formula for the sine function to determine the value of \begin{align*}\sin 15^\circ\end{align*}” rather than \begin{align*}\sin 30^\circ.\end{align*}

## Equations with Inverse Circular Functions

*Solving Trigonometric Equations Using Inverse Notation*

**Introduction**

Students may be prone to mix up the ranges for the various inverse functions, and they may be baffled as to why the ranges are seemingly arbitrarily different. The rationale for choosing those particular ranges is that each of them represents two adjacent quadrants, in one of which the given function is positive and in the other of which it is negative. That way, the whole set of possible values for the trig function is covered, but none of the values are repeated.

For the cosine function, for example, it is convenient to pick the first two quadrants, because the graph of the cosine function over the interval from \begin{align*}0\end{align*} to \begin{align*}\pi\end{align*} hits every possible \begin{align*}y-\end{align*}value exactly once. (The same holds true for the secant function, as it is the reciprocal of cosine.) But for the sine function, that interval won’t work; the graph of the sine function from \begin{align*}0\end{align*} to \begin{align*}\pi\end{align*} hits all the positive \begin{align*}y-\end{align*}values twice, but none of the negative values. Instead, we use the interval from \begin{align*}\frac{-\pi}{2}\end{align*} to \begin{align*}\frac{\pi}{2}\end{align*}, or the fourth and first quadrants, because over that interval the sine function hits all the possible \begin{align*}y-\end{align*}values once each. (We could also use the interval from \begin{align*}\frac{\pi}{2}\end{align*} to \begin{align*}\frac{3 \pi}{2}\end{align*}, or the second and third quadrants, but it’s more convenient to use an interval that includes \begin{align*}0\end{align*}.) And of course we use that same interval for cosecant, the reciprocal of sine. (For the cosecant and secant functions, though, we have to leave out the point in the very center of the range where the sine or cosine equals \begin{align*}0\end{align*} and the cosecant or secant is therefore undefined.)

Finally, for the tangent and cotangent functions, we choose an interval that includes all possible values of the function while avoiding the function’s asymptotes; that is why the endpoints are excluded from the ranges of those inverse functions, while the other functions’ ranges include their endpoints.

Note: In the graphs in this section, the degree signs should actually be \begin{align*}-1\text{'}\mathrm{s};\end{align*} for example, \begin{align*}``y = sin^\circ x\text{''}\end{align*} should read \begin{align*}``y = sin^{-1} x.\text{''}\end{align*}

**Points to Consider**

We haven’t yet covered any identities that will help us express inverse trig functions in other ways. However, the identities we’ve learned so far can help us when we encounter expressions that have both trig functions and inverse trig functions in them, as we can use them to simplify the non-inverse functions.

*Solving Trigonometric Equations Using Inverse Functions*

**Points to Consider**

This isn’t a trick question; we can indeed use identities to solve trig equations, though we won’t always be able to in any particular case.

*Solving Inverse Equations Using Trigonometric Identities*

**Points to Consider**

We will see the applicability to real-world problems in the following review questions. You might encourage students to think of some more cases where these techniques could be applied.

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