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Chapter 5 An Introduction to Trigonometric Functions Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-1 5.1 Angle Measure, Special Triangles and Special Angles Learning Objectives In Section 5.1 you will learn how to: A. Use the vocabulary associated with a study of angles and triangles B. Find fixed ratios of the sides of special triangles C. Use radians for angle measure and compute circular arc length and area using radians D. Convert between degrees and radians for nonstandard angles E. Solve applications involving angular velocity and linear velocity using radians Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or 1-2 5.1 Angle Measure, Special Triangles and Special Angles Vocabulary An angle of meaasuring 90° is a right angle. An angle of measuring 180° is called a straight angle. A half line or all points extended from a single point in a single direction The joining of two rays at a common endpoint. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-3 We know that the measure of angle is degree and 1° = 1 360 of a full rotation. Going a little deeper, are there units smaller or finer than degree? Yes, they are called minutes and seconds. • Each degree can be divided into 60 smaller parts called minutes (1° = 60′ ) • Each minutes can be divided into 60 smaller parts called seconds (1′ = 60′′) Therefore, we also have the following converting ratios: 1 1′ = of a degree 1′′ = 60 1 3600 of a degree Notation: sixty-one degree, eighteen minutes, and forty-five seconds 61° 18′ 45′′ Application Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-4 Some exercises: 1) Determine the measure of each angle described: a. The complement of a 57° angle: 33° angle (why?) b. The supplement of a 132° angle: 48° angle (why?) c. 𝜃 and 39° angle are complement. What is the measure of 𝜃? 𝜃 = 59° (why?) Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-5 5.1 Angle Measure, Special Triangles and Special Angles 1’ (1 minute) = 1 60 11 11 a. 61 18 45 60 3600 60 3600 18 45 61 60 3600 61 0.3 0.0125 61 .3125 1 of a degree and 1”(1 second) = 3600 of a degree b. 142.2075o 142o 0.2075o o 142 0.207560' 142o 12.45' o 142 12'0.4560" o 142 12' 27" Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-6 5.1 Angle Measure, Special Triangles and Special Angles Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-7 Similar triangles: Two triangles are similar if their corresponding angles are equal and/or their corresponding sides are proportional. Link Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-8 5.1 Angle Measure, Special Triangles and Special Angles 5 .5 H 8 44 5.5 ft 8 ft H 44 ft 8H 242 H 30.25 ft H 30 ft 3 in Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-9 5.1 Angle Measure, Special Triangles and Special Angles 45o 45o 5 2 5 5 5 2 45o 45o 5 5 2 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-10 5.1 Angle Measure, Special Triangles and Special Angles 60o 8 4 30o 4 3 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-11 Angle measure can be thought of the amount of rotation from a fixed ray called the initial side to a rotated ray called the terminal side. (this way of thinking allows us to have the concept of positive and negative angles depend on the direction of the rotation). Counterclockwise rotations give positive angle. Clockwise rotations give negative angle. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-12 Coterminal angles: share the same initial and terminal sides Coterminal angles will always differ by multiples of 360°. In other words, for any integer 𝑘, angles 𝜃 and 𝜃 + 360𝑘 will be coterminal. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-13 5.1 Angle Measure, Special Triangles and Special Angles 360o If k is an integer then 60o + 360k is coterminal If k= 1 then 60o + 360(1) = 420o If k= 2 then 60o + 360(2) = 780o If k= -1 then 60o + 360(-1) = -300o If k= -2 then 60o + 360(-2) = -660o Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-14 Sometimes, it might be helpful to imagine the angle on the xy-coordinate plane (we refer that the angle is in standard position). In that case, we may put the vertex of the angle at the origin. A few vocabularies: • A central circle is a circle on the xy-plane with its center at the origin. • A central angle is an angle whose vertex is at the center of the circle. • For a central angle 𝜃 intersecting the circle at points B and C, we say the circular arc BC, denoted 𝐵𝐶, subtends ∠𝐵𝐴𝐶 The letter s is usually used to represent the arc length. If we define one radian (1 rad) to be the measure of an angle subtended by an arc equal in length to the radius, then 𝜃 = 1 𝑟𝑎𝑑 when 𝑠 = 𝑟. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-15 5.1 Angle Measure, Special Triangles and Special Angles Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-16 5.1 Angle Measure, Special Triangles and Special Angles = 3.5 r = 10 cm s=r s = 10 cm *3.5 s = 35 cm Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-17 Using the central angle 𝜃 measured in radians, we develop a formula for the area of a circular sector using a proportion. Recall: the circumference of the circle is 𝐶 = 2𝜋𝑟. This implies that the radius, or an arc length of measurement r, can be wrapped around the circumference of the circle 2𝜋 ≈ 6.28 times. This shows the radian measure of a full 360° rotation is 2𝜋. 2𝜋 𝑟𝑎𝑑 = 360° The ratio of the area of a sector to total area = the ratio of the subtended angle to 2𝜋. Using 𝒜 to represent the area of the sector, we have 𝒜 𝜃 𝒜 2𝜋 =𝜃𝜋𝑟 2 1 2 = 𝒜 = 𝑟 𝜃 𝜋𝑟 2 2𝜋 2 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-18 5.1 Angle Measure, Special Triangles and Special Angles r Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-19 5.1 Angle Measure, Special Triangles and Special Angles 3π =4 r = 72ft 1 2 A r 2 1 2 3 (72) 2 4 15552 8 1944 ft 2 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-20 5.1 Angle Measure, Special Triangles and Special Angles deg rad multiply by 180 o 5 rad 75 o 180 12 o 360o 2 rad rad deg m ultiplyby 180o 180o 180o 7.5o 24 24 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-21 5.1 Angle Measure, Special Triangles and Special Angles 78o 51o o 27 o 180 3 20 s r 3 s 3960 20 s 594 mi 1866mi M Q Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. r 1-22 5.1 Angle Measure, Special Triangles and Special Angles The angular velocity of an object is the amount of rotation per unit time. The linear velocity of an object is defined as a change of position or distance traveled per unit time. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-23 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-24 5.1 Angle Measure, Special Triangles and Special Angles r = 13 in 300rev 3002 600 1 min 1 min 1 min 600 24,504 .4 in V 13 in 1 min 1 min 24,504.4 in 60 min 1 ft 1 m i 23.2m ph V 1 min 1 hr 12in 5280ft Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-25 5.1 Angle Measure, Special Triangles and Special Angles Linear Speed Angular Speed 5.1 Angle Measure, Special Triangles and Special Angles A child is spinning a rock at the end of a 3-foot rope at the rate of 160 revolutions per minute (rpm). Find the linear speed of the rock when it is released. 160 rev 160 2 320 1 min 1 min 1 min 320 V 3 ft 3016ft per min 1 min V 34.3 mi per hr 5.2 Unit Circles and the Trigonometry of Real Numbers 1-28 Circle: a set of points on the plane that are a fixed distance called the radius from a fixed point called the center. The equation of a circle centered at origin, (0,0), and the radius r is given by 𝑥 2 + 𝑦 2 = 𝑟 2 (𝑤ℎ𝑦? ) If the central circle has radius equal 1 (𝑥 2 + 𝑦 2 = 1), then it is known as the unit circle. Observe that (0,1), (1,0), (0,-1), (-1,0) are four particular points on the unit circles. They are known as quadrantal points. 1-29 Some exercises: 1 1. Find a point on the unit circle given 𝑦 = with (x,y) in QII. 2 𝑥2 + 𝑦2 = 1 2 1 𝑥2 + =1 2 1 𝑥2 + = 1 4 3 𝑥2 = 4 3 𝑥=± 2 We want the point in QII so we pick 𝑥 = − Which quadrant would the point 3 1 , 2 2 3 , 2 and the point is − 3 1 , 2 2 . be in? QI FACT: If (𝑎, 𝑏) is on the unit circle, then (−𝑎, 𝑏), (𝑎, −𝑏), (−𝑎, −𝑏) are also on the unit circle because a circle is symmetric to both axes and the origins. So − 3 1 , 2 2 3 1 , 2 2 − 3 5 2. Name the quadrant containing − , − 4 5 3 1 ,− 2 2 3 1 ,− 2 2 are on the unit circle. and verify it’s on the unit circle. Then use symmetry to find three other points on the circle. (Do at home and turn it in) 1-30 1-31 1-32 1-33 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or 1-34 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or 1-35 From the example of 𝑡 = 3𝜋 , 2 it gives rise to the concept of domain. By definition, cosine (cos) and sine (sin) have no restrictions for its domain. For functions with x in the denominator (sec and tan), we want to avoid odd 𝜋 multiples of because the x-coordinate of the related point is zero. 2 𝜋 2 𝑡 ≠ + 𝜋𝑘 for all integers k For functions with y in the denominator (csc and cot), we want to avoid multiples of 𝜋 because the y-coordinate of the related point is zero. 𝑡 ≠ 𝜋𝑘 for all integers k Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-36 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or 1-37 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or 1-38 In the previous example, we were able to determine the values of the trig functions even though t is unknown. In many cases, we need to find the value of t. 3 For instance, what is the value of t given cos 𝑡 = − with t in QII? 2 Exercises of this type fall into two categories: (1) you recognize the given number 1 2 as one of the special values: ±{0, , 2 3 3 , , , 2 2 3 3, 1}; or (2) you don’t. In the first case, you can often name the real number t after a careful consideration of the related quadrant and required sign. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-39 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or 1-40 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-41 5.3 Graph of Since and Cosine; Cosecant and Secant Functions Learning Objectives In Section 5.3 you will learn how to: A. Graph f(t)=sin t using special values and symmetry B. Graph f(t)=cos t using special values and symmetry C. Graph sine and cosine functions with various amplitudes and periods D. Investigate graphs of the reciprocal functions f(t)=csc(Bt) and f(t)=sec(Bt) E. Write the equation for a given graph Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-43 1 2 3 Noting that = 0.5, ≈ 0.71, ≈ 0.87, we plot these points and connect them with a smooth 2 2 2 curve to graph 𝑦 = sin 𝑡 in the interval [0,2𝜋]. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-44 What happens if we expand the table from 2𝜋 to 4𝜋? The function values begin to repeat. One way of verifying that it is true is by using the concept of 13𝜋 𝜋 𝜋 9𝜋 reference angle. For instance, sin = sin since the reference angle 𝜃𝑟 = ; sin = sin 𝜋 4 since 𝜃𝑟 = 𝜋 , 4 6 6 6 4 and so on. Hence sine function is an example of what is called a periodic function. (cycle through a set pattern of values). Definition: a function f is said to be periodic if there is a positive number P such that 𝑓(𝑡 + 𝑃) = 𝑓(𝑡) for all t in the domain. The smallest number P for which this occurs is called the period of f. For the sine function we have sin 𝑡 = sin(𝑡 + 2𝜋), as in sin sin 𝜋 4 13𝜋 6 = sin 𝜋 6 + 2𝜋 and sin 9𝜋 4 = + 2𝜋 . So we conclude that the sine function is periodic with period 𝑃 = 2𝜋. FACT: In general, for all real number t, sin 𝑡 = sin(𝑡 + 2𝜋𝑘) for all integers k. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-45 Observation: both the graph and the unit circle confirm that the range of 𝑦 = sin 𝑡 is −1,1 and that 𝑦 = sin 𝑡 is an odd function. 𝜋 Recall: a function f is odd if 𝑓 −𝑡 = −𝑓(𝑡). A quick verification for the sine function: sin − = − sin 𝜋 2 2 . Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-46 a. 𝑓 𝜋 4 𝜋 4 + 8𝜋 = sin 𝜋 + 8𝜋 = sin 𝜋 𝜋 6 b. 𝑓 − 6 = sin − 6 = − sin c. 𝑓 17𝜋 2 = sin 17𝜋 2 = sin 𝜋 2 𝜋 4 + 4 2𝜋 =− 1 2 = sin 𝜋 4 = 2 2 (IV) (II) + 8𝜋 = sin 𝜋 2 + 4 2𝜋 = sin 𝜋 2 = 1 (I) d. 𝑓 21𝜋 = sin 21𝜋 = sin 𝜋 + 20𝜋 = sin 𝜋 = 0 (V) e. 𝑓 11𝜋 2 = sin 11𝜋 2 = sin 3𝜋 2 + 4𝜋 = sin 3𝜋 2 + 2 2𝜋 = sin 3𝜋 2 = −1 (III) FACT: In general, for all real number t, sin 𝑡 = sin(𝑡 + 2𝜋𝑘) for all integers k. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-47 cos t 3 2 1 cos t 1 1 2 3 2 0 1 2 0 − 1 2 − 2 2 − 3 2 -1 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-48 1 2 There is another way to get the values for cosine: we know that ± , ± 3 2 , ± 3 1 ,± 2 2 , ± 2 2 ,± 2 2 are all points on the unit circle since they satisfy 𝑥 2 + 𝑦 2 = 1. Since cos 𝑡 = 𝑥 and sin 𝑡 = 𝑦, the 1 equation cos2 𝑡 + sin2 𝑡 = 1 can be obtained by direct substitution. This means if sin 𝑡 = ± then cos 𝑡 = ± 3 2 2 and vice versa, with the sign taken from the appropriate quadrant. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-49 The function is decreasing for t in 0, 𝜋 , and increasing for t in (𝜋, 2𝜋). Similar to sine function, cosine is also periodic with a period of 2𝜋. And we note that cosine is an even function, meaning cos(𝑡) = cos(−𝑡) for all t in the domain. 𝜋 Recall: a function f is even if 𝑓 −𝑡 = 𝑓(𝑡). A quick verification for the cosine function: cos − = cos 𝜋 2 2 = 0. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-50 Observing the graph of the functions 𝑦 = sin 𝑥 and 𝑦 = cos 𝑥 again: They are oscillating back and forth, and one question to consider is how far is the distance between the maximum and minimum value. Now half of this distance is what called an amplitude. Another way of thinking of amplitude is the height of the function. With the basic sine and cosine function, it is easy to see that the amplitude is 1. However, how can we tell the amplitude of a more complicated sine and cosine functions? ANSWER: if 𝑦 = 𝐴𝑠𝑖𝑛(𝐵𝑥) or 𝑦 = 𝐴𝑐𝑜𝑠(𝐵𝑥), the amplitude can be found by computing 𝐴 . Quick examples: • 𝑦 = 4𝑠𝑖𝑛(2𝑥) has an amplitude of 𝐴 = 4 = 4 • 𝑦 = −5𝑐𝑜𝑠(3𝑥) has an amplitude of 𝐴 = −5 = 5 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-51 Graphing the basic 𝑠𝑖𝑛 𝑥 function by using its characteristics: Amplitude |1|=1 Period 2𝜋 Starting point 𝑥 = 0 End point 𝑥 = 0 + 2𝜋 = 2𝜋 Mid-point 𝑥= 0+2𝜋 2 =𝜋 Q1 𝑥= 0+𝜋 2 𝜋 2 Q3 𝑥= 𝜋+2𝜋 2 = = Notice that we are focusing in the interval [0,2𝜋]. If we were to aim our focus to the graph of this function on 𝜋 a different interval, say [− , 3𝜋], 2 how would the graph look like? 3𝜋 2 The starting point, end point and mid-point are the zeroes of the sine function while Q1 and Q3 represent the x-coordinates where the maximum and minimum values occur. 52 Let us look at a variation of 3𝑠𝑖𝑛 𝑥 and see how different it is from the basic one. 𝑦 = 3 sin 𝑥 𝑦 = sin 𝑥 We see that the amplitude affects the height of the graph. It alters the maximum and minimum values. These values still occur at the same x-values as the basic function, but the maximum and minimum values are no longer 1 and -1. They depend on the value of the amplitude. 53 Continue with a different variation of graph of 𝑠𝑖𝑛 2𝑥 and see how it differs from the basic one. 𝑦 = sin 𝑥 𝑦 = sin 𝑠𝑖𝑛3𝑥 2𝑥 1 𝑦 = sin 𝑥 2 What did we notice? It seems like multiply the input x by 2 means each cycle will complete twice as fast. What if I multiply x by 3? 1 What if I multiply x by ? 2 54 We conclude to ourselves that changing the input of the sin 𝑥 function will affect its graph horizontally. More importantly, it changes the period and the x-values where the graph intersects the x-axis (remember these are called the zeroes of the function). To find the period P of 𝑦 = 𝐴𝑠𝑖𝑛(𝐵𝑥) and 𝑦 = 𝐴𝑐𝑜𝑠(𝐵𝑥), we do 2𝜋 𝑃= 𝐵 1 2 Example: Sketch the graph of 𝑦 = sin(4𝑥) 1 2 2𝜋 4 1 Amplitude 𝐴= Period 𝑃= Starting point 𝑥=0 End point 𝑥 = 0+2 = Mid-point 0+ 2 Q1 Q3 =2 = 𝜋 2 𝜋 𝜋 𝑥= 𝑥= 𝑥= 2 𝜋 4 0+ 2 𝜋 𝜋 + 4 2 2 𝜋 2 = 𝜋 4 = 𝜋 8 = 3𝜋 8 55 Observe that once we correctly graph one cycle of the function, extending the graph in either direction is not a hard matter because the cycle repeats itself. 56 Another variation that we may see are horizontal and vertical shifting in the function 𝑠𝑖𝑛 𝑥. This is relatively straightforward 𝑦 = sin 𝑥 ± 𝐶 ± 𝐷 means that the function sin 𝑥 shift either left or right C units and up or down D unit (depend on the signs in front of C and D). Example: Sketch the graph of 𝑦 = sin 𝑥 + 𝜋 2 −1 𝑦 = sin 𝑥 𝜋 𝑦 = sin 𝑥 + 2 𝑦 = sin 𝑥 + 𝜋 −1 2 57 One more transformation we might see is the graph of 𝑦 = − sin 𝑥. This is a vertical reflection of the original function 𝑦 = sin 𝑥. Its graph will look like this Notice: sin(−𝑥) = − sin 𝑥 since the sine function is an odd function. 58 Example: Sketch the graph of 𝑦 = 3 sin 1 𝑥 2 + 𝜋 2 Solution: Amplitude: 𝐴 = 3 = 3 Period: 𝑃 = 2𝜋 1 2 = 4𝜋 Phase shift: 1 𝑥 2 1 𝑥 2 + 𝜋 2 =0 =− 𝜋 2 𝑥 = −𝜋 (this is the starting point) Vertical shift: None End point: 𝑥 = −𝜋 + 4𝜋 = 3𝜋 Mid-point: 𝑥 = 𝑄1 = 𝑄3 = −𝜋+3𝜋 2 =𝜋 Amplitude 3 Period 4𝜋 Starting point 𝑥 = −𝜋 End point 𝑥 = 3𝜋 Mid-point 𝑥=𝜋 Q1 𝑥=0 Q3 𝑥 = 2𝜋 −𝜋+𝜋 =0 2 𝜋+3𝜋 = 2𝜋 2 59 60 Example: Sketch the graph of 𝜋 𝑦 = −3 cos 𝑥 − 4 Amplitude: 𝐴 = 3 = 3 2𝜋 Period: 𝑃 = = 2𝜋 1 𝜋 Phase shift: 𝑥 − = 0 ⟹ 𝑥 = 4 starting point) Vertical shift: none 𝜋 9𝜋 End point: 𝑥 = + 2𝜋 = 4 Mid-point: 𝑥 = Q1: 𝑥 = Q2: 𝑥 = 𝜋 5𝜋 + 4 4 2 = 5𝜋 9𝜋 + 4 4 2 𝜋 9𝜋 + 4 4 2 𝜋 4 (this is my 4 = 5𝜋 4 3𝜋 = 4 7𝜋 4 Notice that for cosine function, Q1 and Q2 become the zeroes of the function while the starting point, end point and mid-point become where the minimum/maximum occur. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-61 Once we know how to graph 𝑦 = 𝐴 sin(𝐵𝑥) and 𝑦 = 𝐴 cos(𝐵𝑥), graphing 𝑦 = 𝐴 csc(𝐵𝑥) and 𝑦 = 𝐴 sec(𝐵𝑥) can be done easily. Why? Cosecant and secant function are the reciprocal of sine and cosine function: 1 1 csc 𝑥 = sec 𝑥 = sin 𝑥 cos 𝑥 One of the crucial thing worth noticing is that cosecant and secant will have vertical asymptote at every point where sine and cosine equal to 0, respectively. Example: graph the function 𝑦 = csc 𝑥 for 𝑥 ∈ [0,4𝜋] We follow the same procedure as before to graph 𝑦 = sin 𝑥 first Notice that sin 𝑥 = 0 at 𝑥 = 0, 𝜋, 2𝜋, 3𝜋, 4𝜋. So these are the places where the vertical asymptotes appear. 𝜋 5𝜋 𝜋 5𝜋 Also, sin 2 = sin 2 = 1 implies csc 2 = csc 2 = 1. sin 3𝜋 2 = sin 7𝜋 2 = −1 implies csc 3𝜋 2 = csc 7𝜋 2 = −1. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-62 𝑦 = sec 𝑥 𝑦 = cos 𝑥 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-63 So far we have been dealing with given a function then graph that function. Can we do this conversely, meaning given the graph, can we write the equation that it represents? In this section, we look at the simple cases by examining the amplitude and the period of sine and cosine function. Example: given the following graph, write the equation of the function By looking at the behavior of the graph, this is the graph of some function 𝑦 = 𝐴 sin(𝐵𝑥). Since the 3 3 amplitude of this graph is , 𝐴 = . The period 𝑃 = 8𝜋 . 3 4 Now 𝑃 = 2𝜋 𝐵 3 4 4 so to find B, we do 𝐵 = Therefore, the function is 𝑦 = sin 3 𝑥 4 2𝜋 𝑃 = 2𝜋 8𝜋 3 = 3 4 . Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-64 Exercises: do at home and turn it in my next Thursday, January 30 2013 before the exam 1 starts. 1. 2. Graph the following functions. Make sure to include these information: amplitude, period, starting point, end point, mid-point, Q1, Q3 𝜋 a) 𝑦 = − sin 𝑥 − b) 𝑦 = cos 2𝑥 + c) 𝑦 = −2 csc 4 𝜋 2 1 𝑥 3 (Just graph one period of the function correctly) Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-65 5.4 5.4 Graphs of Tangent and Cotangent Functions Learning Objectives In Section 5.4 you will learn how to: A. Graph y = tan t using asymptotes, zeroes, and the ratio (sin t)/(cos t) B. Graph y = cot t using asymptotes, zeroes, and the ratio (cos t)/(sin t) C. Identify and discuss important characteristics of y = tan t and y = cot t D. Graph y = A tan Bt and y = A cot Bt with various values of A and B E. Solve applications of y = tan t and y = cot t Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-66 5.4 0 1 3 1 3 UND − 3 -1 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. − 1 3 0 1-67 5.4 EXAMPLE 1 CONTINUED Vertical Asymptotes A few good points to know: tan 0 = 0 ⟹ 0,0 is on the graph 𝜋 𝜋 tan = 1 ⟹ , 1 is on the graph Period 𝜋 𝜋 𝜋 2 2 Usually starts at − to for one period 4 4 𝜋 tan(− ) = −1 ⟹ − , −1 is on the graph 4 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 1-68 5.4 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-69 EXAMPLE 2 CONTINUED Vertical Asymptotes 𝜋 ,1 4 𝜋 ,0 2 3𝜋 , −1 4 Period Usually starts at 0 to 𝜋 for one period Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-70 5.4 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-71 5.4 What is the period of 𝑓 𝑡 = tan(𝑡)? 3𝜋 4 −𝜋4 𝜋 4 5𝜋 4 𝜋 7𝜋 13𝜋 , 6 6 6 5𝜋 and − differ? 6 How do , Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-72 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-73 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-74 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-75 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-76 5.4 𝑃𝑒𝑟𝑖𝑜𝑑 𝜋 𝜋 2𝜋 P= − − = 3 3 3 2𝜋 𝜋 P= = 3 𝐵 2𝜋 𝜋 = 3 𝐵 3𝜋 3 𝜋3 2 = 𝐴 𝐵= −2 = 𝐴 tan 𝑦 = 𝐴 tan 𝑡 2𝜋 2 22 3 𝐵= 3𝜋 2 𝒚 −2 = 𝐴 tan 3 4 𝑦 = 𝐴 tan 𝑡 2 −2 = 𝐴(−1) 𝟑 = 𝟐 𝒕𝒂𝒏 𝒕 𝟐 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-77 5.5 Transformations and Applications of Trigonometric Graphs Learning Objectives In Section 5.5 you will learn how to: A. Apply vertical translations in context B. Apply horizontal translations in context C. Solve applications involving harmonic motion Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-78 Sinusoidal pattern • • A pattern that can be modeled by a sine (or cosine) function. One particular example is outdoor temperatures 𝑦 = 𝐴𝑠𝑖𝑛(𝐵𝑡) 𝜋 • We take 𝑡 = 0 corresponds to 12:00 noon. Note that 𝐴 = 15 and 𝑃 = 24 ⟹ 𝐵 = • • If a vertical shift is applied, then the function would have the form 𝑦 = 𝐴𝑠𝑖𝑛 𝐵𝑡 + D. D can be thought of as the average value. If we let M be the maximum value and m be the minimum value, we have M−m 2 𝐴+𝐷 = 𝐷−𝐴 = M−m 2 M+m 2 + − M+m 2 M−m 2 = = 2𝑀 2 2m 2 =𝐴 M+m 2 . 12 =𝐷 = 𝑀 (amplitude + average value = maximum value) = m (average value – amplitude = minimum value) Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-79 Note: 𝑃 = 24 and 𝐵 = 𝜋 as stated before. Hence, 𝜋 𝑦 = 𝐴 sin 𝑡 +𝐷 12 We know 𝑀 = 85 and 𝑚 = 61, thus we can compute the amplitude A and the average value D M − m 85 − 61 𝐴= = = 12 2 2 M + m 85 + 61 𝐷= = = 73 2 2 𝜋 Therefore, 𝑦 = 12 sin 𝑡 + 73 is the sinusoidal 12 12 equation model for the daily temperature. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-80 𝜋 2𝜋 5 5 Since 𝐵 = , we can compute the period 𝑃 = 𝜋 = 10. This means the population of this species rises and falls over a 10-yr cycle. amplitude + average value = maximum value So 𝑀 = 1200 + 9000 = 10,200 average value – amplitude = minimum value So 𝑚 = 9000 – 1200 = 7800 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-81 Horizontal Translation: 𝑦 = 𝐴 sin(𝐵𝑡 + 𝐶) + 𝐷 𝜋 𝑃 𝑡 = 1200 cos 𝑡 − 2.5 + 9000 5 𝜋 𝜋 = 1200 cos 𝑡 − + 9000 5 2 So both graph represents the same function. The only difference is that one of them is in shifted form and the other is in standard form. Standard form: 𝑦 = 𝐴 sin(𝐵𝑡 ± 𝐶) + 𝐷 𝐶 Shifted form: 𝑦 = 𝐴 sin 𝐵 𝑡 ± +𝐷 𝐵 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-82 Amplitude: 𝐴 = 2.5 = 2.5 2𝜋 2𝜋 Period: 𝑃 = = 𝜋 =8 𝐵 4 Average value: 𝐷 = 6 Horizontal shift: we write the equation in shifted form: 𝑦 = 2.5 sin For the endpoints of the primary interval, we solve 0 ≤ 𝜋 4 𝜋 4 𝑡+3 + 6. So it shifts 3 units to the left. 𝑡 + 3 < 2𝜋, which gives −3 ≤ 𝑡 < 5. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-83 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-84 Simple Harmonic Motion • The periodic motion of springs, tides, sound exhibit what is known as harmonic motion, which can be modeled using sinusoidal functions. • For object in harmonic motion, the input variable t is always a time unit. Amplitude and period are defined as before. Frequency: the number of cycles it completes per unit time. Frequency can be computed as 1 𝐵 𝑓= = 𝑃 2𝜋 • Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-85 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-86 5.6 The Trigonometry of Right Triangles Learning Objectives In Section 5.6 you will learn how to: A. Find values of the six trigonometric functions from their ratio definitions B. Solve a right triangle given one angle and one side C. Solve a right triangle given two sides D. Use cofunctions and complements to write equivalent expressions E. Solve applications involving angles of elevation and depression F. Solve general applications of right triangles Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-87 Trig functions using right triangle instead of the unit circle SOH CAH TOA Using Pythagorean theorem, we can figure out the length of the adj side 𝑎𝑑𝑗 = 72 − 42 = 33 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-89 SOLVING RIGHT TRIANGLES GIVEN ONE ANGLE AND ONE SIDE 𝜽 𝟎 𝝅 (𝟑𝟎°) 𝟔 𝝅 (𝟒𝟓°) 𝟒 𝝅 (𝟔𝟎°) 𝟑 𝝅 (𝟗𝟎°) 𝟐 sin 𝜃 0 1 2 1 3 2 3 2 1 2 1 cos 𝜃 tan 𝜃 0 2 2 2 2 1 3 undefined csc 𝜃 undefined 2 3 = 3 3 1 1 3 = 3 3 2 2 2 s𝑒𝑐 𝜃 1 cot 𝜃 undefined 2 3 3 3 = 2 2 1 0 2 2 undefined 3 3 0 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-90 To solve for a right triangle means to find the measure of all three angles and all three sides To solve for c, we do sin 30° = 𝑐= 17.9 𝑐 17.9 17.9 = = 35.8 1 sin 30° 2 To solve for b, we can apply the Pythagorean theorem to get 𝑏 ≈ 31 Since ∠𝐴 and ∠𝐵 are complements, that means 𝐵 = 60°. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-91 Tips for solving right triangle: 1. Angles must sum up to 180° 2. Pythagorean Theorem 3. Special triangles 4. SOH CAH TOA Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-92 Using calculator can help to find the value of trig functions of angles that are less familiar. We see that 𝐵 = 58° because angles A and B are complements. To solve for b, we do tan 32° = 𝑏= 24 𝑏 24 24 ≈ ≈ 38.41 tan 32° 0.624869 Knowing b now, we can apply to Pythagorean theorem or use another trig ratio to figure out c. Here I will use a trig ratio: 24 sin 32° = 𝑐 𝑐= 24 ≈ 45.29 sin 32° Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-93 SOLVING RIGHT TRIANGLES GIVEN TWO SIDES In some case, you are given the value of the trig function at an angle but not the angle itself. (i.e. sin 𝜃 = 0.7604, what is 𝜃?) An easy case: sin 𝜃 = 2 , 2 what is 𝜃? From the table, it is obvious that 𝜃 = 45°. Rigorously speaking, 𝜃 = sin−1 2 2 = 45° NOTATION: 𝜃 = sin−1 𝑥 or 𝜃 = arcsin 𝑥 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-94 From the example above, we see that 𝛼 + 𝛽 = 90° ⟹ 𝛽 = 90° − 𝛼 𝑎 Now, sin 𝛼 = 𝑐 ; cos 𝛽 = 𝑎 𝑐 So sin 𝛼 = cos 𝛽 = cos(90° − 𝛼) In words: the sine of an angle is equal to cosine of its complement. Therefore, sine and cosine are called cofunctions. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-96 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-99 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-100 QUICK REVIEW Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1101 What does it mean to solve a right triangle? A: finding all the angles’ measurement and the length of each side. Cofunctions: Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1102 APPLICATIONS USING ANGLE OF ELEVATION/DEPRESSION Terminology: An angle of elevation is defined to be the acute angle form by a horizontal line of orientation (parallel to level ground) and a line of sight. An angle of depression is likewise defined but involves a line of sight that is below the horizontal line of orientation. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1103 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1104 Singh is in there Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1105 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1106 5.7 Trigonometry and the Coordinate Plane Learning Objectives In Section 5.7 you will learn how to: A. Define the trigonometric functions using the coordinates of a point B. Using reference angles to evaluate the trig functions for any angle C. Solve applications using the trig functions of any angle Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-107 Given a 30-60-90 triangle with hypotenuse of 10, we can use the ratio in the 30-60-90 triangle to figure out the length of the two legs: 5 and 5 3. If we place it in QI with the 30° angle at the origin and the longer leg along the x-axis, we get sin 30° = 𝑦 5 1 = = 𝑟 10 2 𝑥 5 3 3 cos 30° = = = 𝑟 10 2 5 3 tan 30° = = = 3 5 3 3 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 1108 Therefore, we can define the six trig functions in terms of 𝑥, 𝑦, and 𝑟, where 𝑟 = 𝑥2 + 𝑦2 The slope of the line coincident with the hypotenuse is 5 5 = 3 3 . 3 Since the line goes through the origin, using the point-slope formula, we get the equation of that line 3 𝑦= 𝑥 3 Notice: no matter what point we pick on that line, it will be at the 60° vertex of a right triangle if we draw a perpendicular line from that point to the x-axis. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1109 For convenience, we pick 𝑥 = 6 (you can pick any value x you want). Hence, 𝑦 = 2 3, which leads to the point (6,2 3) being on the line and at the vertex of a 60° triangle. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1110 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-111 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-112 𝑦 5 2 sin 𝜃 = = = 𝑟 5 2 2 𝑟 csc 𝜃 = = 2 𝑦 𝑥 5 2 cos 𝜃 = = − =− 𝑟 2 5 2 𝑟 sec 𝜃 = = − 2 𝑥 𝑦 5 tan 𝜃 = = = −1 𝑥 −5 𝑥 cot 𝜃 = = −1 𝑦 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-113 REFERENCE ANGLE REVISIT To find reference angle, it is not enough to just plus/minus 360° from the original angle. One technique to find reference angle: 1. Plus/minus 360° to make your original angle is within 0° to 360° (if necessary) 2. Draw out the new angle to see which quadrant it is in 3. Use the above to help figure out the reference angle Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-114 Angle 315° is in QIV. So to find the reference angle in this case, we compute: 𝜃𝑟 = 360° − 315° = 45° b. Angle 150° is in QII. So to find the reference angle in this case, we compute: 𝜃𝑟 = 180° − 150° = 30° c. Angle −121° is not within 0° to 360°. So I’m going to add 360° until the new angle is within 0° to 360° for the first time: −121° + 360° = 239° Since 0° ≤ 239° ≤ 360° and it is in QIII, hence the reference angle 𝜃𝑟 = 239° − 180° = 59° d. Once again, 425° is not within 0° to 360°. So in this case, I’m going to subtract 360° until the new angle is within 0° to 360° for the first time: 425° − 360° = 65° Since 0° ≤ 65° ≤ 360° and it is in QI, hence the reference angle 𝜃𝑟 = 65° a. Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-115 The terminal side is in QIV where x is positive and y is negative. With 𝜃 = 45°, we have 2 sin 315° = − 2 2 cos 315° = 2 tan 315° = −1 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-116 sin 𝜃 is positive in QI and QII, while cos 𝜃 is negative in QII and QIII. Both conditions are satisfied in QII only. 𝑦 Since sin 𝜃 = , we have that 𝑦 = 5 and 𝑟 = 13. By using Pythagorean theorem, 𝑟 we have 𝑥 = 132 − 52 = 144 = ±12. To be in QII, we pick 𝑥 = −12 Hence, cos 𝜃 = − 12 13 Moreover, csc 𝜃 = and tan 𝜃 = 13 , 5 sec 𝜃 = − 13 −12 13 , 12 and cot 𝜃 = − 12 5 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-117 Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-118