Chapter 4 Angles and Their Measure

Chapter 4 Angles and Their Measure

Angles An angle is formed by two rays that have a common endpoint called the vertex. One ray is called the initial side and the other the terminal side. The arrow near the vertex shows the direction and the amount of rotation from the initial side to the terminal side. A è B C Terminal Side Initial Side Vertex

Angles of the Rectangular Coordinate System
An angle is in standard position if its vertex is at the origin of a rectangular coordinate system and its initial side lies along the positive x-axis. x y Terminal Side Initial Side Vertex   is positive Positive angles rotate counterclockwise. x y Terminal Side Initial Side Vertex   is negative Negative angles rotate clockwise.

Measuring Angles Using Degrees
The figures below show angles classified by their degree measurement. An acute angle measures less than 90º. A right angle, one quarter of a complete rotation, measures 90º and can be identified by a small square at the vertex. An obtuse angle measures more than 90º but less than 180º. A straight angle has measure 180º. 90º Right angle 1/4 rotation  Acute angle 0º <  < 90º  Obtuse angle 90º <  < 180º 180º Straight angle 1/2 rotation

Coterminal Angles An angle of xº is coterminal with angles of
xº + k · 360º where k is an integer.

Text Example Assume the following angles are in standard position. Find a positive angle less than 360º that is coterminal with: a. a 420º angle b. a –120º angle. Solution We obtain the coterminal angle by adding or subtracting 360º. Our need to obtain a positive angle less than 360º determines whether we should add or subtract. a. For a 420º angle, subtract 360º to find a positive coterminal angle. 420º – 360º = 60º A 60º angle is coterminal with a 420º angle. These angles, shown on the next slide, have the same initial and terminal sides.

Text Example cont. Solution
y 420º 60º x y 240º -120º Solution b. For a –120º angle, add 360º to find a positive coterminal angle. -120º + 360º = 240º A 240º angle is coterminal with a –120º angle. These angles have the same initial and terminal sides.

Finding Complements and Supplements
For an xº angle, the complement is a 90º – xº angle. Thus, the complement’s measure is found by subtracting the angle’s measure from 90º. For an xº angle, the supplement is a 180º – xº angle. Thus, the supplement’s measure is found by subtracting the angle’s measure from 180º.

Definition of a Radian One radian is the measure of the central angle of a circle that intercepts an arc equal in length to the radius of the circle.

Radian Measure Consider an arc of length s on a circle or radius r.
The measure of the central angle that intercepts the arc is  = s/r radians.  O r s

Conversion between Degrees and Radians
Using the basic relationship  radians = 180º, To convert degrees to radians, multiply degrees by ( radians) / 180 To convert radians to degrees, multiply radians by 180 / ( radians)

Example Convert each angle in degrees to radians 40º 75º -160º

Example cont. Solution: 40º = 40*/180 = 2  /9
75º = 75*  /180 = 5  /12 -160º = -160*  /180 = -8  /9

Length of a Circular Arc
Let r be the radius of a circle and  the non- negative radian measure of a central angle of the circle. The length of the arc intercepted by the central angle is s = r   O s r

Example A circle has a radius of 7 inches. Find the length of the arc intercepted by a central angle of 2/3 Solution: s = (7 inches)*(2  /3) =14  /3 inches

Definitions of Linear and Angular Speed
If a point is in motion on a circle of radius r through an angle of  radians in time t, then its linear speed is v = s/t where s is the arc length given by s = r , and its angular speed is  = /t

Linear Speed in Terms of Angular Speed
The linear speed, v, of a point a distance r from the center of rotation is given by v=r where  is the angular speed in radians per unit of time.

Trigonometric Functions: The Unit Circle

Definitions of the Trigonometric Functions in Terms of a Unit Circle
If t is a real number and P = (x, y) is a point on the unit circle that corresponds to t, then

Text Example Use the Figure to find the values of the
trigonometric functions at t=/2. Solution: The point P on the unit circle that Corresponds to t= /2 has coordinates (0,1). We use x=0 and y=1 to find the Values of the trigonometric functions

The Domain and Range of the Sine and Cosine Functions
The domain of the sine function and the cosine function is the set of all real numbers The range of these functions is the set of all real numbers from -1 to 1, inclusive.

Trigonometric Functions at /4.

Even and Odd Trigonometric Functions
The cosine and secant functions are even. cos(-t) = cos t sec(-t) = sec t The sine, cosecant, tangent, and cotangent functions are odd. sin(-t) = -sin t csc(-t) = -csc t tan(-t) = -tan t cot(-t) = -cot t

Example Use the value of the trigonometric function at t = /4 to find sin (- /4 ). Solution: sin(-t) = -sin t, so sin(- /4 ) = -sin(/4 ) = -2/2

Reciprocal Identities

Quotient Identities

Pythagorean Identities

Definition of a Periodic Function
A function f is periodic if there exists a positive number p such that f(t + p) = f(t) For all t in the domain of f. The smallest number p for which f is periodic is called the period of f.

Periodic Properties of the Sine and Cosine Functions
sin(t + 2) = sin t and cos(t + 2) = cos t The sine and cosine functions are periodic functions and have period 2.

Periodic Properties of the Tangent and Cotangent Functions
tan(t + ) = tan t and cot(t + ) = cot t The tangent and cotangent functions are periodic functions and have period .

Repetitive Behavior of the Sine, Cosine, and Tangent Functions
For any integer n and real number t, sin(t + 2n) = sin t, cos(t + 2n) = cos t, and tan(t + n) = tan t

Right Triangle Trigonometry

The Six Trigonometric Functions
The figure below shows a right triangle with one of its acute angles labeled . The side opposite the right angle is known as the hypotenuse. The other sides of triangle are described by the position relative to the acute angle . One side is opposite  and one is adjacent to .  Hypotenuse Side opposite . Side adjacent to .

Right Triangle Definitions of Trigonometric Functions
Hyp Opp  Adj

Example

Sines, Cosines, and Tangents of Special Angles

Cofunction Identities
The value of a trigonometric function of  is equal to the cofunction of the complement of . sin  = cos (90º – ) cos  = sin (90º – ) tan  = cot (90º – ) cot  = tan (90º – ) sec  = csc (90º – ) csc  =sec (90º – ) If  is in radians, replace 90º with /2.

Text Example Sighting the top of a building, a surveyor measured the angle of elevation to be 22º. The transit is 5 feet above the ground and 300 feet from the building. Find the building’s height. Solution Let a be the height of the portion of the building that lies above the transit in the figure shown. The height of the building is the transit’s height, 5 feet, plus a. Thus, we need to identify a trigonometric function that will make it possible to find a. In terms of the 22º angle, we are looking for the side opposite the angle. 22º Line of sight 300 feet a 5 feet h Transit

The transit is 300 feet from the building, so the side adjacent to the 22º angle is 300 feet. Because we have a known angle, an unknown opposite side, and a known adjacent side, we select the tangent function. Solution 22º Line of sight 300 feet a 5 feet h tan 22º = 300 Length of side opposite the 22º angle Length of side adjacent to the 22º angle a = 300 tan 22º (0.4040) Multiply both sides of the equation by 300. The height of the part of the building above the transit is approximately 121 feet. If we add the height of the transit, 5 feet, the building’s height is approximately 126 feet. Transit

Trigonometric Functions of Any Angle

Definitions of Trigonometric Functions of Any Angle
Let  is be any angle in standard position, and let P = (x, y) be a point on the terminal side of . If r = x2 + y2 is the distance from (0, 0) to (x, y), the six trigonometric functions of  are defined by the following ratios.

Text Example Let P = (-3, -4) be a point on the terminal side of . Find each of the six trigonometric functions of . Solution The situation is shown below. We need values for x, y, and r to evaluate all six trigonometric functions. We are given the values of x and y. Because P = (-3, -4) is a point on the terminal side of , x = -3 and y = -4. Furthermore, r x = -3 y = -4 P = (-3, -4)  x y -5 5

The bottom row shows the reciprocals of the row above.
Text Example Cont. Solution Now that we know x, y, and r, we can find the six trigonometric functions of . The bottom row shows the reciprocals of the row above.

The Signs of the Trigonometric Functions
x y Quadrant II Sine and cosecant positive Quadrant I All functions positive Quadrant III tangent and cotangent positive Quadrant IV cosine and secant positive

Definition of a Reference Angle
Let  be a nonacute angle in standard position that lies in a quadrant. Its reference angle is the positive acute angle ´ prime formed by the terminal side or  and the x-axis.

Example Find the reference angle , for the following angle:  =315º
Solution:  =360º - 315º = 45º a b 315 45 P ( , )

Using Reference Angles to Evaluate Trigonometric Functions
The values of a trigonometric functions of a given angle, , are the same as the values for the trigonometric functions of the reference angle, ´, except possibly for the sign. A function value of the acute angle, ´, is always positive. However, the same functions value for  may be positive or negative.

A Procedure for Using Reference Angles to Evaluate Trigonometric Functions
The value of a trigonometric function of any angle  is found as follows: Find the associated reference angle, ´, and the function value for ´. Use the quadrant in which  lies to prefix the appropriate sign to the function value in step 1.

Text Example Use reference angles to find the exact value of the following trigonometric functions. a. sin 135° Solution a. We use our two-step procedure to find sin 135°. Step 1 Find the reference angle,  ´, and sin  ´. 135º terminates in quadrant II with a reference angle  ´ = 180º – 135º = 45º. x y 135° 45°

The function value for the reference angle is sin 45º = 2 / 2. Step 2 Use the quadrant in which è lies to prefix the appropriate sign to the function value in step 1. The angle 135º lies in quadrant II. Because the sine is positive in quadrant II, we put a + sign before the function value of the reference angle. Thus, sin135= +sin45=2 / 2

Graphs of Sine and Cosine Functions

The Graph of y=sinx The trigonometric functions can be graphed in a rectangular coordinate system by plotting points whose coordinates belong to the function. Thus, we graph y = sin x by listing some points on the graph. Because the period of the sine function is 2 we will graph the function on the interval [0, 2]. The rest of the graph is made up of repetitions of this portion. 2  ˝ 1 -1 Period: 2 x y 6  3 2 3  y = sin x, 0 < x < 2

Graphing Variations of y=sinx
Identify the amplitude and the period. Find the values of x for the five key points – the three x-intercepts, the maximum point, and the minimum point. Start with the value of x where the cycle begins and add quarter-periods – that is, period/4 – to find successive values of x. Find the values of y for the five key points by evaluating the function at each value of x from step 2. Connect the five key points with a smooth curve and graph one complete cycle of the given function. Extend the graph in step 4 to the left or right as desired.

Text Example Determine the amplitude of y = 1/2 sin x. Then graph y = sin x and y = 1/2 sin x for 0 < x < 2. Solution Step 1 Identify the amplitude and the period. The equation y = 1/2 sin x is of the form y = A sin x with A = 1/2. Thus, the amplitude |A| = 1/2. This means that the maximum value of y is 1/2 and the minimum value of y is -1/2. The period for both y = 1/2 sin x and y = sin x is 2.

Step 2 Find the values of x for the five key points. We need to find the three x-intercepts, the maximum point, and the minimum point on the interval [0, 2]. To do so, we begin by dividing the period, 2, by 4. Period/4 = 2 /4 = /2 We start with the value of x where the cycle begins: x = 0. Now we add quarter periods,  /2, to generate x-values for each of the key points. The five x-values are x = 0, x = 0 + /2 = /2, x = /2 + /2 = , x =  + /2 = 3 /2, x= 3 /2 + /2 = 2 

Step 3 Find the values of y for the five key points. We evaluate the function at each value of x from step 2. y = 1/2 sin0 = 0 (0,0) y = 1/2sin/2 = 1/2*1 = 1/2 (/2, 1/2) y = 1/2sin = 1/2*0 = 0 (,0) y = 1/2 sin 3/2 = 1/2(-1) = -1/2 (3 /2, -1/2) y = 1/2sin2 = 1/2*0 = 0 (2, 0) There are x-intercepts at 0,  and 2 . The maximum and minimum points are (/2, 1/2) and (3 /2, -1/2)

Step 4 Connect the five key points with a smooth curve and graph one complete cycle of the given function. The five key points for y = 1/2sin x are shown below. By connecting the points with a smooth curve, the figure shows one complete cycle of y = 1/2sin x. Also shown is graph of y = sin x. The graph of y = 1/2sin x shrinks the graph of y = sin x. x y 2˝ ˝ -1 1 y = sin x y = 1/2sinx

Amplitudes and Periods
The graph of y = A sin Bx has amplitude = | A| period = 2 /B. x y y = A sin Bx Amplitude: | A| Period: 2 /B

The Graph of y = Asin(Bx - C)
The graph of y = A sin (Bx – C) is obtained by horizontally shifting the graph of y = A sin Bx so that the starting point of the cycle is shifted from x = 0 to x = C/B. The number C/B is called the phase shift. amplitude = | A| period = 2 /B. y y = A sin Bx Amplitude: | A| x Starting point: x = C/B Period: 2/B

Example Determine the amplitude, period, and phase shift of y = 2sin(3x-) Solution: Amplitude = |A| = 2 period = 2/B = 2/3 phase shift = C/B = /3

Example cont. y = 2sin(3x- )

The Graph of y = AcosBx We graph y = cos x by listing some points on the graph. Because the period of the cosine function is 2, we will concentrate on the graph of the basic cosine curve on the interval [0, 2 ]. The rest of the graph is made up of repetitions of this portion. . /2  3/2 2 5/2 - -/2 -3 /2 1 -1 y x Period: 2

The Graph of y = AcosBx The graph of y = A cos Bx has amplitude = | A|
period = 2/B. x y y = A cos Bx Amplitude: | A| Period: 2/B 2/B

The Graph of y = Acos(Bx - C)
The graph of y = A cos (Bx – C) is obtained by horizontally shifting the graph of y = A cos Bx so that the starting point of the cycle is shifted from x = 0 to x = C/B. The number C/B is called the phase shift. amplitude = | A| period = 2 /B. x y y = A cos( Bx-C) Amplitude: | A| Period: 2/B

Graphs of Other Trigonometric Functions

The Tangent Curve: The Graph of y=tanx and Its Characteristics
Period:  Domain: All real numbers except  /2 + k , k an integer Range: All real numbers Symmetric with respect to the origin Vertical asymptotes at odd multiples of  /2 1 –2  –   2  x 5  3    3  5  – – – 2 2 2 2 2 2 –1

Graphing y = A tan(Bx – C)
1. Find two consecutive asymptotes by setting the variable expression in the tangent equal to -/ and /2 and solving Bx – C = -/2 and Bx – C = /2 2. Identify an x-intercept, midway between consecutive asymptotes. 3. Find the points on the graph 1/4 and 3/4 of the way between and x-intercept and the asymptotes. These points have y-coordinates of –A and A. 4. Use steps 1-3 to graph one full period of the function. Add additional cycles to the left or right as needed. y = A tan (Bx – C) Bx – C =  /2 Bx – C = - /2 x x-intercept between asymptotes

Text Example Graph y = 2 tan x/2 for – < x < 3  Solution
Step 1 Find two consecutive asymptotes. Thus, two consecutive asymptotes occur at x = -  and x = . Step 2 Identify any x-intercepts, midway between consecutive asymptotes. Midway between x = - and x =  is x = 0. An x-intercept is 0 and the graph passes through (0, 0).

Step 3 Find points on the graph 1/4 and 1/4 of the way between an x-intercept and the asymptotes. These points have y-coordinates of –A and A. Because A, the coefficient of the tangent, is 2, these points have y-coordinates of -2 and 2. Step 4 Use steps 1-3 to graph one full period of the function. We use the two consecutive asymptotes, x = - and x = , an x-intercept of 0, and points midway between the x-intercept and asymptotes with y-coordinates of –2 and 2. We graph one full period of y = 2 tan x/2 from – to . In order to graph for – < x < 3 , we continue the pattern and extend the graph another full period on the right. y -4 -2 2 4 ˝ x -˝ 3˝ y = 2 tan x/2

The Cotangent Curve: The Graph of y = cotx and Its Characteristics
Period:  Domain: All real numbers except integral multiples of  Range: All real numbers Vertical asymptotes: at integral multiples of  n x-intercept occurs midway between each pair of consecutive asymptotes. Odd function with origin symmetry Points on the graph 1/4 and 3/4 of the way between consecutive asymptotes have y- coordinates of –1 and 1. y -4 -2 2 4 x -  -  /2  /2 ˝ 3  /2 2 

Graphing y=Acot(Bx-C)
1. Find two consecutive asymptotes by setting the variable expression in the cotangent equal to and ˝ and solving Bx – C = 0 and Bx – C =  2. Identify an x-intercept, midway between consecutive asymptotes. 3. Find the points on the graph 1/4 and 3/4 of the way between an x-intercept and the asymptotes. These points have y-coordinates of –A and A. 4. Use steps 1-3 to graph one full period of the function. Add additional cycles to the left or right as needed. y = A cot (Bx – C) Bx – C =  y-coord-inate is A. x x-intercept between asymptotes y-coord-inate is -A. Bx – C = 0

Example Graph y = 2 cot 3x Solution: 3x=0 and 3x=
x=0 and x = /3 are vertical asymptotes An x-intercepts occurs between 0 and /3 so an x-intercepts is at (/6,0) The point on the graph midway between the asymptotes and intercept are /12 and 3/12. These points have y-coordinates of -A and A or -2 and 2 Graph one period and extend as needed

Example cont Graph y = 2 cot 3x

The Cosecant Curve: The Graph of y = cscx and Its Characteristics
Period: 2 Domain: All real numbers except integral multiples of  Range: All real numbers y such that y < -1 or y > 1 Vertical asymptotes: at integral multiples of  Odd function with origin symmetry y -  /2 1 3  /2 x -2 -3  /2 -˝  /2 ˝ 2  -1

The Secant Curve: The Graph of y=secx and Its Characteristics
Period: 2 Domain: All real numbers except odd multiples of  /2 Range: All real numbers y such that y < -1 or y > 1 Vertical asymptotes: at odd multiples of / Even function with origin symmetry y 1 x -2 -3  /2 -  -˝ /2 ˝ /2  3  /2 2 -1

Text Example Use the graph of y = 2 sin 2x to obtain the graph of y = 2 csc 2x. y -2 2 x ˝ -˝ y -2 2 x ˝ Solution The x-intercepts of y = 2 sin 2x correspond to the vertical asymptotes of y = 2 csc 2x. Thus, we draw vertical asymptotes through the x-intercepts. Using the asymptotes as guides, we sketch the graph of y = 2 csc 2x.

Graphs of Other Trigonometric Functions

Inverse Trigonometric Functions

The Inverse Sine Function
The inverse sine function, denoted by sin-1, is the inverse of the restricted sine function y = sin x, -  /2 < x <  / 2. Thus, y = sin-1 x means sin y = x, where -  /2 < y <  /2 and –1 < x < 1. We read y = sin-1 x as “ y equals the inverse sine at x.” y -1 1  /2 x -  /2 y = sin x -  /2 < x < /2 Domain: [-  /2,  /2] Range: [-1, 1]

Finding Exact Values of sin-1x
Let  = sin-1 x. Rewrite step 1 as sin  = x. Use the exact values in the table to find the value of  in [-/2 , /2] that satisfies sin  = x.

Example Find the exact value of sin-1(1/2)

The Inverse Cosine Function
The inverse cosine function, denoted by cos-1, is the inverse of the restricted cosine function y = cos x, 0 < x < . Thus, y = cos-1 x means cos y = x, where 0 < y <  and –1 < x < 1.

Text Example Find the exact value of cos-1 (-3 /2) Solution
Step 1 Let  = cos-1 x. Thus =cos-1 (-3 /2) We must find the angle , 0 <  < , whose cosine equals -3 /2 Step 2 Rewrite  = cos-1 x as cos  = x. We obtain cos  = (-3 /2)

Text Example cont. Find the exact value of cos-1 (-3 /2) Solution
Step 3 Use the exact values in the table to find the value of  in [0, ] that satisfies cos  = x. The table on the previous slide shows that the only angle in the interval [0, ] that satisfies cos  = (-3 /2) is 5/6. Thus,  = 5/6

The Inverse Tangent Function
The inverse tangent function, denoted by tan-1, is the inverse of the restricted tangent function y = tan x, -/2 < x < /2. Thus, y = tan-1 x means tan y = x, where -  /2 < y <  /2 and –  < x < .

Inverse Properties The Sine Function and Its Inverse
sin (sin-1 x) = x for every x in the interval [-1, 1]. sin-1(sin x) = x for every x in the interval [-/2,/2]. The Cosine Function and Its Inverse cos (cos-1 x) = x for every x in the interval [-1, 1]. cos-1(cos x) = x for every x in the interval [0, ]. The Tangent Function and Its Inverse tan (tan-1 x) = x for every real number x tan-1(tan x) = x for every x in the interval (-/2,/2).

Applications of Trigonometric Functions

Solving Right Triangles
Solving a right triangle means finding the missing lengths of its sides and the measurements of its angles. We will label right triangles so that side a is opposite angle A, side b is opposite angle B, and side c is the hypotenuse opposite right angle C. C A B a b c

Text Example Solve the right triangle shown.
C A B a b = 10.5 c 34.5º Solve the right triangle shown. Solution We begin by finding the the measure of angle B. We do not need a trigonometric function to do so. Because C = 90º and the sum of a triangle’s angles is 180, we see that A + B = 90º. Thus, B = 90º – A = 90º – 34.5º = 55.5º. Now we need to find a. Because we have a known angle, and unknown opposite side, and a known adjacent side, we use the tangent function. tan34.5º = a/10.5 Now we solve for a. A = 10.5tan34.5=7.22

Text Example cont. Solve the right triangle shown. Solution
B Solve the right triangle shown. Solution Finally, we need to find c. Because we have a known angle, a known adjacent side, and an unknown hypotenuse, we use the cosine function. os34.5 = 10.5/c c=10.5/cos34.5 = 12.74 In summary, B = 55.5º, a = 7.22, and c = c 55.5º 7.22 34.5º A b = 10.5 C

Text Example Use the figure to find: a. the bearing from O to B. b. the bearing from O to A. W N E S A B C D O 40º 75º 25º 20º Solution a. To find the bearing from O to B, we need the acute angle between the ray OB and the north-south line through O. The measurement of this angle is given to be 40º. The figure shows that the angle is measured from the north side of the north-south line and lies west of the north-south line. Thus, the bearing from O to B is N 40º W.

Text Example cont. Use the figure to find: a. the bearing from O to B. b. the bearing from O to A. N 40º B A 20º W E O 75º D C 25º Solution S b. To find the bearing from O to A, we need the acute angle between the ray OA and the north-south line through O. This angle is specified by the voice balloon in the figure. The figure shows that this angle measures 90º – 20º, or 70º. This angle is measured from the north side of the north-south line. This angle is also east of the north-south line. This angle is also east of the north-south line. Thus the bearing from O to A is N 70º E.

Simple Harmonic Motion
An object that moves on a coordinate axis is in simple harmonic motion if its distance from the origin, d, at time t is given by either d = a cos  t or d = a sin  t. The motion has amplitude |a|, the maximum displacement of the object from its rest position. The period of the motion is 2/ , where  > 0. The period gives the time it takes for the motion to go through one complete cycle.

Example An object in simple harmonic motion has a frequency of 1/4 oscillation per minute and an amplitude of 8 ft. Write an equation in the form for the objects simple harmonic motion.

Example cont. Solution:
a = 8 and the period is 4 minutes since it travels 1/4 oscillation per minute

Frequency of an Object in Simple Harmonic Motion
An object in simple harmonic motion given by d=acost or d = asint has frequency f given by f = /2, >0. Equivalently, f = 1/period.

Example A mass moves in simple harmonic motion described by the following equation, with t measured in seconds and d in centimeters. Find the maximum displacement, the frequency, and the time required for one cycle.

Example cont. Since a=8, the maximum displacement is 8 cm.

Example cont. The frequency is 1/6 cm per second.

Example cont. The time required for one cycle is 6 seconds.

Chapter 4 Angles and Their Measure

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