Chapter 4: Inverse Trigonometric Functions
Chapter Outline
 4.1. Basic Inverse Trigonometric Functions
 4.2. Graphing Inverse Trigonometric Functions
 4.3. Inverse Trigonometric Properties
 4.4. Applications & Models
Chapter Summary
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 reallife 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[−π2,π2],y≠0 .
 Arccosine

Read “cosine inverse” and also written
cos−1 . The domain of this function is [1, 1]. The range is[0,π] .
 Arccotangent

Read “cotangent inverse” and also written
cot−1 . The domain of this function is all reals. The range is(0,π) .
 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,π],y≠π2 .
 Arcsine

Read “sine inverse” and also written
sin−1 . The domain of this function is [1, 1]. The range is[−π2,π2] .
 Arctangent

Read “tangent inverse” and also written
tan−1 . The domain of this function is all reals. The range is(−π2,π2) .
 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)=Acos(2πft+φ) whereA is the amplitude,f is the frequency, andφ 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 liney=x . The points(a,b) and(b,a) are reflections of each other across the liney=x .
 Invertible
 If a function has an inverse, it is invertible.
 OnetoOne Function

A function, where, for every
x value, there is EXACTLY oney− value. These are the only invertible functions.
Review Questions
 Find the exact value of the following expressions:

csc−1(−2) 
cos−13√2 
cot−1(−3√3) 
sec−1(−2√) 
arcsin0 
arctan1

 Use your calculator to find the value of each of the following expressions:

arccos35 
csc−12.25 
tan−18 
arcsin(−0.98) 
cot−1(−940) 
sec−165

 Find the exact value of the following expressions:

cos(sin−12√2) 
tan(cot−11) 
csc(sec−123√3) 
sin(arccos1213) 
tan(arcsin57) 
sec−1(cscπ6)

 Find the inverse of each of the following:

f(x)=5+cos(2x−1) 
g(x)=−4sin−1(x+3)

 Sketch a graph of each of the following:

y=3−arcsin(12x+1) 
f(x)=2tan−1(3x−4) 
h(x)=sec−1(x−1)+2 
y=1+2arccos2x

 Using the triangles from Section 4.3, find the following:

sin(cos−1x3) 
tan2(sin−1x23) 
cos4(arctan(2x)2)

 A ship leaves port and travels due west 20 nautical miles, then changes course to
E40∘S and travels 65 more nautical miles. Find the bearing to the port of departure.  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∘ on the12th of May.  Find the inverse of
sin(x±y)=sinxcosy±cosxsiny . HINT: Seta=sinx andb=siny and rewritecosx andcosy in terms of sine.  Find the inverse of
cos(x±y)=cosxcosy∓sinxsiny . HINT: Seta=cosx andb=cosy and rewritesinx andsiny in terms of sine.
Review Answers


−π6 
π6 
−π3 
3π4 
0 
π4
 0.927
 0.461
 1.446
 1.37
 1.792
 0.586

2√2  1
 2

513 
526√ or56√12 
π3

f(x)yxx−5cos−1(x−5)1+cos−1(x−5)1+cos−1(x−5)2=5+cos(2x−1)=5+cos(2x−1)=5+cos(2y−1)=cos(2y−1)=2y−1=2y=y 
g(x)yx−x4sin(−x4)sin(−x4)−3=−4sin−1(x+3)=−4sin−1(x+3)=−4sin−1(y+3)=sin−1(y+3)=y+3=y

sin(cos−1x3)=1−(x3)2−−−−−−−√=1−x6−−−−−√ 
tan2(sin−1x23)=⎛⎝⎜⎜⎜x231−(x23)2−−−−−−−−√⎞⎠⎟⎟⎟2=x491−(x49)=x49(1−x49)=x49−x4 
cos4(arctan(2x)2)=cos4(tan−14x2)=⎛⎝⎜1(4x2)2+1−−−−−−−−√⎞⎠⎟4=116x4+1−−−−−−−√4=1(16x4+1)2


x∘ can help us find our final answer, but we need to findy andz first.Now, the bearing from the ship to the point of departure is north, and thensin40∘cos40∘20+ztanxx=y65→y=65sin40∘=41.78=20+z65→20+z=65cos40∘=49.79→z=29.79=41.7829.79→x=tan−141.7829.79=54.51∘ (90−54.51)∘ west, which is written asN35.49∘W . 
36∘ on the12th of May=90∘−36∘−23.5∘cos[(132+10)360365]=72.02∘ 
sin(x±y)=sinxcosy±cosxsiny,a=sinx andb=siny→x=sin−1a andy=sin−1b sin(x±y)sin(x±y)x±ysin−1a±sin−1b=a1−sin2y−−−−−−−√±b1−sin2x−−−−−−−−√=a1−b2−−−−−√±b1−a2−−−−−√=sin−1(a1−b2−−−−−√±b1−a2−−−−−√)=sin−1(a1−b2−−−−−√±b1−a2−−−−−√) 
cos(x±y)=cosxcosy∓sinxsiny,a=cosx andb=cosy→x=cos−1a andy=cos−1b cos(x±y)cos(x±y)x±ycos−1a±cos−1b=ab∓(1−cos2x)(1−cos2y)−−−−−−−−−−−−−−−−−−√=ab∓(1−a2)(1−b2)−−−−−−−−−−−−√=cos−1(ab∓(1−a2)(1−b2)−−−−−−−−−−−−√)=cos−1(ab∓(1−a2)(1−b2)−−−−−−−−−−−−√)
Texas Instruments Resources
In the CK12 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.