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Chapter 7 The Trigonometric Functions 1 Chapter 7 Topics • Radian Measure • Trig Functions of angles • Evaluating Trig Functions • Algebra and Trig Functions • Right Triangle Trig 2 Chapter 7.1 Real Numbers 3 Overview • Radians for Angle Measurement • Radian Measure of Standard Angles • Converting 4 Why Radians? • Degrees are a bit artificial, radians refer to an actual measure • Definition: If we put the vertex of an angle at the center of a circle of radius r and let s be the arc length intercepted, then the angle, is defined as = s/r • Because of the definition, angle is now related to a measureable quantity, that is the ratio of two lengths, s and r 5 Radians • Now, lets make some sense of this: – The arc whose length is the entire circle is the circumference – The circumference is 2 r – Therefore, the angle that encompasses the whole circle is 2 , which in degree measure is 360 • Bottom line, the angle of a whole circle is 2 ; from that you can determine all other angle measures 6 Therefore: • Find the angle measure in radians of the following: Degree Radian 360 2 180 90 270 60 30 45 7 Angle Measures Degree Radian 360 2 180 90 /2 270 3/2 60 /3 30 /6 45 /4 8 Converting: • Let d be an angle in degrees and be its measure in radians • 180 degrees is • Need to have 180/d be the same as / , or d/180 = / • This is the same as /180 = /d • 9 Example • Find 36 deg in radians: 36/180 = / 36/180 = 1/5 = 0.2 So = 0.2 • We will not ask you to do much converting, but you will need to learn the common angles in radians 10 Angles in Degrees and Radians Angle 0 30 45 60 90 120 135 150 180 Radian 0 /6 /4 /3 /2 2/3 3/4 5/6 Angle 180 210 225 240 270 300 315 330 360 Radian 7/6 5/4 4/3 3/2 5/3 7/4 11/6 2 11 Example, Finding Angle Measure • Find the radian measure of the following angle: • 120= • -225= • 270= 12 Solutions • Find the radian measure of the following angle: • 120= 2(60 ) = 2 (/3) = 2/3 • -225= - 5(45 ) = -5(/4) = -5/4 [= 2 - 5/4 = 3/4] • 270= 3(90 ) = 3(/2) 13 More Examples Find the following angles in radians • 180 = • 30 = • 330 = • -210 = 14 Solutions Find the following angles in radians • 180 = rad • 30 =/6 rad • 330 =11 /6 rad • -210 = -180-30 = -7 /6 rad 15 Example Convert to radians: -75 = 250 12’ = 16 Example (Solution) Convert to radians: -75 = -45 - 30 = - /4 - /6 = - 5/12 250 12’ = 250.2 = 259.2 ( /180 ) = 1.39 17 Example (Solution) Convert from radians to degrees / 24 = -5 = 18 Example Convert from radians to degrees / 24 = /24 (180 / ) = 180 / 24 = 7.5 -5 = -5 (180 / ) = 900 / 19 More Examples Convert to radians: • 230 = • -35 = • 200 48’ = 20 Example (Solutions) Convert to radians: •230 = 23/18 •-35 =7 /6 •200 48’ = ? For this one, you need to convert 48’ to degrees. 48’ = 48/60= 0.8 degrees. Then do the multiplication: (200.8/180) 21 Convert to Degrees: • /4 = • /2 = • 5 /6 = • 6= 22 Solutions Convert to Degrees: •/4 = 45 •/2 =90 •5 /6 =150 •6 = 1080 23 More Examples Find the terminal side quadrant: • = 3.9 • = 5.4 • = -4.3 24 Solutions Find the terminal side quadrant: • = 3.9, Q 3 • = 5.4, Q 4 • = - 4.3, Q 2 25 Arc Length Example If a circle has radius 8 cm, and the arc length is 18 cm, what is the radian measure of an angle ? = s/r = 18/8 = 9/4 = 2.25 rad 26 Some More Examples Find the radian measure of the angle with arc s and circle radius r • s = 24 m, r = 4 m • s = 10 ft, r = 10 ft • s = 5.2 mm, r = 2.6 mm • s = 11 cm, r = 20 m 27 Example (Solutions) Find the radian measure of the angle with arc s and circle radius r • s = 24 m, r = 4 m; = 6 • s = 10 ft, r = 10 ft; = 1 • s = 5.2 mm, r = 2.6 mm; = 2 • s = 11 cm, r = 20 m; equivalent to 11 cm and 2000 cm: = 11/2000 28 Sector Area • The area of a sector is the portion of a circle with angle and radius r – A whole circle has area r2 – The area of a sector of angle is r2/2 • Think about it. This formula gives the following for a circle of radius 1: – Angle , half circle, area = r2/2 – Angle 2 , whole circle, area = r2 – Angle /4, 1/8 of a circle, area = r2/8 29 Summary • We can convert from radians to degrees and vice versa: The key is that 360 degrees is 2 radians • Arc length is r , where is in radians • Sector area is r2/2 30 Angular Velocity Angular velocity equals the amount of rotation per unit time • Often designated as (omega) • = / t, the angle per time For example, a Ferris wheel rotating at 10 revolutions per minute has angular velocity of =/t = 10 revolutions/min each revolution is 2 rad = 10 (2 ) / 1 min = 20 rad/min 31 Linear Velocity Linear velocity is distance traveled (or change in position) per unit time D = r t, r = D / t For angular motion, the distance D is the length of the arc, s rate = v = s / t Since s = r , v = r / t = r ( / t ) = r Linear velocity v = r 32 Example A point P rotates around the circumference of a circle with radius r = 2 ft at a constant rate. If it takes 5 sec to rotate through an angle of 510, a. What is the angular velocity of P b. What is the linear velocity of P 33 Solution a. 510 = 510 ( / 180 ) = (510/180) = 17/6 = / t = (17/6) / 5 = (17/30) rad/sec b. V = r = 2ft (17/30 ) = (17/15) ft/ sec 34 More Examples Point P passes through angle in time t as it travels around the circle. Find its angular velocity in radians • = 540 , t = 9 yr • = 270 , t = 12 min • = 690 , t = 5 sec • = 300 , t = 5 hr 35 Solutions Covert to radians, = angle / time • = 540 , t = 9 yr; 540 = 540 / 180 = 3 , = /3 rad/yr • = 270 , t = 12 min; = /8 rad/min • = 690 , t = 5 sec; = 23 / 30 rad/sec • = 300 , t = 5 hr; = /3 rad/hr 36 Examples Point P travels around a circle of radius r, find its linear velocity 1. = 12 rad/min, r = 15 ft 2. = 2312 rad/sec, r = 0.01 km 3. = 282, t = 4.1 min, r = 1.2 yd 4. = 45, t = 3 hr, r = 2 mi 37 Solutions V=r 1. 180 ft/min 2. 72.6 mph 3. 1.44 yd/min 4. 30 mph 38 Example Using Angular Velocity to Determine Linear Velocity The wheels on a bicycle have a radius of 13 in. How fast, mph, is the cyclist going if the wheels turn at 300 rpm? = 300 rev/min = 300 (2 ) / min = 600 /min V = r = 13 in (600 /min) 1 mile = 5280 ft x 12 in /ft, 1 hour = 60 min V = 13 in (600 /min) (60 min/hr) ( 1 mi / (5280 x 12 in) V = 23.2 mph 39 Examples • A belt goes around two wheels • The radius of the smaller is 5cm and the larger 15 cm • The angular speed of the larger wheel is 1800 RPM • Find the angular speed of the larger in radians, the linear speed of a point on the edge of the larger, the angular speed of the smaller, and the angular speed of the smaller in rpm 40 Solution • The radius of the smaller is 5cm and the larger 15 cm • The angular speed of the larger wheel is 1800 RPM • Find the angular speed of the larger in radians, the linear speed of a point on the edge of the larger, the angular speed of the smaller, and the angular speed of the smaller in rpm • 3000 pi rad/min • 54000 pi cm/min • 10800 pi cm/min; the linear velocity has to be the same; v=r. 54000/5 = 10800 • Rpm = /2 pi, so10800 pi/2 pi = 5400 rpm 41 Summary • Arc Length: s = r , in radians • Angular Velocity: = / t, in radians • Linear Velocity: v = r • Area of a circular sector subtended by angle : r2/2, in radians All assume is in radians!!!! 42 Chapter 7.2 Trig Functions of Angles Using Radians 43 Overview • Locate points on a unit circle • Use special triangles to find points on a unit circle • Define the 6 trig functions in in terms of points on the unit circle 44 Unit Circle • Circle of radius 1 • Center at the origin (-1,0) y (0,1) x (0,0) (1,0) (0,1) 45 Exercise • Find a point of the unit circle if y = 1/2 46 Solution x2 + y2 = 1 x = ± sqrt (1 – ¼) = ± sqrt(3)/2 47 Symmetry • Find the quadrant containing (-3/5, -4/5) and verify it is on the unit circle 48 Solution • Quadrant : x and y < 0, so is in quadrant 3 • Check: x2 + y2 = 1 (3/5) 2 + (4/5) 2 = (9 + 16)/25 = 1 49 More on Symmetry • If the point (a,b) is on the unit circle, so are (-a, b) (a, -b) (-a, -b) 50 Special Triangles Find points on the unit circle associated with /4: /4: /2 triangle 51 Solution • In quadrant 1, the triangle has x and y values sqrt(2)/2 • In quadrant 2 it is (- sqrt(2)/2, sqrt(2)/2) • In quadrant 3, both signs are – • In quadrant 4, x is positive, y is negative 52 More Examples • Do the same for a /6: /3: /4 triangle 53 Solutions (± sqrt(3)/2, ± 1/2) (± 1/2, ± sqrt(3)/2) 54 Points Associated with Rotation Find the points on the unit circle associated with a. 5/6 b. 4 /3 c. 7 /4 55 Solution a. (-sqrt(3)/2, 1/2) b. (-1/2, - sqrt(3)/2) c. (sqrt(2)/2, -sqrt(2)/2) 56 Trig Functions Find the six trig functions for = 5 /4 57 Solution Quadrant 3 • cos = -1/sqrt(2) • sin = -1/sqrt(2) • tan = 1 • sec = -sqrt(2) • csc = -sqrt(2) • cot = 1 58 Summary • Can find points on the unit circle for angles in radians • Can calculate trig functions for points on the unit circle 59 Chapter 7.3 Evaluating Trig Fucntions 60 Overview • Define trig functions in terms of a real number t • Find the number associated with special values of the trig functions • Find the real number t associated with any trig value • In this section we cover reference angles 61 Real Numbers • Integers … -2, -1, 0, 1, 2, … • Rationals: any number that can be expressed as a fraction Rationals include integers • Reals: the non-imaginary numbers. Includes sqrt (2) and all rational numbers 62 Trigonometry of Real Numbers • Work with the unit circle: r=1. if r = 1, then arc length s = r = • That way, any function of is a function of arc length, s 63 Reference Angle • A reference angle is an acute angle associated with an angle formed by the x axis and the terminal ray of the angle. It is, necessarily, /2 64 Give the Reference Angles • 70 deg • 115 deg • 200 deg • 400 deg • - 80 deg • - 325 deg 65 Solution • 70 deg; 70 • 115 deg; 65 • 200 deg; 20 • 400 deg; 40 • - 80 deg; 80 • - 325 deg; 55 66 Give the reference angle • 4/3 • 5 • 5/6 • 7/8 67 Solution • 4/3; /3 • 5 ; • 5/6 ; /6 • 7/8 ; /8 • 21/10 ; /10 68 Examples Give the six trig functions of: a. 11 / 6 b. 3 / 2 69 Solution a. Quadrant 4: cos = sqrt(3)/2 sin = -1/2 tan = -1/sqrt(3) b. Along y axis cos = 0 sin = -1 tan = undefined (-) Given these, can find csc, sec, cot 70 More Examples • csc (/6) = • csc (5/6) = • csc (11/6) = • csc (- /6) = • csc (-17/6) = 71 Solutions • csc (/6) = 2 • csc (5/6) = 2 ; /6 away from 180 • csc (11/6) = - 2 ; /6 away from 360 • csc (- /6) = - 2 ; /6 away from 0 • csc (-17/6) = -2 ; /6 away from -18 /6 = -3 = -180 72 Examples • cot = • cot 0 = • cot /2 = • cot 3 /2 73 Solutions • cot = undefined • cot 0 =undefined • cot /2 = 0 • cot 3 /2 = 0 74 Special Angles Find t such that a. cos t = -1/sqrt(2) in quadrant 2 b. tan t = sqrt (3) in quadrant 3 75 Solutions Find t such that a. cos t = -1/sqrt(2); Q2 (like a 45 deg angle) t = 3 /4 b. tan t = sqrt (3); Q3(like a 60 deg angle) t = 4 /3 76 What you need to know • Determine any reference angle • Know how to find special angles on the unit circle • Calculate the trig functions of these special angles 77 Chapter 7.4 Algebra and the Trig Functions 78 Notation • sin of the quantity a is written sin a or sin (a); parentheses are not required • When the argument of sin is an expression parentheses are required: sin of the quantity (a + b) is written sin(a+b) • The expression sin2a means (sin a)(sin a) • The expression sin2 (a+b) means (sin(a+b))(sin(a+b)) • The expression sin (a+b) 2 means sin of the quantity (a+b)2 , the function is not squared, only its argument 79 Manipulating Trig Functions • Add sin a and 1/cos a • Simplify (Sec a + 1)/ (sin + tan) 80 Solution • Add sin a and 1/cos a find a common denominator: (sin a cos a + 1)/cos a • Simplify (Sec a + 1)/ (sin a+ tan a) Replace sec and tan by sin and cos (1/cos a + 1) / (sin a + sin a / cos a) Multiply top and bottom by cos a (1 + cos a)/(sin a cos a + sin a) This becomes 1+cos 𝑎 (sin 𝑎)(1+cos 𝑎) = 1/sin a or csc a 81 Using identities to calculate functions • Sin a = 2/3 in the second 2nd quad, find cos a and tan a 82 Solution • Sin a = 2/3 in the second 2nd quad, find cos a and tan a • 1 – sin2 a = cos2 a = 1-4/9 = 5/9 • Cos2 a = 5/9 or cos a = 5 /3 83 Example • Simplify sin a/cos a + cos a/sin a 84 Solution • Simplify sin a/cos a + cos a/sin a • Get a common denominator, sin a cos a • (sin2 a + cos2 a )/(sin a cos a) = 1/(sin a cos a) 85 Example • cos a/(1-sin a) = (1+sin a)/cos a 86 Solution • cos a/(1-sin a) = (1+sin a)/cos a • Multiply both sides by cos a (1 – sin a) to remove the denominators • cos2 a = 1 – sin2 a which is the Pythagorean identity • Note, in this case we manipulated both sides of the equation. We could have worked with just one. 87 Examples • 1/sin a – sin a = cot a cos a 88 Solution • 1/sin a – sin a = cot a cos a • Get rid of cot and get the left a common denominator • (1 – sin2 a)/sin a = cos2 a / sin a • Holds by Pythagorean identity 89 Example • (cos a – sin a)2 + 2sin a cos a = 1 90 Solution • (cos a – sin a)2 + 2sin a cos a = 1 • Multiply out parentheses • Cos2 a + sin2 a – 2 sin a cos a + 2 sin a cos a = 1 • By Pythagoras becomes: • 1+0=1 91 Example • Tan a tan b = (tan a + tan b)/(cot a + cot b) 92 Solution • Tan a tan b = (tan a + tan b)/(cot a + cot b) • Eliminate the denominator tan a tan b(cot a + cot b) = tan a + tan b • Multiply left side out, remembering tan a cot a = 1 the left becomes tan b + tan a, which is the right 93 How to prove isn’t identity • Show cos A + cos B = cos (A+B) is not an identity • To prove something isn’t an identity, only need to show that it doesn’t work for one value! 94 Solution • Show cos A + cos B = cos (A+B) is not an identity • Try A = B = 60 deg, cos 60 = ½ • Does ½ + ½ =1 = cos 120? Cos 120 is – cos 60 = - ½ NO 95 Which is an identity? • (Csc a – cot a) = (1 + cos a)(1-cos a) • Cot a + sin a/(1+cos a) = csc a 96 Solution • (Csc a – cot a) = (1 + cos a)(1-cos a) The right side becomes 1 – cos2a = 1, the left is (1- cos a)/sin a; there are many values for which that is not = 1; Not an identity • Cot a + sin a/(1+cos a) = csc a Simplify by multiplying by sin a cos a + sin2 a/(1 + cos a) = 1 • Replace sin2 a by (1-cos2 a) • We have cos a + 1 – cos a on the left which = 1 97 What you need to know • Notation: sin2 a vs sin a2 • Need to be comfortable in manipulating trig functions to prove identities • Need to be able to disapprove identities by substituting a value 98 Chapter 7.5 Right Angle Trigonometry 99 • We have already completed most of this section • The section uses a given trig value to find the remainder, e.g., if sin = 3/5, find cos and tan 100 Cofunctions • Sin (90 – A) = cos A; If two acute angles, A and B, are complementary, then sin A = cos B • We have the three angles of the triangle as A, B, 90 • Clearly 90 – A = B if they are complementary B c A a b • From the picture, sin A = a/c = cos B, etc. • Additionally, cos (90 – A ) = sin B 101 Using Cofunctions • Given cos (75) = a find sin (15) 102 Solution • Given cos (75) = a find sin (15) • We know that cos (90 – x) = sin x, so sin (15) = a 103 Trig values in other quadrants • We have done these 104 More Examples 105 • Give the reference angle for: 7 /6 24 /3 106 Solution • Give the reference angle for: 7 /6 : / 6 24 /3: 8 or 0 107 Find • Sin /3 = • Cos 2 /3 = • Tan /3 = • Cos 7 /4 = • Tan 7/4 = 108 Solution 109 • Verify that (1/3, - 2 sqrt(2) / 3) is a point on the unit circle. • Find the value of all the trig functions associated with this point 110 Solution 111 • A camera crew rids a cart on a circular arc. The radius of the arc is 75 ft and can sweep an angle of 172.5º in 20 sec. – Find the length of the track in feet – Find the angular velocity of the cart – Find the linear velocity of the cart. 112 Solution • A camera crew rids a cart on a circular arc. The radius of the arc is 75 ft and can sweep an angle of 172.5º in 20 sec. – Find the length of the track in feet • S=r = 75 ft (172.5)(/180) = 225.8 ft – Find the angular velocity of the cart • ω = /time = 172.5(/180) /20 = 0.15 rad/sec – Find the linear velocity of the cart • V = r ω = (0.15 rad/sec) (75ft) = 11.29 ft/sec 113 Example • Find t, between 0 and 2 if Sin t = -1/2 in Q3 Sec t = 2 sqrt(3)/3 in Q4 Tan t = -1 in Q2 114 Solution • Find t, between 0 and 2 if Sin t = -1/2 in Q3: t= 7 /6 Sec t = 2 sqrt(3)/3 in Q4: t = 11 /6 Tan t = -1 in Q2: t = 3 /4 115 Example • If (20/29, 21/29) is a point on the central unit circle, use symmetry to find 3 other points on the circle. 116 Solution • If (20/29, 21/29) is a point on the central unit circle, use symmetry to find 3 other points on the circle. (-20/29, 21/29) (-20/29, -21/29) (20/29, -21/29) 117 • Convert to radians: 300 -7239’ 118 • Convert to radians: 300: 5 /3 -7239’= -1.26; First change the minutes to seconds = 0.0548 119 • Convert to degrees: 9.29 -3 /2 45 120 Solution • Convert to degrees: 9.29 = 532.3 -3 /2 = -270 45 = 2578.3 Remember, to convert to degrees, multiply by 180 / 121 Example • Assume Memphis is in directly north of New Orleans, at 90 º W longitude. Find the distance between cities, in km, if the radius of the earth is 6000 km, Memphis is at 35º north, and New Orleans is at 29.6º. 122 Solution • Assume Memphis is in directly north of New Orleans, at 90 º W longitude. Find the distance between cities, in km, if the radius of the earth is 6000 km, Memphis is at 35º north, and New Orleans is at 29.6º. Angle is 35º - 29.6 º = 5.4º 5.4º (/180) = 0.094 rad Radius is 6000 km, so arclength is 6000 ( 0.094) = 565 km 123