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Calculus Volume 1

1.3 Trigonometric Functions

Calculus Volume 11.3 Trigonometric Functions
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  1. Preface
  2. 1 Functions and Graphs
    1. Introduction
    2. 1.1 Review of Functions
    3. 1.2 Basic Classes of Functions
    4. 1.3 Trigonometric Functions
    5. 1.4 Inverse Functions
    6. 1.5 Exponential and Logarithmic Functions
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Limits
    1. Introduction
    2. 2.1 A Preview of Calculus
    3. 2.2 The Limit of a Function
    4. 2.3 The Limit Laws
    5. 2.4 Continuity
    6. 2.5 The Precise Definition of a Limit
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  4. 3 Derivatives
    1. Introduction
    2. 3.1 Defining the Derivative
    3. 3.2 The Derivative as a Function
    4. 3.3 Differentiation Rules
    5. 3.4 Derivatives as Rates of Change
    6. 3.5 Derivatives of Trigonometric Functions
    7. 3.6 The Chain Rule
    8. 3.7 Derivatives of Inverse Functions
    9. 3.8 Implicit Differentiation
    10. 3.9 Derivatives of Exponential and Logarithmic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  5. 4 Applications of Derivatives
    1. Introduction
    2. 4.1 Related Rates
    3. 4.2 Linear Approximations and Differentials
    4. 4.3 Maxima and Minima
    5. 4.4 The Mean Value Theorem
    6. 4.5 Derivatives and the Shape of a Graph
    7. 4.6 Limits at Infinity and Asymptotes
    8. 4.7 Applied Optimization Problems
    9. 4.8 L’Hôpital’s Rule
    10. 4.9 Newton’s Method
    11. 4.10 Antiderivatives
    12. Key Terms
    13. Key Equations
    14. Key Concepts
    15. Chapter Review Exercises
  6. 5 Integration
    1. Introduction
    2. 5.1 Approximating Areas
    3. 5.2 The Definite Integral
    4. 5.3 The Fundamental Theorem of Calculus
    5. 5.4 Integration Formulas and the Net Change Theorem
    6. 5.5 Substitution
    7. 5.6 Integrals Involving Exponential and Logarithmic Functions
    8. 5.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Applications of Integration
    1. Introduction
    2. 6.1 Areas between Curves
    3. 6.2 Determining Volumes by Slicing
    4. 6.3 Volumes of Revolution: Cylindrical Shells
    5. 6.4 Arc Length of a Curve and Surface Area
    6. 6.5 Physical Applications
    7. 6.6 Moments and Centers of Mass
    8. 6.7 Integrals, Exponential Functions, and Logarithms
    9. 6.8 Exponential Growth and Decay
    10. 6.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  8. A | Table of Integrals
  9. B | Table of Derivatives
  10. C | Review of Pre-Calculus
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
  12. Index

Learning Objectives

  • 1.3.1. Convert angle measures between degrees and radians.
  • 1.3.2. Recognize the triangular and circular definitions of the basic trigonometric functions.
  • 1.3.3. Write the basic trigonometric identities.
  • 1.3.4. Identify the graphs and periods of the trigonometric functions.
  • 1.3.5. Describe the shift of a sine or cosine graph from the equation of the function.

Trigonometric functions are used to model many phenomena, including sound waves, vibrations of strings, alternating electrical current, and the motion of pendulums. In fact, almost any repetitive, or cyclical, motion can be modeled by some combination of trigonometric functions. In this section, we define the six basic trigonometric functions and look at some of the main identities involving these functions.

Radian Measure

To use trigonometric functions, we first must understand how to measure the angles. Although we can use both radians and degrees, radians are a more natural measurement because they are related directly to the unit circle, a circle with radius 1. The radian measure of an angle is defined as follows. Given an angle θ,θ, let ss be the length of the corresponding arc on the unit circle (Figure 1.30). We say the angle corresponding to the arc of length 1 has radian measure 1.

An image of a circle. At the exact center of the circle there is a point. From this point, there is one line segment that extends horizontally to the right a point on the edge of the circle and another line segment that extends diagonally upwards and to the right to another point on the edge of the circle. These line segments have a length of 1 unit. The curved segment on the edge of the circle that connects the two points at the end of the line segments is labeled “s”. Inside the circle, there is an arrow that points from the horizontal line segment to the diagonal line segment. This arrow has the label “theta = s radians”.
Figure 1.30 The radian measure of an angle θθ is the arc length ss of the associated arc on the unit circle.

Since an angle of 360°360° corresponds to the circumference of a circle, or an arc of length 2π,2π, we conclude that an angle with a degree measure of 360°360° has a radian measure of 2π.2π. Similarly, we see that 180°180° is equivalent to ππ radians. Table 1.8 shows the relationship between common degree and radian values.

Degrees Radians Degrees Radians
0 0 120 2π/32π/3
30 π/6π/6 135 3π/43π/4
45 π/4π/4 150 5π/65π/6
60 π/3π/3 180 ππ
90 π/2π/2
Table 1.8 Common Angles Expressed in Degrees and Radians

Example 1.22

Converting between Radians and Degrees

  1. Express 225°225° using radians.
  2. Express 5π/35π/3 rad using degrees.

Solution

Use the fact that 180°180° is equivalent to ππ radians as a conversion factor: 1=πrad180°=180°πrad.1=πrad180°=180°πrad.

  1. 225°=225°·π180°=5π4225°=225°·π180°=5π4 rad
  2. 5π35π3 rad = 5π3·180°π=300°5π3·180°π=300°
Checkpoint 1.17

Express 210°210° using radians. Express 11π/611π/6 rad using degrees.

The Six Basic Trigonometric Functions

Trigonometric functions allow us to use angle measures, in radians or degrees, to find the coordinates of a point on any circle—not only on a unit circle—or to find an angle given a point on a circle. They also define the relationship among the sides and angles of a triangle.

To define the trigonometric functions, first consider the unit circle centered at the origin and a point P=(x,y)P=(x,y) on the unit circle. Let θθ be an angle with an initial side that lies along the positive xx-axis and with a terminal side that is the line segment OP.OP. An angle in this position is said to be in standard position (Figure 1.31). We can then define the values of the six trigonometric functions for θθ in terms of the coordinates xx and y.y.

An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment that extends diagonally upwards and to the right to another point on the edge of the circle. This point is labeled “P = (x, y)”. These line segments have a length of 1 unit. From the point “P”, there is a dotted vertical line that extends downwards until it hits the x axis and thus the horizontal line segment. Inside the circle, there is an arrow that points from the horizontal line segment to the diagonal line segment. This arrow has the label “theta”.
Figure 1.31 The angle θθ is in standard position. The values of the trigonometric functions for θθ are defined in terms of the coordinates xx and y.y.

Definition

Let P=(x,y)P=(x,y) be a point on the unit circle centered at the origin O.O. Let θθ be an angle with an initial side along the positive xx-axis and a terminal side given by the line segment OP.OP. The trigonometric functions are then defined as

sinθ=ycscθ=1ycosθ=xsecθ=1xtanθ=yxcotθ=xysinθ=ycscθ=1ycosθ=xsecθ=1xtanθ=yxcotθ=xy
1.9

If x=0,secθx=0,secθ and tanθtanθ are undefined. If y=0,y=0, then cotθcotθ and cscθcscθ are undefined.

We can see that for a point P=(x,y)P=(x,y) on a circle of radius rr with a corresponding angle θ,θ, the coordinates xx and yy satisfy

cosθ=xrx=rcosθcosθ=xrx=rcosθ
sinθ=yry=rsinθ.sinθ=yry=rsinθ.

The values of the other trigonometric functions can be expressed in terms of x,y,x,y, and rr (Figure 1.32).

An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one blue line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another blue line segment that extends diagonally upwards and to the right to another point on the edge of the circle. This point is labeled “P = (x, y)”. These line segments have a length of “r” units. Between these line segments within the circle is the label “theta”, representing the angle between the segments. From the point “P”, there is a blue vertical line that extends downwards until it hits the x axis and thus hits the horizontal line segment, at a point labeled “x”. At the intersection horizontal line segment and vertical line segment at the point x, there is a right triangle symbol. From the point “P”, there is a dotted horizontal line segment that extends left until it hits the y axis at a point labeled “y”.
Figure 1.32 For a point P=(x,y)P=(x,y) on a circle of radius r,r, the coordinates xx and yy satisfy x=rcosθx=rcosθ and y=rsinθ.y=rsinθ.

Table 1.9 shows the values of sine and cosine at the major angles in the first quadrant. From this table, we can determine the values of sine and cosine at the corresponding angles in the other quadrants. The values of the other trigonometric functions are calculated easily from the values of sinθsinθ and cosθ.cosθ.

θθ sinθsinθ cosθcosθ
00 00 11
π6π6 1212 3232
π4π4 2222 2222
π3π3 3232 1212
π2π2 11 00
Table 1.9 Values of sinθsinθ and cosθcosθ at Major Angles θθ in the First Quadrant

Example 1.23

Evaluating Trigonometric Functions

Evaluate each of the following expressions.

  1. sin(2π3)sin(2π3)
  2. cos(5π6)cos(5π6)
  3. tan(15π4)tan(15π4)

Solution

  1. On the unit circle, the angle θ=2π3θ=2π3 corresponds to the point (12,32).(12,32). Therefore, sin(2π3)=y=32.sin(2π3)=y=32.
    An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment that extends diagonally upwards and to the left to another point on the edge of the circle. This point is labeled “(-(1/2), ((square root of 3)/2))”. These line segments have a length of 1 unit. From the point “(-(1/2), ((square root of 3)/2))”, there is a vertical line that extends downwards until it hits the x axis. Inside the circle, there is a curved arrow that starts at the horizontal line segment and travels counterclockwise until it hits the diagonal line segment. This arrow has the label “theta = (2 pi)/3”.
  2. An angle θ=5π6θ=5π6 corresponds to a revolution in the negative direction, as shown. Therefore, cos(5π6)=x=32.cos(5π6)=x=32.
    An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment that extends diagonally downwards and to the left to another point on the edge of the circle. This point is labeled “(-((square root of 3)/2)), -(1/2))”. These line segments have a length of 1 unit. From the point “(-((square root of 3)/2)), -(1/2))”, there is a vertical line that extends upwards until it hits the x axis. Inside the circle, there is a curved arrow that starts at the horizontal line segment and travels clockwise until it hits the diagonal line segment. This arrow has the label “theta = -(5 pi)/6”.
  3. An angle θ=15π4=2π+7π4.θ=15π4=2π+7π4. Therefore, this angle corresponds to more than one revolution, as shown. Knowing the fact that an angle of 7π47π4 corresponds to the point (22,22),(22,22), we can conclude that tan(15π4)=yx=−1.tan(15π4)=yx=−1.
    An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment that extends diagonally downwards and to the right to another point on the edge of the circle. This point is labeled “(((square root of 2)/2), -((square root of 2)/2))”. These line segments have a length of 1 unit. From the point “(((square root of 2)/2), -((square root of 2)/2))”, there is a vertical line that extends upwards until it hits the x axis and thus the horizontal line segment. Inside the circle, there is a curved arrow that starts at the horizontal line segment and travels counterclockwise. The arrow makes one full rotation around the circle and then keeps traveling until it hits the diagonal line segment. This arrow has the label “theta = (15 pi)/4”.
Checkpoint 1.18

Evaluate cos(3π/4)cos(3π/4) and sin(π/6).sin(π/6).

As mentioned earlier, the ratios of the side lengths of a right triangle can be expressed in terms of the trigonometric functions evaluated at either of the acute angles of the triangle. Let θθ be one of the acute angles. Let AA be the length of the adjacent leg, OO be the length of the opposite leg, and HH be the length of the hypotenuse. By inscribing the triangle into a circle of radius H,H, as shown in Figure 1.33, we see that A,H,A,H, and OO satisfy the following relationships with θ:θ:

sinθ=OHcscθ=HOcosθ=AHsecθ=HAtanθ=OAcotθ=AOsinθ=OHcscθ=HOcosθ=AHsecθ=HAtanθ=OAcotθ=AO
An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment with length labeled “H” that extends diagonally upwards and to the right to another point on the edge of the circle. From the point, there is vertical line with a length labeled “O” that extends downwards until it hits the x axis and thus the horizontal line segment at a point with a right triangle symbol. The distance from this point to the center of the circle is labeled “A”. Inside the circle, there is an arrow that points from the horizontal line segment to the diagonal line segment. This arrow has the label “theta”.
Figure 1.33 By inscribing a right triangle in a circle, we can express the ratios of the side lengths in terms of the trigonometric functions evaluated at θ.θ.

Example 1.24

Constructing a Wooden Ramp

A wooden ramp is to be built with one end on the ground and the other end at the top of a short staircase. If the top of the staircase is 44 ft from the ground and the angle between the ground and the ramp is to be 10°,10°, how long does the ramp need to be?

Solution

Let xx denote the length of the ramp. In the following image, we see that xx needs to satisfy the equation sin(10°)=4/x.sin(10°)=4/x. Solving this equation for x,x, we see that x=4/sin(10°)23.035x=4/sin(10°)23.035 ft.

An image of a ramp and a staircase. The ramp starts at a point and increases diagonally upwards and to the right at an angle of 10 degrees for x feet. At the end of the ramp, which is 4 feet off the ground, a staircase descends downwards and to the right.
Checkpoint 1.19

A house painter wants to lean a 2020-ft ladder against a house. If the angle between the base of the ladder and the ground is to be 60°,60°, how far from the house should she place the base of the ladder?

Trigonometric Identities

A trigonometric identity is an equation involving trigonometric functions that is true for all angles θθ for which the functions are defined. We can use the identities to help us solve or simplify equations. The main trigonometric identities are listed next.

Rule: Trigonometric Identities

Reciprocal identities

tanθ=sinθcosθcotθ=cosθsinθcscθ=1sinθsecθ=1cosθtanθ=sinθcosθcotθ=cosθsinθcscθ=1sinθsecθ=1cosθ

Pythagorean identities

sin2θ+cos2θ=11+tan2θ=sec2θ1+cot2θ=csc2θsin2θ+cos2θ=11+tan2θ=sec2θ1+cot2θ=csc2θ

Addition and subtraction formulas

sin(α±β)=sinαcosβ±cosαsinβsin(α±β)=sinαcosβ±cosαsinβ
cos(α±β)=cosαcosβsinαsinβcos(α±β)=cosαcosβsinαsinβ

Double-angle formulas

sin(2θ)=2sinθcosθsin(2θ)=2sinθcosθ
cos(2θ)=2cos2θ1=12sin2θ=cos2θsin2θcos(2θ)=2cos2θ1=12sin2θ=cos2θsin2θ

Example 1.25

Solving Trigonometric Equations

For each of the following equations, use a trigonometric identity to find all solutions.

  1. 1+cos(2θ)=cosθ1+cos(2θ)=cosθ
  2. sin(2θ)=tanθsin(2θ)=tanθ

Solution

  1. Using the double-angle formula for cos(2θ),cos(2θ), we see that θθ is a solution of
    1+cos(2θ)=cosθ1+cos(2θ)=cosθ

    if and only if
    1+2cos2θ1=cosθ,1+2cos2θ1=cosθ,

    which is true if and only if
    2cos2θcosθ=0.2cos2θcosθ=0.

    To solve this equation, it is important to note that we need to factor the left-hand side and not divide both sides of the equation by cosθ.cosθ. The problem with dividing by cosθcosθ is that it is possible that cosθcosθ is zero. In fact, if we did divide both sides of the equation by cosθ,cosθ, we would miss some of the solutions of the original equation. Factoring the left-hand side of the equation, we see that θθ is a solution of this equation if and only if
    cosθ(2cosθ1)=0.cosθ(2cosθ1)=0.

    Since cosθ=0cosθ=0 when
    θ=π2,π2±π,π2±2π,…,θ=π2,π2±π,π2±2π,…,

    and cosθ=1/2cosθ=1/2 when
    θ=π3,π3±2π,…orθ=π3,π3±2π,…,θ=π3,π3±2π,…orθ=π3,π3±2π,…,

    we conclude that the set of solutions to this equation is
    θ=π2+nπ,θ=π3+2nπ,andθ=π3+2nπ,n=0,±1,±2,.θ=π2+nπ,θ=π3+2nπ,andθ=π3+2nπ,n=0,±1,±2,.
  2. Using the double-angle formula for sin(2θ)sin(2θ) and the reciprocal identity for tan(θ),tan(θ), the equation can be written as
    2sinθcosθ=sinθcosθ.2sinθcosθ=sinθcosθ.

    To solve this equation, we multiply both sides by cosθcosθ to eliminate the denominator, and say that if θθ satisfies this equation, then θθ satisfies the equation
    2sinθcos2θsinθ=0.2sinθcos2θsinθ=0.

    However, we need to be a little careful here. Even if θθ satisfies this new equation, it may not satisfy the original equation because, to satisfy the original equation, we would need to be able to divide both sides of the equation by cosθ.cosθ. However, if cosθ=0,cosθ=0, we cannot divide both sides of the equation by cosθ.cosθ. Therefore, it is possible that we may arrive at extraneous solutions. So, at the end, it is important to check for extraneous solutions. Returning to the equation, it is important that we factor sinθsinθ out of both terms on the left-hand side instead of dividing both sides of the equation by sinθ.sinθ. Factoring the left-hand side of the equation, we can rewrite this equation as
    sinθ(2cos2θ1)=0.sinθ(2cos2θ1)=0.

    Therefore, the solutions are given by the angles θθ such that sinθ=0sinθ=0 or cos2θ=1/2.cos2θ=1/2. The solutions of the first equation are θ=0,±π,±2π,….θ=0,±π,±2π,…. The solutions of the second equation are θ=π/4,(π/4)±(π/2),(π/4)±π,….θ=π/4,(π/4)±(π/2),(π/4)±π,…. After checking for extraneous solutions, the set of solutions to the equation is
    θ=nπandθ=π4+nπ2,n=0,±1,±2,.θ=nπandθ=π4+nπ2,n=0,±1,±2,.
Checkpoint 1.20

Find all solutions to the equation cos(2θ)=sinθ.cos(2θ)=sinθ.

Example 1.26

Proving a Trigonometric Identity

Prove the trigonometric identity 1+tan2θ=sec2θ.1+tan2θ=sec2θ.

Solution

We start with the identity

sin2θ+cos2θ=1.sin2θ+cos2θ=1.

Dividing both sides of this equation by cos2θ,cos2θ, we obtain

sin2θcos2θ+1=1cos2θ.sin2θcos2θ+1=1cos2θ.

Since sinθ/cosθ=tanθsinθ/cosθ=tanθ and 1/cosθ=secθ,1/cosθ=secθ, we conclude that

tan2θ+1=sec2θ.tan2θ+1=sec2θ.
Checkpoint 1.21

Prove the trigonometric identity 1+cot2θ=csc2θ.1+cot2θ=csc2θ.

Graphs and Periods of the Trigonometric Functions

We have seen that as we travel around the unit circle, the values of the trigonometric functions repeat. We can see this pattern in the graphs of the functions. Let P=(x,y)P=(x,y) be a point on the unit circle and let θθ be the corresponding angle .. Since the angle θθ and θ+2πθ+2π correspond to the same point P,P, the values of the trigonometric functions at θθ and at θ+2πθ+2π are the same. Consequently, the trigonometric functions are periodic functions. The period of a function ff is defined to be the smallest positive value pp such that f(x+p)=f(x)f(x+p)=f(x) for all values xx in the domain of f.f. The sine, cosine, secant, and cosecant functions have a period of 2π.2π. Since the tangent and cotangent functions repeat on an interval of length π,π, their period is ππ (Figure 1.34).

An image of six graphs. Each graph has an x axis that runs from -2 pi to 2 pi and a y axis that runs from -2 to 2. The first graph is of the function “f(x) = sin(x)”, which is a curved wave function. The graph of the function starts at the point (-2 pi, 0) and increases until the point (-((3 pi)/2), 1). After this point, the function decreases until the point (-(pi/2), -1). After this point, the function increases until the point ((pi/2), 1). After this point, the function decreases until the point (((3 pi)/2), -1). After this point, the function begins to increase again. The x intercepts shown on the graph are at the points (-2 pi, 0), (-pi, 0), (0, 0), (pi, 0), and (2 pi, 0). The y intercept is at the origin. The second graph is of the function “f(x) = cos(x)”, which is a curved wave function. The graph of the function starts at the point (-2 pi, 1) and decreases until the point (-pi, -1). After this point, the function increases until the point (0, 1). After this point, the function decreases until the point (pi, -1). After this point, the function increases again. The x intercepts shown on the graph are at the points (-((3 pi)/2), 0), (-(pi/2), 0), ((pi/2), 0), and (((3 pi)/2), 0). The y intercept is at the point (0, 1). The graph of cos(x) is the same as the graph of sin(x), except it is shifted to the left by a distance of (pi/2). On the next four graphs there are dotted vertical lines which are not a part of the function, but act as boundaries for the function, boundaries the function will never touch. They are known as vertical asymptotes. There are infinite vertical asymptotes for all of these functions, but these graphs only show a few. The third graph is of the function “f(x) = csc(x)”. The vertical asymptotes for “f(x) = csc(x)” on this graph occur at “x = -2 pi”, “x = -pi”, “x = 0”, “x = pi”, and “x = 2 pi”. Between the “x = -2 pi” and “x = -pi” asymptotes, the function looks like an upward facing “U”, with a minimum at the point (-((3 pi)/2), 1). Between the “x = -pi” and “x = 0” asymptotes, the function looks like an downward facing “U”, with a maximum at the point (-(pi/2), -1). Between the “x = 0” and “x = pi” asymptotes, the function looks like an upward facing “U”, with a minimum at the point ((pi/2), 1). Between the “x = pi” and “x = 2 pi” asymptotes, the function looks like an downward facing “U”, with a maximum at the point (((3 pi)/2), -1). The fourth graph is of the function “f(x) = sec(x)”. The vertical asymptotes for this function on this graph are at “x = -((3 pi)/2)”, “x = -(pi/2)”, “x = (pi/2)”, and “x = ((3 pi)/2)”. Between the “x = -((3 pi)/2)” and “x = -(pi/2)” asymptotes, the function looks like an downward facing “U”, with a maximum at the point (-pi, -1). Between the “x = -(pi/2)” and “x = (pi/2)” asymptotes, the function looks like an upward facing “U”, with a minimum at the point (0, 1). Between the “x = (pi/2)” and “x = (3pi/2)” asymptotes, the function looks like an downward facing “U”, with a maximum at the point (pi, -1). The graph of sec(x) is the same as the graph of csc(x), except it is shifted to the left by a distance of (pi/2). The fifth graph is of the function “f(x) = tan(x)”. The vertical asymptotes of this function on this graph occur at “x = -((3 pi)/2)”, “x = -(pi/2)”, “x = (pi/2)”, and “x = ((3 pi)/2)”. In between all of the vertical asymptotes, the function is always increasing but it never touches the asymptotes. The x intercepts on this graph occur at the points (-2 pi, 0), (-pi, 0), (0, 0), (pi, 0), and (2 pi, 0). The y intercept is at the origin. The sixth graph is of the function “f(x) = cot(x)”. The vertical asymptotes of this function on this graph occur at “x = -2 pi”, “x = -pi”, “x = 0”, “x = pi”, and “x = 2 pi”. In between all of the vertical asymptotes, the function is always decreasing but it never touches the asymptotes. The x intercepts on this graph occur at the points (-((3 pi)/2), 0), (-(pi/2), 0), ((pi/2), 0), and (((3 pi)/2), 0) and there is no y intercept.
Figure 1.34 The six trigonometric functions are periodic.

Just as with algebraic functions, we can apply transformations to trigonometric functions. In particular, consider the following function:

f(x)=Asin(B(xα))+C.f(x)=Asin(B(xα))+C.
1.10

In Figure 1.35, the constant αα causes a horizontal or phase shift. The factor BB changes the period. This transformed sine function will have a period 2π/|B|.2π/|B|. The factor AA results in a vertical stretch by a factor of |A|.|A|. We say |A||A| is the “amplitude of f.f.” The constant CC causes a vertical shift.

An image of a graph. The graph is of the function “f(x) = Asin(B(x - alpha)) + C”. Along the y axis, there are 3 hash marks: starting from the bottom and moving up, the hash marks are at the values “C - A”, “C”, and “C + A”. The distance from the origin to “C” is labeled “vertical shift”. The distance from “C - A” to “A” and the distance from “A” to “C + A” is “A”, which is labeled “amplitude”. On the x axis is a hash mark at the value “alpha” and the distance between the origin and “alpha” is labeled “horizontal shift”. The distance between two successive minimum values of the function (in other words, the distance between two bottom parts of the wave that are next to each other) is “(2 pi)/(absolute value of B)” is labeled the period. The period is also the distance between two successive maximum values of the function.
Figure 1.35 A graph of a general sine function.

Notice in Figure 1.34 that the graph of y=cosxy=cosx is the graph of y=sinxy=sinx shifted to the left π/2π/2 units. Therefore, we can write cosx=sin(x+π/2).cosx=sin(x+π/2). Similarly, we can view the graph of y=sinxy=sinx as the graph of y=cosxy=cosx shifted right π/2π/2 units, and state that sinx=cos(xπ/2).sinx=cos(xπ/2).

A shifted sine curve arises naturally when graphing the number of hours of daylight in a given location as a function of the day of the year. For example, suppose a city reports that June 21 is the longest day of the year with 15.715.7 hours and December 21 is the shortest day of the year with 8.38.3 hours. It can be shown that the function

h(t)=3.7sin(2π365(x80.5))+12h(t)=3.7sin(2π365(x80.5))+12

is a model for the number of hours of daylight hh as a function of day of the year tt (Figure 1.36).

An image of a graph. The x axis runs from 0 to 365 and is labeled “t, day of the year”. The y axis runs from 0 to 20 and is labeled “h, number of daylight hours”. The graph is of the function “h(t) = 3.7sin(((2 pi)/365)(t - 80.5)) + 12”, which is a curved wave function. The function starts at the approximate point (0, 8.4) and begins increasing until the approximate point (171.8, 15.7). After this point, the function decreases until the approximate point (354.3, 8.3). After this point, the function begins increasing again.
Figure 1.36 The hours of daylight as a function of day of the year can be modeled by a shifted sine curve.

Example 1.27

Sketching the Graph of a Transformed Sine Curve

Sketch a graph of f(x)=3sin(2(xπ4))+1.f(x)=3sin(2(xπ4))+1.

Solution

This graph is a phase shift of y=sin(x)y=sin(x) to the right by π/4π/4 units, followed by a horizontal compression by a factor of 2, a vertical stretch by a factor of 3, and then a vertical shift by 1 unit. The period of ff is π.π.

An image of a graph. The x axis runs from -((3 pi)/2) to 2 pi and the y axis runs from -3 to 5. The graph is of the function “f(x) = 3sin(2(x-(pi/4))) + 1”, which is a curved wave function. The function starts decreasing from the point (-((3 pi)/2), 4) until it hits the point (-pi, -2). At this point, the function begins increasing until it hits the point (-(pi/2), 4). After this point, the function begins decreasing until it hits the point (0, -2). After this point, the function increases until it hits the point ((pi/2), 4). After this point, the function decreases until it hits the point (pi, -2). After this point, the function increases until it hits the point (((3 pi)/2), 4). After this point, the function decreases again.
Checkpoint 1.22

Describe the relationship between the graph of f(x)=3sin(4x)5f(x)=3sin(4x)5 and the graph of y=sin(x).y=sin(x).

Section 1.3 Exercises

For the following exercises, convert each angle in degrees to radians. Write the answer as a multiple of π.π.

113.

240°240°

114.

15°15°

115.

−60°−60°

116.

−225°−225°

117.

330°330°

For the following exercises, convert each angle in radians to degrees.

118.

π2radπ2rad

119.

7π6rad7π6rad

120.

11π2rad11π2rad

121.

−3πrad−3πrad

122.

5π12rad5π12rad

Evaluate the following functional values.

123.

cos(4π3)cos(4π3)

124.

tan(19π4)tan(19π4)

125.

sin(3π4)sin(3π4)

126.

sec(π6)sec(π6)

127.

sin(π12)sin(π12)

128.

cos(5π12)cos(5π12)

For the following exercises, consider triangle ABC, a right triangle with a right angle at C. a. Find the missing side of the triangle. b. Find the six trigonometric function values for the angle at A. Where necessary, round to one decimal place.

An image of a triangle. The three corners of the triangle are labeled “A”, “B”, and “C”. Between the corner A and corner C is the side b. Between corner C and corner B is the side a. Between corner B and corner A is the side c. The angle of corner C is marked with a right triangle symbol. The angle of corner A is marked with an angle symbol.
129.

a=4,c=7a=4,c=7

130.

a=21,c=29a=21,c=29

131.

a=85.3,b=125.5a=85.3,b=125.5

132.

b=40,c=41b=40,c=41

133.

a=84,b=13a=84,b=13

134.

b=28,c=35b=28,c=35

For the following exercises, PP is a point on the unit circle. a. Find the (exact) missing coordinate value of each point and b. find the values of the six trigonometric functions for the angle θθ with a terminal side that passes through point P.P. Rationalize denominators.

135.

P(725,y),y>0P(725,y),y>0

136.

P(−1517,y),y<0P(−1517,y),y<0

137.

P(x,73),x<0P(x,73),x<0

138.

P(x,154),x>0P(x,154),x>0

For the following exercises, simplify each expression by writing it in terms of sines and cosines, then simplify. The final answer does not have to be in terms of sine and cosine only.

139.

tan2x+sinxcscxtan2x+sinxcscx

140.

secxsinxcotxsecxsinxcotx

141.

tan2xsec2xtan2xsec2x

142.

secxcosxsecxcosx

143.

(1+tanθ)22tanθ(1+tanθ)22tanθ

144.

sinx(cscxsinx)sinx(cscxsinx)

145.

costsint+sint1+costcostsint+sint1+cost

146.

1+tan2α1+cot2α1+tan2α1+cot2α

For the following exercises, verify that each equation is an identity.

147.

tanθcotθcscθ=sinθtanθcotθcscθ=sinθ

148.

sec2θtanθ=secθcscθsec2θtanθ=secθcscθ

149.

sintcsct+costsect=1sintcsct+costsect=1

150.

sinxcosx+1+cosx1sinx=0sinxcosx+1+cosx1sinx=0

151.

cotγ+tanγ=secγcscγcotγ+tanγ=secγcscγ

152.

sin2β+tan2β+cos2β=sec2βsin2β+tan2β+cos2β=sec2β

153.

11sinα+11+sinα=2sec2α11sinα+11+sinα=2sec2α

154.

tanθcotθsinθcosθ=sec2θcsc2θtanθcotθsinθcosθ=sec2θcsc2θ

For the following exercises, solve the trigonometric equations on the interval 0θ<2π.0θ<2π.

155.

2sinθ1=02sinθ1=0

156.

1+cosθ=121+cosθ=12

157.

2tan2θ=22tan2θ=2

158.

4sin2θ2=04sin2θ2=0

159.

3cotθ+1=03cotθ+1=0

160.

3secθ23=03secθ23=0

161.

2cosθsinθ=sinθ2cosθsinθ=sinθ

162.

csc2θ+2cscθ+1=0csc2θ+2cscθ+1=0

For the following exercises, each graph is of the form y=AsinBxy=AsinBx or y=AcosBx,y=AcosBx, where B>0.B>0. Write the equation of the graph.

163.
An image of a graph. The x axis runs from -4 to 4 and the y axis runs from -5 to 5. The graph is of a curved wave function that starts at the point (-4, 0) and decreases until the point (-2, 4). After this point the function begins increasing until it hits the point (2, 4). After this point the function begins decreasing again. The x intercepts of the function on this graph are at (-4, 0), (0, 0), and (4, 0). The y intercept is at the origin.
164.
An image of a graph. The x axis runs from -4 to 4 and the y axis runs from -5 to 5. The graph is of a curved wave function that starts at the point (-4, -2) and increases until the point (-3, 2). After this point the function decreases until it hits the point (-2, -2). After this point the function increases until it hits the point (-1, 2). After this point the function decreases until it hits the point (0, -2). After this point the function increases until it hits the point (1, 2). After this point the function decreases until it hits the point (2, -2). After this point the function increases until it hits the point (3, 2). After this point the function begins decreasing again. The x intercepts of the function on this graph are at (-3.5, 0), (-2.5, 0), (-1.5, 0), (-0.5, 0), (0.5, 0), (1.5, 0), (2.5, 0), and (3.5, 0). The y intercept is at the (0, -2).
165.
An image of a graph. The x axis runs from -4 to 4 and the y axis runs from -5 to 5. The graph is of a curved wave function. There are many periods and only a few will be explained. The function begins decreasing at the point (-1, 1) and decreases until the point (-0.5, -1). After this point the function increases until it hits the point (0, 1). After this point the function decreases until it hits the point (0.5, -1). After this point the function increases until it hits the point (1, 1). After this point the function decreases again. The x intercepts of the function on this graph are at (-0.75, 0), (-0.25, 0), (0.25, 0), and (0.75, 0). The y intercept is at (0, 1).
166.
An image of a graph. The x axis runs from -4 to 4 and the y axis runs from -5 to 5. The graph is of a curved wave function. There are many periods and only a few will be explained. The function begins decreasing at the point (-1.25, 0.75) and decreases until the point (-0.75, -0.75). After this point the function increases until it hits the point (0.25, 0.75). After this point the function decreases until it hits the point (0.25, -0.75). After this point the function increases until it hits the point (0.75, 0.75). After this point the function decreases again. The x intercepts of the function on this graph are at (-1, 0), (-0.5, 0), (0, 0), and (0.5, 0). The y intercept is at the origin.

For the following exercises, find a. the amplitude, b. the period, and c. the phase shift with direction for each function.

167.

y=sin(xπ4)y=sin(xπ4)

168.

y=3cos(2x+3)y=3cos(2x+3)

169.

y=−12sin(14x)y=−12sin(14x)

170.

y=2cos(xπ3)y=2cos(xπ3)

171.

y=−3sin(πx+2)y=−3sin(πx+2)

172.

y=4cos(2xπ2)y=4cos(2xπ2)

173.

[T] The diameter of a wheel rolling on the ground is 40 in. If the wheel rotates through an angle of 120°,120°, how many inches does it move? Approximate to the nearest whole inch.

174.

[T] Find the length of the arc intercepted by central angle θθ in a circle of radius r. Round to the nearest hundredth.

a. r=12.8r=12.8 cm, θ=5π6θ=5π6 rad b. r=4.378r=4.378 cm, θ=7π6θ=7π6 rad c. r=0.964r=0.964 cm, θ=50°θ=50° d. r=8.55r=8.55 cm, θ=325°θ=325°

175.

[T] As a point P moves around a circle, the measure of the angle changes. The measure of how fast the angle is changing is called angular speed, ω,ω, and is given by ω=θ/t,ω=θ/t, where θθ is in radians and t is time. Find the angular speed for the given data. Round to the nearest thousandth.

a. θ=7π4rad,t=10θ=7π4rad,t=10 sec b. θ=3π5rad,t=8θ=3π5rad,t=8 sec c. θ=2π9rad,t=1θ=2π9rad,t=1 min d. θ=23.76rad,t=14θ=23.76rad,t=14 min

176.

[T] A total of 250,000 m2 of land is needed to build a nuclear power plant. Suppose it is decided that the area on which the power plant is to be built should be circular.

  1. Find the radius of the circular land area.
  2. If the land area is to form a 45°45° sector of a circle instead of a whole circle, find the length of the curved side.
177.

[T] The area of an isosceles triangle with equal sides of length x is

12x2sinθ,12x2sinθ,

where θθ is the angle formed by the two sides. Find the area of an isosceles triangle with equal sides of length 8 in. and angle θ=5π/12θ=5π/12 rad.

178.

[T] A particle travels in a circular path at a constant angular speed ω.ω. The angular speed is modeled by the function ω=9|cos(πtπ/12)|.ω=9|cos(πtπ/12)|. Determine the angular speed at t=9t=9 sec.

179.

[T] An alternating current for outlets in a home has voltage given by the function

V(t)=150cos368t,V(t)=150cos368t,

where V is the voltage in volts at time t in seconds.

  1. Find the period of the function and interpret its meaning.
  2. Determine the number of periods that occur when 1 sec has passed.
180.

[T] The number of hours of daylight in a northeast city is modeled by the function

N(t)=12+3sin[2π365(t79)],N(t)=12+3sin[2π365(t79)],

where t is the number of days after January 1.

  1. Find the amplitude and period.
  2. Determine the number of hours of daylight on the longest day of the year.
  3. Determine the number of hours of daylight on the shortest day of the year.
  4. Determine the number of hours of daylight 90 days after January 1.
  5. Sketch the graph of the function for one period starting on January 1.
181.

[T] Suppose that T=50+10sin[π12(t8)]T=50+10sin[π12(t8)] is a mathematical model of the temperature (in degrees Fahrenheit) at t hours after midnight on a certain day of the week.

  1. Determine the amplitude and period.
  2. Find the temperature 7 hours after midnight.
  3. At what time does T=60°?T=60°?
  4. Sketch the graph of TT over 0t24.0t24.
182.

[T] The function H(t)=8sin(π6t)H(t)=8sin(π6t) models the height H (in feet) of the tide t hours after midnight. Assume that t=0t=0 is midnight.

  1. Find the amplitude and period.
  2. Graph the function over one period.
  3. What is the height of the tide at 4:30 a.m.?
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