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# Chapter 4: Inverse Trigonometric Functions

Difficulty Level: At Grade Created by: CK-12

## Chapter Summary

In this chapter, we studied all aspects of inverse trigonometric functions. First, we defined the function by finding inverses algebraically. Second, we analyzed the graphs of inverse functions. We needed to restrict the domain of the trigonometric functions in order to take the inverse of each of them. This is because they are periodic and did not pass the horizontal line test. Then, we learned about the properties of the inverse functions, mostly composing a trig function and an inverse. Finally, we applied the principles of inverse trig functions to real-life situations.

## Chapter Vocabulary

Arccosecant
Read “cosecant inverse” and also written $\csc^{-1}$. The domain of this function is all reals, excluding the interval (-1, 1). The range is all reals in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}], y \ne 0$.
Arccosine
Read “cosine inverse” and also written $\cos^{-1}$. The domain of this function is [-1, 1]. The range is $[0, \pi]$.
Arccotangent
Read “cotangent inverse” and also written $\cot^{-1}$. The domain of this function is all reals. The range is $(0, \pi)$.
Arcsecant
Read “secant inverse” and also written $\sec^{-1}$. The domain of this function is all reals, excluding the interval (-1, 1). The range is all reals in the interval $[0, \pi], y \ne \frac{\pi}{2}$.
Arcsine
Read “sine inverse” and also written $\sin^{-1}$. The domain of this function is [-1, 1]. The range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Arctangent
Read “tangent inverse” and also written $\tan^{-1}$. The domain of this function is all reals. The range is $(-\frac{\pi}{2}, \frac{\pi}{2})$.
Composite Function
The final result from when one function is plugged into another, $f(g(x))$.
Harmonic Motion
A motion that is consistent and periodic, in a sinusoidal pattern. The general equation is $x(t) = A \cos(2\pi ft + \varphi)$ where $A$ is the amplitude, $f$ is the frequency, and $\varphi$ is the phase shift.
Horizontal Line Test
The test applied to a function to see if it has an inverse. Continually draw horizontal lines across the function and if a horizontal line touches the function more than once, it does not have an inverse.
Inverse Function
Two functions that are symmetric over the line $y = x$.
Inverse Reflection Principle
The points $(a, b)$ and $(b, a)$ in the coordinate plane are symmetric with respect to the line $y = x$. The points $(a, b)$ and $(b, a)$ are reflections of each other across the line $y = x$.
Invertible
If a function has an inverse, it is invertible.
One-to-One Function
A function, where, for every $x$ value, there is EXACTLY one $y-$value. These are the only invertible functions.

## Review Questions

1. Find the exact value of the following expressions:
1. $\csc^{-1} (-2)$
2. $\cos^{-1} \frac{\sqrt{3}}{2}$
3. $\cot^{-1} \left ( -\frac{\sqrt{3}}{3} \right )$
4. $\sec^{-1} \left ( -\sqrt{2} \right )$
5. $\arcsin 0$
6. $\arctan 1$
2. Use your calculator to find the value of each of the following expressions:
1. $\arccos \frac{3}{5}$
2. $\csc^{-1} 2.25$
3. $\tan^{-1} 8$
4. $\arcsin (-0.98)$
5. $\cot^{-1} \left ( -\frac{9}{40} \right )$
6. $\sec^{-1} \frac{6}{5}$
3. Find the exact value of the following expressions:
1. $\cos \left ( \sin^{-1} \frac{\sqrt{2}}{2} \right )$
2. $\tan ( \cot^{-1} 1)$
3. $\csc \left ( \sec^{-1} \frac{2\sqrt{3}}{3} \right )$
4. $\sin \left ( \arccos \frac{12}{13} \right )$
5. $\tan \left ( \arcsin \frac{5}{7} \right )$
6. $\sec^{-1} \left ( \csc \frac{\pi}{6} \right )$
4. Find the inverse of each of the following:
1. $f(x) = 5+ \cos (2x-1)$
2. $g(x) = -4 \sin^{-1} (x+3)$
5. Sketch a graph of each of the following:
1. $y = 3-\arcsin \left ( \frac{1}{2}x + 1 \right )$
2. $f(x) = 2 \tan^{-1} (3x-4)$
3. $h(x) = \sec^{-1} (x-1)+2$
4. $y = 1 + 2 \arccos 2x$
6. Using the triangles from Section 4.3, find the following:
1. $\sin(\cos^{-1} x^3)$
2. $\tan^2 \left ( \sin^{-1} \frac{x^2}{3} \right )$
3. $\cos^4(\arctan (2x)^2)$
7. A ship leaves port and travels due west 20 nautical miles, then changes course to $E 40^\circ S$ and travels 65 more nautical miles. Find the bearing to the port of departure.
8. Using the formula from Example 1 in Section 4.4, determine the measurement of the sun’s angle of inclination for a building located at a latitude of $36^\circ$ on the $12^{th}$ of May.
9. Find the inverse of $\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y$. HINT: Set $a= \sin x$ and $b = \sin y$ and rewrite $\cos x$ and $\cos y$ in terms of sine.
10. Find the inverse of $\cos(x \pm y) = \cos x \cos y \mp \sin x \sin y$. HINT: Set $a = \cos x$ and $b= \cos y$ and rewrite $\sin x$ and $\sin y$ in terms of sine.

1. $-\frac{\pi}{6}$
2. $\frac{\pi}{6}$
3. $-\frac{\pi}{3}$
4. $\frac{3\pi}{4}$
5. $0$
6. $\frac{\pi}{4}$
1. 0.927
2. 0.461
3. 1.446
4. -1.37
5. 1.792
6. 0.586
1. $\frac{\sqrt{2}}{2}$
2. 1
3. 2
4. $\frac{5}{13}$
5. $\frac{5}{2\sqrt{6}}$ or $\frac{5\sqrt{6}}{12}$
6. $\frac{\pi}{3}$
1. $f(x) & = 5 + \cos (2x-1)\\y & = 5 + \cos (2x-1)\\x & = 5 + \cos (2y-1)\\x-5 & = \cos (2y-1)\\\cos^{-1} (x-5) & = 2y-1\\1 + \cos^{-1} (x-5) & = 2y\\\frac{1 + \cos^{-1} (x-5)}{2} & = y$
2. $g(x) & = -4\sin^{-1} (x+3)\\y & = -4\sin^{-1} (x+3)\\x & = -4\sin^{-1} (y+3)\\-\frac{x}{4} & = \sin^{-1} (y+3)\\\sin \left ( -\frac{x}{4} \right ) & = y+3\\\sin \left ( -\frac{x}{4} \right )-3 & = y$
1. $\sin (\cos^{-1}x^3) = \sqrt{1-(x^3)^2} = \sqrt{1-x^6}$
2. $\tan^2 \left ( \sin^{-1} \frac{x^2}{3} \right ) = \left ( \frac{\frac{x^2}{3}}{\sqrt{1-\left ( \frac{x^2}{3} \right )^2}} \right )^2 = \frac{\frac{x^4}{9}}{1-\left ( \frac{x^4}{9} \right )} = \frac{x^4}{9 \left ( 1-\frac{x^4}{9} \right )} = \frac{x^4}{9-x^4}$
3. $\cos^4 (\arctan(2x)^2) = \cos^4 (\tan^{-1} 4x^2) = \left ( \frac{1}{\sqrt{(4x^2)^2 + 1}} \right )^4 = \frac{1}{\sqrt{16x^4 + 1}^4} = \frac{1}{(16x^4 + 1)^2}$
1. $x^\circ$ can help us find our final answer, but we need to find $y$ and $z$ first. $\sin 40^\circ & = \frac{y}{65} \rightarrow y = 65\sin 40^\circ = 41.78\\\cos 40^\circ & = \frac{20+z}{65} \rightarrow 20+z = 65\cos 40^\circ\\20 + z & = 49.79 \rightarrow z = 29.79\\\tan x & = \frac{41.78}{29.79} \rightarrow x = \tan^{-1} \frac{41.78}{29.79}\\x & = 54.51^\circ$ Now, the bearing from the ship to the point of departure is north, and then $(90-54.51)^\circ$ west, which is written as $N 35.49^\circ W$.
2. $36^\circ$ on the $12^{th}$ of May $= 90^\circ - 36^\circ - 23.5^\circ \cos \left [ (132+10) \frac{360}{365} \right ]= 72.02^\circ$
3. $\sin (x \pm y) = \sin x \cos y \pm \cos x \sin y, a= \sin x$ and $b= \sin y \rightarrow x = \sin^{-1} a$ and $y = \sin^{-1}b$ $\sin (x \pm y) & = a \sqrt{1-\sin^2 y} \pm b \sqrt{1-\sin^2 x}\\\sin (x \pm y) & = a \sqrt{1-b^2} \pm b \sqrt{1-a^2}\\x \pm y & = \sin^{-1} \left ( a \sqrt{1-b^2} \pm b \sqrt{1-a^2} \right )\\\sin^{-1} a \pm \sin^{-1} b & = \sin^{-1} \left ( a \sqrt{1-b^2} \pm b \sqrt{1-a^2} \right )$
4. $\cos (x \pm y) = \cos x \cos y \mp \sin x \sin y , a = \cos x$ and $b= \cos y \rightarrow x = \cos^{-1} a$ and $y = \cos^{-1} b$ $\cos (x \pm y) & = ab \mp \sqrt{(1-\cos^2 x)(1-\cos^2 y)}\\\cos (x \pm y) & = ab \mp \sqrt{(1-a^2)(1-b^2)}\\x \pm y & = \cos^{-1} \left ( ab \mp \sqrt{(1-a^2)(1-b^2)} \right )\\\cos^{-1} a \pm \cos^{-1} b & = \cos^{-1} \left ( ab \mp \sqrt{(1-a^2)(1-b^2)} \right )$

## Texas Instruments Resources

In the CK-12 Texas Instruments Trigonometry FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9702.

Sep 26, 2013