# 2.4: Inverse Functions and Trigonometric Equations

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

## Inverse Trigonometric Functions

*General Definitions of Inverse Trigonometric Functions*

**In-Text Examples**

1) Students who don’t refer to the illustration may get mixed up about which angle is \begin{align*}\theta\end{align*}. If they start with \begin{align*}\tan \theta = \frac{16 \ \text{feet}} {10 \ \text{feet}}\end{align*}, they will end up with \begin{align*}58^\circ\end{align*} as their answer, which is the angle the boards will make with the short side of the deck rather than the long side.

Students may also be using the wrong key on their calculators to find the inverse tangent; they may be just using the “tan” key by itself, or they may be using a combination of the “tan” key and the reciprocal \begin{align*}(x^{-1})\end{align*} key.

**Review Questions**

1) Some students will still be confusing the vertical and horizontal line tests here, or otherwise be fuzzy on the idea of what makes a relation a function. (Also, the inverse of relation \begin{align*}i\end{align*} is in fact a function.)

2) Many students will try to make side \begin{align*}BC\end{align*} in the diagram, rather than \begin{align*}AC\end{align*}, equal \begin{align*}9\;\mathrm{feet}\end{align*}, thinking that the ladder is \begin{align*}9\;\mathrm{feet}\end{align*} up the wall rather than 9 feet long itself.

**Additional Problems**

1) While walking home, you decide to take a shortcut across an empty lot. From one corner of the lot, you cannot see the opposite corner, but you know the lot is \begin{align*}30\;\mathrm{yards}\end{align*} long and \begin{align*}20\;\mathrm{yards}\end{align*} wide. At what angle should you set off across the lot in order to aim directly for the opposite corner?

**Answers to Additional Problems**

1) \begin{align*}\tan^{-1} \left (\frac{30}{20}\right ) \approx 56.31^\circ\end{align*}, so you should set off at a \begin{align*}56.31^\circ\end{align*} angle from the shorter side (or a \begin{align*}33.69^\circ\end{align*} angle from the longer side).

*Using the “Inverse” Notation*

**In-Text Examples**

1) Students may need a refresher on the unit circle at this point; in particular, they may have forgotten how to figure out what quadrant the reference triangle should be in.

**Review Questions**

1) This is a trick question; students may get confused by the fact that \begin{align*}\frac{\pi}{2}\end{align*} is a commonplace angle and may try to take the sine of it instead of the inverse sine, or they may try to take the inverse sine of it but be confused about how to find an angle whose sine is \begin{align*}\frac{\pi}{2}\end{align*} (which they should be, since there is no such angle).

## Exact Values of Inverse Functions

*Exact Values of Special Inverse Circular Functions*

**Review Questions**

Using special triangles may be the wrong approach for these problems, as the angles here are not acute ones and therefore can’t be directly found in triangles. Triangles may still be useful for figuring out the reference angles for the angles given, but that’s after students figure out what quadrant the angles are in.

When using the unit circle, though, some students may need to be reminded that they can simply read off the sine and cosine values from the coordinates of the points on the circle, as the sine values is equal to the \begin{align*}y-\end{align*}coordinate and the cosine value is the \begin{align*}x-\end{align*}coordinate.

*Range of the Outside Function, Domain of the Inside Function*

**In-Text Examples**

1) Students may not realize yet that the easiest way to prove two functions are each other’s inverses is to compose the two functions and prove that the composition always equals \begin{align*}x\end{align*}; they may instead try to derive the one function from the other by swapping the variables, which will work but is not as easy.

**Review Questions**

1) Both of these functions are actually not invertible simply because their inverses are not functions.

*Applications, Technological Tools*

**Additional Problems**

1) Find the inverse of the following function: \begin{align*}f(x) = 3 + 5 \sin (2x - 7)\end{align*}.

**Answers to Additional Problems**

1)

\begin{align*}f(x) & = 3 + 5 \sin (2x - 7)\\ \therefore x & = 3 + 5 \sin (2y - 7)\\ \therefore x - 3 & = 5 \sin (2y - 7)\\ \therefore \frac{x - 3}{5} & = \sin (2y - 7)\\ \therefore \sin^{-1} \left (\frac{x - 3}{5}\right ) + 7 & = 2y\\ \therefore \frac{1}{2} \left (\sin^{-1} \left ( \frac{x - 3}{5}\right ) + 7 \right ) & = y\\ \therefore f^{-1} (x) & = \frac{1}{2} \left (\sin^{-1} \left (\frac{x - 3}{5}\right ) + 7 \right )\end{align*}

## Properties of Inverse Circular Functions

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

**In-Text Examples**

1) Remind students to label which angle is \begin{align*}\theta\end{align*} when they are drawing triangles to solve these problems; otherwise they may end up taking the sine or cosine of the wrong angle.

(Of course, they can simply plug the values of \begin{align*}x\end{align*} into the set of equations above to solve these problems without drawing triangles. This method is faster, but may mask a lack of understanding of the actual principles involved.)

2) On part a, students may miss the part about the bottom of the screen not being directly on the ground. On part b, they may be confused by the fact that they don’t have enough information to find exact angle measures. Instead, they must find a formula for the angle measure in terms of \begin{align*}x\end{align*}, and then figure out what value of \begin{align*}x\end{align*} maximizes this angle.

**Review Questions**

2) After all the previous discussion of how \begin{align*}f^{-1}(f(x)) = x\end{align*}, students are likely to assume that \begin{align*}\tan^{-1} (\tan x) = x\end{align*}. The only reason they’re wrong is that the tangent function has to have its domain restricted in order to be invertible; inside this limited domain, \begin{align*}\tan^{-1}(\tan x)\end{align*} is indeed \begin{align*}x\end{align*}, but outside of it, the values of \begin{align*}\tan^{-1}(\tan x)\end{align*} just repeat themselves, as students will see if they use their calculators to graph it. (Also, those vertical lines on the graph represent vertical asymptotes; that is, they are not actually places where the graph is vertical, but rather where the graph is undefined.)

**Additional Problems**

1) Express the function \begin{align*}2 \cos (\tan^{-1} 3x)\end{align*} as an algebraic expression involving no trigonometric functions.

**Answers to Additional Problems**

1) \begin{align*}2\ \cos (\tan^{-1} 3x) = 2 \ \cos \theta = \frac{2} {\sqrt{(3x)^2 + 1}} = \frac{2} {\sqrt{9x^2 + 1}}\end{align*}

*Derive Inverse Cofunction Properties*

**Review Questions**

Remembering to get the domain restrictions right should be the only tricky part here.

*Find Exact Values of Functions of Inverse Functions Using Pythagorean Triples*

**In-Text Examples**

3) The most likely error here is getting the quadrant wrong.

## Applications of Inverse Circular Functions

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

**In-Text Examples**

Students may make the same errors here that they made when they originally covered this topic, particularly getting the various shifts and stretches mixed up and getting the sign of the phase shift backwards.

**Additional Problems**

1) How would you express the equation from problem 1 above as a transformation of \begin{align*}y = \sin x\end{align*}?

**Answers to Additional Problems**

1) \begin{align*}y = 3 \ \sin(2(x + 15^\circ)) + 2\end{align*}, or \begin{align*}y = 3 \ \sin (2(x - 165^\circ)) + 2\end{align*}

*Solving for Particular Values in Trigonometric Equations*

**Review Questions**

1) Students may still be unsure how to derive an equation from just the two points given. The trick is to realize that the horizontal distance between the two points is half the period (and the frequency is \begin{align*}2 \pi\end{align*} divided by the period); the vertical distance between them is the amplitude; the average of the two heights is the vertical shift; and the horizontal distance between the maximum point and zero (divided by \begin{align*}2 \pi\end{align*}) is the phase shift (if we are using a cosine function to model the graph).

**Additional Problems**

1) You are riding a Ferris wheel at an amusement park. \begin{align*}15\;\mathrm{seconds}\end{align*} after the wheel starts turning, you are at the top of the wheel, \begin{align*}100\;\mathrm{feet}\end{align*} high. \begin{align*}20\;\mathrm{seconds}\end{align*} later, you are at the bottom of the wheel, \begin{align*}10\;\mathrm{feet}\end{align*} off the ground. At what time did you first reach a height of \begin{align*}70\;\mathrm{feet}\end{align*}?

**Answers to Additional Problems**

1) The equation that models this problem is \begin{align*}55 + 45 \ \cos \frac{\pi}{20} \left (x - \frac{15}{2 \pi}\right )\end{align*}. Solving for \begin{align*}x\end{align*} in terms of \begin{align*}y\end{align*} gives us:

\begin{align*}x & = \frac{\cos^{-1} \left [\frac{y - c} {a}\right ]} {b} + d \ \ y = 70, \ \ c = 55, \ \ a = 45, \ \ b = \frac{\pi}{20}, \ \ d = \frac{15} {2 \pi}\\ \therefore x & = \frac{\cos^{-1} \left [\frac{70 - 55} {45}\right ]} { \frac{x} {20}} + \frac{15}{2 \pi}\end{align*}

Using a calculator gives us an answer of approximately \begin{align*}10.22\;\mathrm{seconds}\end{align*} after the Ferris wheel began moving.

## Trigonometric Equations

*Solving Trigonometric Equations Analytically*

**In-Text Examples**

1) Students may get stuck on these if they don’t remember the basic identities.

2) Make sure to point out on part b that there are a total of four solutions on the given interval.

**Review Questions**

1) Of course, students are likely to miss the fact that they need to double the interval of possible solutions (as explained in the solution key). Even if they do figure this out, they may not realize that this means the two solutions within the interval \begin{align*}[0, 2 \pi)\end{align*} must each have \begin{align*}2 \pi\end{align*} added to them to make the other two solutions. And finally, they may forget to divide the solutions by \begin{align*}2\end{align*} at the end to get the values for \begin{align*}\theta\end{align*} rather than \begin{align*}2 \theta.\end{align*}

2) As on other problems, students may forget to look for solutions in more than one quadrant. They may also forget to consider the cosines of both \begin{align*}\frac{1}{4}\end{align*} and \begin{align*}- \frac{1}{4}\end{align*}.

3) The trick here is to treat \begin{align*}4 \theta\end{align*} as a double angle and \begin{align*}2 \theta\end{align*} as the corresponding single angle; students may not think of this right away. Also, they may make the same mistakes as in problem 1.

4) The same errors as on problem 1 apply.

*Solve Trig Equations (Factoring)*

**In-Text Examples**

2) Students will of course be tempted to divide both sides of the equation by tan \begin{align*}x + 1;\end{align*} the only problem with this is that it will eliminate possible solutions because tan \begin{align*}x + 1\end{align*} can equal zero. The way to get around this is to find the solutions for \begin{align*}2\ \sin x = 1\end{align*}, but then also find the solutions for \begin{align*}\tan x + 1 = 0\end{align*}. The solution presented in the text is another way of doing this.

**Review Questions**

1) Upon finding that \begin{align*}\sin x\end{align*} can equal either \begin{align*}3\end{align*} or \begin{align*}-1\end{align*}, students may be stumped by the \begin{align*}3\end{align*}, since that isn’t a possible value for \begin{align*}\sin x\end{align*}. It doesn’t mean they’ve done anything wrong, it just means they should discard that solution.

2) Again, beware of dividing through by \begin{align*}\tan x\end{align*}, as this will eliminate the solution \begin{align*}x = 0.\end{align*}

*Solve Equations (Using Identities)*

**Review Questions**

1) Getting the signs reversed when factoring is the most likely error here.

2) Students are likely to forget the “for all values of \begin{align*}x\end{align*}” part; they may also forget the principal solution \begin{align*}\frac{3 \pi}{2}\end{align*}.

## Trigonometric Equations with Multiple Angles

*Solve Equations (with Double Angles)*

**In-Text Examples**

1) Again, beware of dividing both sides by \begin{align*}\cos x\end{align*}.

4) …or by \begin{align*}\sin x\end{align*}.

**Review Questions**

1) Finding the actual measure of angle \begin{align*}x\end{align*} is tempting but not necessary.

2) The solution to part b involves realizing that \begin{align*}\cos^4 \theta - \sin^4 \theta\end{align*} can be treated as a difference of squares, which may escape students at first.

3) This equation is actually true for all values of \begin{align*}\theta\end{align*}.

*Solving Trigonometric Equations Using Half Angle Formulas*

**In-Text Examples**

3) Students may be momentarily stumped by the appearance of the unfamiliar-looking angle \begin{align*}\frac{7 \pi}{6}\end{align*}. Rewriting it as \begin{align*}\pi + \frac{\pi}{6}\end{align*} may help them see that it is simply a third-quadrant angle with a reference angle of \begin{align*}\frac{\pi}{6}\end{align*}, whose sine and cosine they should be able to figure out easily.

**Review Questions**

3) Students will get three solutions to this equation if they follow the procedure outlined in the solution key, but not all of those solutions will in fact be correct. Squaring both sides of an equation introduces extraneous solutions, so they must check their answers afterward to see which ones need to be discarded. This is true any time both sides of an algebraic equation are squared.

*Solving Trigonometric Equations with Multiple Angles*

**In-Text Examples**

2) Once again, remind students that they must look for values of \begin{align*}2 \theta\end{align*} between and \begin{align*}4 \pi\end{align*} in order to find values of \begin{align*}\theta\end{align*} between and \begin{align*}2 \pi\end{align*}.

**Review Questions**

2) Students may get stuck trying to figure out how to express \begin{align*}\cos 3x\end{align*} in terms of \begin{align*}\cos x\end{align*}, perhaps by trying to refer back to the “triple-angle formula” derived in chapter 3, or trying to derive one themselves from the double- and half-angle formulas they already know. This isn’t necessary, though; they only need to solve for \begin{align*}\cos 3x\end{align*} and then find possible values for \begin{align*}3x\end{align*} that lie between and \begin{align*}6 \pi\end{align*} in order to find the possible values of \begin{align*}x\end{align*} that lie between and \begin{align*}2 \pi\end{align*}.

3) After finding the principal solutions to the equation, students may try to derive the other solutions by adding \begin{align*}2 \pi k\end{align*}, rather than \begin{align*}\pi k\end{align*}. (If they were dealing with the sine or cosine functions instead of the tangent function, they’d be right, but the tangent function repeats its values every \begin{align*}\pi\end{align*} rather than \begin{align*}2 \pi \;\mathrm{units}\end{align*}.)

## Equations with Inverse Circular Functions

*Solving Trigonometric Equations Using Inverse Notation*

**In-Text Examples**

2) Students may be momentarily confused by the lack of “arc”-function buttons on their calculators, and may need to be reminded to use the \begin{align*}``\sin^{-1},\text{''} ``\cos^{-1},\text{''},\end{align*} and \begin{align*}``\tan^{-1}\text{''}\end{align*} functions instead.

**Review Questions**

3-4) Students are likely to stop when they have found the one solution within the restricted range, and forget that they are looking for solutions between and \begin{align*}2 \pi\end{align*} rather than just between and \begin{align*}\pi\end{align*}.

*Solving Trigonometric Equations Using Inverse Functions*

**In-Text Examples**

3) The most likely error here is for students to try “unwrapping” the right-hand side of the equation in the wrong order, for example by taking the inverse cosine before dividing by \begin{align*}A\end{align*}, or by subtracting \begin{align*}\phi\end{align*} before it is properly isolated. This sort of error is often more likely when working with expressions that have a large number of variables, as students can get a little bit lost when they are not working with concrete numbers.

**Review Questions**

1) This problem is subject to the same error as example 3 above; also, students who do not recognize that the problem relies on an angle sum identity will approach it wrong from the beginning and get an answer that makes no sense, if they get any answer at all.

*Solving Inverse Equations Using Trigonometric Identities*

**In-Text Examples**

1) This example is much trickier than it seems at first, because of the double angle. Watch out for students forgetting to account for the \begin{align*}2\end{align*} and just solving for \begin{align*}\cos (\sin^{-1} x)\end{align*}, which is \begin{align*}\sqrt{1 - x^2}\end{align*}.

**Review Questions**

1) Students may get their double- and half-angle identities mixed up here, or may forget that if they are applying a double-angle identity to \begin{align*}\sin 2 \theta \cos 2 \theta\end{align*}, they will end up with \begin{align*}\sin 4 \theta\end{align*} rather than \begin{align*}\sin 2 \theta.\end{align*}

2) (Note: \begin{align*}10\;\mathrm{feet}\end{align*} is actually the distance from one end to the other, rather than the width of the one end.)

Students may think they don’t have enough information to solve this problem, but they do if they express the legs of the triangles as \begin{align*}\sin \theta\end{align*} and \begin{align*}\cos \theta\end{align*}; then they can set up an equation for the area of one end (and thus the volume) in terms of \begin{align*}\theta\end{align*}. (They may need calculators, though, to find the equation’s maximum value.)

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