Principles of Mathematics 1 Contents 1 Module : Trigonometr Section 1 Trigonometric Functions 3 Lesson 1 The Trigonometric Values for θ, 0 θ 360 5 Lesson Solving Trigonometric Equations, 0 θ 360 9 Lesson 3 Special Trigonometric Values 17 Lesson 4 The Unit Circle 5 Lesson 5 The Circular Functions 33 Lesson 6 Radian Measures 45 Summar of Section 1 55 Review 57 Section Assignment.1 59 Section : Graphs of Trigonometric Functions 65 Lesson 1 Sine and Cosine Graphs 67 Lesson Transformations of the Sine and Cosine Functions 75 Lesson 3 Graphs of the Remaining Circular Functions 85 Summar of Section 9 Review 95 Section Assignment. 97 Section 3: Trigonometric Identities 105 Lesson 1 Elementar Identities 107 Lesson Using Elementar Identities 117 Lesson 3 Sum and Difference Identities 13 Lesson 4 Double Angle Identities 131 Summar of Section 3 137 Review 139 Section Assignment.3 141 Module
Contents Principles of Mathematics 1 Section 4: Problem Solving 151 Lesson 1 Modelling with Trigonometric Functions 153 Lesson Solving Trigonometric Equations Using Your Graphing Calculator 161 Lesson 3 Solving Trigonometric Equations : Finding the General Solution 161 Summar of Section 4 165 Review 167 Section Assignment.4 171 Module : Answer Ke 177 Module
Principles of Mathematics 1 Section 1, Introduction 3 Module, Section 1 Trigonometric Functions Introduction You have studied trigonometric ratios since Grade 9 Mathematics. In this module ou will stud those ratios as functions. You will look at tables of values of these functions, generate their graphs, and list their properties. It is important for ou to do all the questions in the assignments because some new concepts are developed within the assignments. Using techniques that ou developed in Module 1, ou will stud transformations of these circular functions. In order to find the zeros of the functions, ou will develop skills in circular function equation solving and in relationships amongst these functions which are called identities. Section 1 Outline Lesson 1 The Trigonometric Values for θ, 0 θ 360 Lesson Solving Trigonometric Equations for 0 θ 360 Lesson 3 Lesson 4 Lesson 5 Lesson 6 Lesson 7 Review Special Trigonometric Values The Unit Circle The Circular Functions Radian Measures Sine and Cosine Graphs Module
4 Section 1, Introduction Principles of Mathematics 1 Notes Module
Principles of Mathematics 1 Section 1, Lesson 1 5 Lesson 1 The Trigonometric Values for θ, 0 θ 360 To this point, ou have determined the values of the trigonometric ratios for angles between 0 and 180. In this short introductor lesson, we will eplore the trigonometric ratios for an angle θ between 0 and 360. First we will re-define the primar trig ratios sine, cosine, and tangent. Suppose θ is an angle in standard position (its initial arm lies on the positive -ais). Further, let P(,) be an point on θ s terminal arm and r be the distance from P to the origin (0,0). Since r is a distance, it will alwas be considered positive. Now, using Pthagorean Theorem, we can determine the value of r: r = + and we define: opposite sinθ = = r hpotenuse adjacent cosθ = = r hpotenuse opposite tanθ = = adjacent θ (hpotenuse) r (adjacent) terminal arm P(,) (opposite) initial arm These definitions can be applied to determine the trigonometric ratios for an angle, θ, in standard position if the coordinates of a point on its terminal arm are known. Eample 1 Determine the three primar trig ratios for the angle θ in standard position, with the point P( 4,5) on its terminal arm. Module
6 Section 1, Lesson 1 Principles of Mathematics 1 Solution θ terminates in Quadrant II and we will be using our new definitions to solve this problem, but first we need to determine r. r = + = ( 4) (5) 41 5 sinθ = = 0.781 r 41 4 cosθ = = 0.65 r 41 5 tan θ = = = 1.50 4 P( 4,5) r = 41 5 θ 3 3 As ou can see, for an angle θ terminating in Quadrant II, as in Eample 1, the onl one of the three primar trig ratios which is positive is sin θ. We will come back to this result shortl. Eample An angle θ in standard position has the point Q( 1, 5) on its terminal arm. Determine its primar trigonometric ratios. For θ in Quadrant III: 3 r = + = ( 1) ( 5) 13 5 sinθ = = 0.385 r 13 1 cosθ = = 0.93 r 13 5 tanθ = = =+ 0.417 1 10 5 r = 13 3 θ P( 1, 5) 5 Module
Principles of Mathematics 1 Section 1, Lesson 1 7 Note that the onl positive one of the primar trigonometric ratios for an angle θ terminating in Quadrant III is tan θ. If we eamine the signs of our and -coordinates in the other two quadrants, we see the following. Quadrant IV In Quadrant IV, is positive and is negative. () () + () + () + () () + So sinθ = = = r () cosθ = = = + r () tanθ = = = () θ r (alwas positive) P(+, ) So for θ terminating in Quadrant IV onl cos θ is positive. Quadrant I In Quadrant I, is positive and is positive. () + () + () + () + () + () + So sinθ = = = + r () cosθ = = = + r () tanθ = = = + () (alwas positive) r θ P(+,+) So for θ terminating in Quadrant I, all three of the primar trigonometric ratios are positive. Module
8 Section 1, Lesson 1 Principles of Mathematics 1 Now we are able to summarize what we know about the trig ratios of an angle in standard position. Quadrant II Sine onl is positive Quadrant III Tangent onl is positive Quadrant I All trig ratios are positive Quadrant IV Cosine onl is positive From the first letter of each sentence, this is called the CAST diagram and it will prove ver helpful to ou in the future! Now for a little practice. Guided Practice Using the new definitions for sin θ, cos θ, and tan θ, determine the eact values for these trigonometric ratios. Please leave our answers in fractional and/or radical form if the are not integer values. 1. P(1,11) is on the terminal arm.. Q( 6,8) is on the terminal arm. 3. R(5, 1) is on the terminal arm. 4. S( 6, 3) is on the terminal arm. 5. M(5,0) is on the terminal arm. 6. N(0,4) is on the terminal arm. 7. L( 3,0) is on the terminal arm. 8. K(0, ) is on the terminal arm. Check our answers in the Module Answer Ke. Module
Principles of Mathematics 1 Section 1, Lesson 9 Lesson Solving Trigonometric Equations, 0 θ 360 Outcomes Upon completing this lesson, ou will be able to: determine the three primar trigonometric ratios for an angle θ, in standard position 0 θ 360 solve trigonometric equations for 0 θ < 360 Overview In Lesson 1 we learned new definitions for sin θ, cos θ, and tan θ. Using these new definitions, we were able to determine the trigonometric ratios for angles terminating in an one of the four quadrants and make observations about their signs (the CAST diagram below). S T A C In this lesson we will be using the CAST diagram along with a given angle s reference angle (defined below) to solve simple trigonometric equations. Definition: The reference angle for an angle θ in standard position is the acute angle between its terminal arm and the nearest part of the - ais. Eample 1 Determine the quadrant where each of the following terminate and the reference angle. a) 55 b) 169 c) 8 d) 310 Solution a) Angles which terminate in QI QI (Quadrant 1) are identical in measure to their reference angle. 55º Reference Angle = 55º Module
10 Section 1, Lesson Principles of Mathematics 1 b) The reference angle for θ terminating in QII is given b: QII 180 θ. This is because the θ = 169º nearest part of the -ais to the terminal arm of θ is the negative -ais, where the angle is 180 from the initial arm. Reference Angle = 180º 169º = 11º c) The reference angle for θ θ = 0º terminating in QIII is given b: θ 180. θ comes before 180 to make the reference angle positive. QIII Reference Angle = 0º 180º = 40º d) The reference angle for θ terminating in QIV is given b: 360 θ. Note: our abbreviation for reference angle is θ R. QIV θ R = 360º 310º = 50º sin 310º = 0.766 cos 310º = 0.643 tan 310º = 1.19 θ = 310º Angles having the same reference angles have ver similar trigonometric ratios. The differences are in the signs (positive or negative) of each. Definition: Coterminal angles are angles with the same terminal arm. Coterminal angles have the same reference angle Module
Principles of Mathematics 1 Section 1, Lesson 11 Eample Find an angle coterminal with: a) 30 b) 0 Solution a) 30º 390º 30 and 390 are coterminal because the both end up in the same position or 30º 330º 30 and 330 are coterminal because the both end up in the same position b) 0º 580º 0 and 580 are coterminal because the both end up in the same position or 0º 140 0 and 140 are coterminal because the both end up in the same position
1 Section 1, Lesson Principles of Mathematics 1 Eample 3 Use our calculator to determine the three primar trigonometric ratios for each of the following: a) 50 b) 130 c) 30 d) 310 Solution a) b) 50º 130º θ R = 50º sin 50º = 0.766 cos 50º = 0.643 tan 50º = +1.19 θ R = 180º 130º = 50º sin 130º = 0.766 cos 130º = 0.643 tan 130º = 1.19 c) d) 30º θ R = 30º 180º = 50º sin 30º = 0.766 cos 30º = 0.643 tan 30º = 1.19 θ R = 360º 310º = 50º sin 310º = 0.766 cos 310º = 0.643 tan 310º = 1.19 θ = 310º You can check these results against the CAST diagram, too. In Quadrant I(a) all trig ratios are positive, in Quadrant II (b) onl sin is positive, in Quadrant III (c) onl tan is positive, and in Quadrant IV (d) onl cos is positive. Now we will turn things around a little and solve some equations for θ, 0 θ < 360. Module
Principles of Mathematics 1 Section 1, Lesson 13 Eample 4 Solve for θ if sin θ 0.58 = 0. Epress θ to the nearest whole degree. 0 θ < 360 Solution If sin θ 0.58 = 0 sin θ = 0.58 (positive) Now using the CAST diagram below... S T A C we see that sine is positive in both Quadrants I and II, so there are two solutions; θ, in Quadrant I and θ in Quadrant II. Both θ 1 and θ will have the same reference angle which we will determine first. 1 ( ) θ R = sin 0.58 [Calculator] 3º. This is the reference angle. θ = 3º and θ = 180º 3º = 148º 1 3º = θ 1 θ = 148º Eample 5 Solve for θ if 1 cos θ + 5 = cos θ, 0 θ < 360 Solution Rearrange the equation first. 10 cos = 5 5 cos θ = = 0.500 negative 10 ( ) Module
14 Section 1, Lesson Principles of Mathematics 1 If cosine is negative, then θ must terminate in Quadrant S A II or Quadrant III. (Using CAST we see that cosine is T C positive in the other two Quadrants.) [We ignore signs when determining the 1 Now θr = cos ( 0.500) reference angle because reference angles θr = 60 alwas lie between 0 and 90.] θ = 180 60 = 10 θ = 180 + 60 = 40 3 θ = 10º Eample 6 3tanθ 4 Solve for θ: =1 0 θ < 360. Answer to nearest degree. Solution Rearrange the equation first. 3tanθ 4 =1 3tanθ 4 = 3tanθ = 6 tanθ =.000 positive S T A C ( ) tan θ positive means that θ terminates in Quadrant I or Quadrant III. (Using CAST.) 1 ( ) θ 3 = 40º θr = tan.000 63 θ = 63 and θ = 180 + 63 = 43 1 3 Module
Principles of Mathematics 1 Section 1, Lesson 15 θ 1 = 60º θ 3 = 43º Eample 7 Solve for θ if cos θ + 3 cosθ = 0 0 θ < 360 Solution If cos θ + 3cosθ = 0 ( θ )( θ ) cos 1 cos + = 0 [Factoring the Trinomial] 1 cos θ = or cos θ = cosθ = 0.500 or cos θ = Module
16 Section 1, Lesson Principles of Mathematics 1 Case 1: If cos θ = 0.500 (positive) S T A C then θ terminates in Quadrant I or Quadrant IV. ( ) θ = cos 0.500 = 60 R 1 θ = 60 or θ = 360 60 = 300 1 4 θ 1 = 60º θ 4 = 300º Case : cos θ = 1 ( ) = cos No solution θ R This is because sin θ and cos θ range between 1 and 1. So the solution set in Eample 5 is {60, 300 }. Now it s our turn to tr a few. Module
Principles of Mathematics 1 Section 1, Lesson 17 Guided Practice Solve for θ, 0 θ < 360, in each of the following. Answer to the nearest degree, or state an eact solution where possible. 1. a) Find the reference angles for 116, 65, 89, 33. b) Evaluate the three trigonometric ratios for the angles above. c) Find a coterminal angle with the angles above.. 3 sin θ = 3. 5 tan θ + 4 = tan θ 4. 3 sin θ + 1 = 0 5. 3 cos θ + 14 cos θ = 5 6. 7 cos θ 8 = 4 7. sin θ = 3 sinθ 8. 4 tan 13 tan = 1 Check our answers in the Module Answer Ke. Module
18 Section 1, Lesson 3 Principles of Mathematics 1 Lesson 3 Special Trigonometric Values Outcomes Upon completing this lesson, ou will be able to: state the trigonometric ratios for an Quadrantal angle θ, 0 θ 360 state the trigonometric ratios of an multiple of 30, 45, or 60 without the use of a calculator or trigonometric table solve linear and quadratic trigonometric equations with solutions which are multiples of 30, 45, or 60 (these equations will be solved without the use of an aids) Overview In Lesson ou determined the trigonometric ratios for an θ, where 0 θ 360. You also solved various trigonometric equations for θ, 0 θ < 360, using a calculator and the CAST diagram. In this lesson ou will be solving the same tpe of problems ecept without the use of a calculator. This will require some recall of Lesson 1 and the establishment of some geometric facts. In the Lesson 1, Guided Practice 5 8, ou determined the three primar trigonometric ratios for four ver special angles which we call Quadrantal. The are angles which terminate on the positive or negative or -ais, thereb dividing the coordinate plane into the four quadrants. Let s look at them once more so that we can summarize the results. Module
Principles of Mathematics 1 Section 1, Lesson 3 19 0, 360 4 The point (5,0) is on the terminal arm of the angle θ, which represents a rotation of either 0 or 360 from its initial arm. Therefore: 4 = 5 M(5,0) }r θ = 0º or θ = 360º sin 0 = sin 360 0 = = 0 r 5 cos 0 = cos 360 5 = = 1 r 5 tan 0 = tan 360 0 = = 0 5 90 4{ 5 N(0,4) r = 4 θ=90º The point (0,4) is on the terminal arm of the angle θ, which represents a rotation of 90 from its initial arm. Therefore: 4 sin 90 = = = 1 r 4 0 cos 90 = = = 0 r 4 4 tan 90 = = undefined 0 Module
0 Section 1, Lesson 3 Principles of Mathematics 1 180 L( 3,0) r = 3{ θ=180º 3 The point ( 3,0) is on the terminal arm of the angle θ, which represents a rotation of 180 from its initial arm. Therefore: 0 sin180 = = = 0 r 3 3 cos180 = = = 1 r 3 0 tan180 = = = 0 3 70 θ=70º } r = K(0, ) 4 The point (0, ) is on the terminal arm of the angle θ, which represents a rotation of 70 from its initial arm. Therefore: sin 70 = = = 1 r 0 cos 70 = = = 1 r tan 70 = = undefined 0 The following table summarizes the trigonometric values for the Quadrantal angles. θ 0 /360 90 180 70 sin θ 0 1 0 1 cos θ 1 0 1 0 tan θ 0 0 Module
Principles of Mathematics 1 Section 1, Lesson 3 1 Now for some geometric facts. Fact 1: 45-45 - 90 Triangle A B 45 45 C Since ABC is isosceles it follows that AB = AC. Let these two equal lengths be and use the Pthagorean Theorem to find the length of side BC. b g b BC = AB + AC bbcg = bg + bg BC = b g BC = b g g b g Therefore, the ratios of a 45-45 - 90 triangle are alwas : :, or 1:1:, respectivel. This fact enables ou to find the trigonometric ratios of a 45 angle. A 1 1 B 45 45 C Therefore, opposite side 1 sin 45 = = hpotenuse adjacent side 1 cos45 = = hpotenuse opposite side 1 tan45 = = = 1 adjacent side 1 Module
Section 1, Lesson 3 Principles of Mathematics 1 You should memorize these ratios. When ou are answering a question based on the 45-45 -90 triangle, ou ma find that sketching the above reference triangle, with its ratios, is useful. Eample 1 Without the use of a calculator: 1 1 1 1 a) 1 = 1 = b) sinθ sin θ = 0 ( θ) ( ) a) simplif sin45 cos45 tan45 b) solve for, if 0 360 : sin sin θ θ θ= θ sinθ 1 sin = 0 (factor) Solution 1 Either sinθ = 0 or sin θ = sin θ is positive in both cases, so θ lies in Quadrants 1 and onl. θ = 0, 180, 45, 135 Fact : 30-60 -90 Triangle Beginning with an equilateral ABC each side of length, ou form two 30-60 -90 triangles b dropping a perpendicular, AD, from verte A. A B 60 30 30º D 60º C Triangles ABD and ACD are congruent b ASA. Hence, 1 BD = BC =. Note: (ASA: Angle-Side-Angle: the two triangles have equal 30 angles, equal 60 angles, and sides of equal length which lie between the two angle(s). Module
Principles of Mathematics 1 Section 1, Lesson 3 3 You use the Pthagorean Theorem to find the third side AD. ( AD) = ( AB) ( BD) ( AD) = ( ) ( ) ( AD) = 3 AD = 3 Therefore, the ratios of a 30-60 -90 triangle are alwas : 3:, or 1: 3:, respectivel. This fact enables ou to find the trigonometric ratios of a 30 or a 60 angle. A 30 3 B 60 1 D opposite side 1 3 sin30 = = sin 60 = hpotenuse adjacent side 3 1 cos30 = = cos60 = hpotenuse opposite side 1 3 tan30 = = tan60 = = 3 adjacent side 3 1 You should memorize these values! When ou are answering a question based on a 30-60 - 90 triangle, ou ma find that sketching the above reference triangle, with its ratios, is useful. Observe the length of the longest side it s, not 3. Man students overlook that and put 3 on the wrong side of the triangle. Eample Without the use of a calculator: a) find the value of cos 30 + sin60 tan30 b) solve for if 0 < 360 : sin tan 3 sin = 0 Module
4 Section 1, Lesson 3 Principles of Mathematics 1 Solution 3 3 1 1 3 1 a) + + 3 = + = 3 ( ) b) sin tan 3 = 0 sin = 0 or tan 3 = 0 sin = 0 or tan = 3 ( ) = 0, 180, 60, 180 + 60 = 0, 180, 60, 40 The following activit should serve as a review of our previous trigonometr eperiences. The onl new work is that ou are to tr to do this work without using a calculator. Guided Practice 1. Complete the following table with eact values. Do as much as ou can without a calculator. You will be required to be able to state these values without a calculator, whenever the question asks for eact values. sin θ cos θ tan θ 0 30 45 60 90 10 135 150 180 sin θ cos θ tan θ 10 5 40 70 300 315 330 360. Determine the eact value for each epression. a) cos 45 sin 45 sin 60 b) tan 30 tan 60 + sin 90 ( ( ) ) c) sin 10 + cos 10 Recall that sin = sin. Module
Principles of Mathematics 1 Section 1, Lesson 3 5 tan 60 sin 60 d) cos 135 e) sin 10 cos 10 tan 10 f) cos 150 sin 150 + cos 300 g) sin 150 cos 60 + cos 150 sin 60 sin 10 3. Find the value(s) of θ, where 0 < θ <360, that satisf the following equations: 1 a) sin θ = b) tan θ = 3 3 1 c) cos θ = d) cos θ = and tan θ > 0 3 e) sin θ = and cos θ < 0 f) cos θ = g) sin θ + sinθ = 0 h) 4 cos θ = 1 i) tan θ + 3 = 0 j) sin θ sin θ = 1 k) cos θ cos θ = 0 Check our answers in the Module Answer Ke.
6 Section 1, Lesson 4 Principles of Mathematics 1 Lesson 4 The Unit Circle Outcomes Upon completing this lesson, ou will be able to: locate the position of the terminal point P(θ) on the unit circle for special values of the real number θ state a value of θ which would produce a given position of a special circular point P(θ) Overview This lesson is a lead-in to the stud of circular functions. However, our abilit to locate special points on the unit circle is etremel important for the rest of this module. Definition: A unit circle is the circle with centre at the origin and with a one-unit radius. + (1, 0) The circumference of this circle is r = (1) =. We now set up a circular function game in which ou move certain distances around this circle governed b the following rules: 1. The point with coordinates (1,0) is the starting point for all of our trips around the circle. It corresponds to the initial arm defined in Lesson 1. The arc is then said to be in standard position.. If the distance to be moved is positive ou will move in a counterclockwise direction along the unit circle. 3. If the distance to be moved is negative ou will move in a clockwise direction along the unit circle. Module
Principles of Mathematics 1 Section 1, Lesson 4 7 4. The notation P(θ) will be used to indicate where our trip has terminated after ou have moved a distance of θ units. P(θ) will be called the circular, or terminal, point. We use θ to represent a distance here rather than an angle, but do not be concerned. We will connect this distance to an angle soon! The following diagram will be used to label some special points on the unit circle so that ou can tr some trips: A B S C Figure A Where is P(0)? This notation asks, What is our terminal point if ou do not move from the starting point? Answer S, since ou alwas begin our trips from the point with coordinates (1,0). Where is P()? This asks for the circular point of a counterclockwise trip halfwa around the circle. Answer B Note: Since the circumference of the unit circle is it follows that the distance: once around the unit circle is Therefore, the distance halfwa around the circle must be. Man students feel that going once around the circle should be. Well it is not! You ma wish to remember the following fact: halfwa around the unit circle is Module
8 Section 1, Lesson 4 Principles of Mathematics 1 Eercise 1 Use Figure A (previous page) to fill in the table. θ 3 3 5 7 101 100 P(θ) Solution θ 3 3 5 7 101 100 P(θ) B A C C A S A A S B S To make the game more interesting, four additional points are inserted at the midpoints of each quarter arc. A L K B M C S N Figure B The midpoints subdivide a semi-circle of arc length into four equal arcs, each of length 4. ll ll ll ll Module
ll Principles of Mathematics 1 Section 1, Lesson 4 9 Eercise Use Figure B to fill in the table. θ 4 3 4 4 3 4 5 7 5 4 4 4 7 4 11 57 83 4 4 4 P(θ) Solution θ 4 3 4 4 3 4 5 7 5 4 4 4 7 4 11 57 83 4 4 4 P(θ) K L N M L N M K L N L To make the game still more interesting, we subdivide the semicircle into si arcs of equal length b trisecting each of the quarter circles. ll ll ll ll ll Each small arc has a length of. Since reduces to, 6 6 3 it follows that ou will have multiples of and multiples of 6 3 as ou travel around the circle. Module
30 Section 1, Lesson 4 Principles of Mathematics 1 Eercise 3 Use the diagram below to fill in the table. (Assume that all the little arcs are equal.) D C B E A F S G K H I J θ 6 3 3 3 5 7 5 6 6 3 7 11 7 6 6 3 13 6 P(θ) Solution θ 6 3 3 3 5 7 5 6 6 3 7 11 7 6 6 3 13 6 P(θ) A B J H G G J E K B K Module
Principles of Mathematics 1 Section 1, Lesson 4 31 Guided Practice If we include the midpoints and the trisection points on one unit circle, the following diagram results. Use this diagram to answer the questions in the assignment. D F E G C B A H S I Q J K N M L 1. Find the letter corresponding to the following value of θ: 7 a) b) 3 6 c) 7 d) 5 4 e) f) 14 6 5 g) h) 53 11 17 i) j) 6 3 k) l) 4 3 m) n) 3 5 15 o) p) 6 4 7 q) r) 4 3 Module
3 Section 1, Lesson 4 Principles of Mathematics 1. Find the letter(s) corresponding to the following value of θ: a) k if k is an even integer (e.g.,,, etc....) b) k if k is an odd integer (e.g., 1, 1, 3, etc....) k 3 c) if k is an integer (e.g.,,,, etc...) 3. Find the letter(s) corresponding to the following value of θ: k a) if k is an odd integer 4 k b) if k is an even integer (e.g.,, 4 = = etc....) 4 4 4 4. State the smallest positive value of θ which will terminate at the point with letter: a) B b) H c) M d) L e) A f) C g) S h) F i) I j) D 5. State the smallest negative value of θ which will terminate at the point with letter: a) B b) H c) M d) L e) A f) K g) S h) F i) I j) N 3 (e.g.,,,, etc...) 4 4 4 Check our answers in the Module Answer Ke. Module
Principles of Mathematics 1 Section 1, Lesson 5 33 Lesson 5 The Circular Functions Outcomes Upon completing this lesson, ou will be able to: state the coordinates of special values of the circular points P(θ) define the trigonometric functions as circular functions define the three reciprocal functions in terms of their reciprocals solve circular function equations, which have real number solutions Overview For ever arc length θ on the unit circle there is onl one point, P(θ), determined b this arc. Hence, ou have a function which maps ever arc into a unique point. Now our emphasis will switch from the point to the coordinates of the point. We redefine P(θ) as the ordered pair of numbers rather than just the location. For eample, P(0) = (1,0), our starting point; P = (0,1), the coordinates at the top of the unit circle; and so on. For each real number θ there corresponds an ordered pair of numbers (, ), which are the coordinates of P(θ). The above two eamples were rather eas because the circular point terminated on the ais. What happens when the point is not on an ais? We will fill in the coordinates of the special terminal points in the first quadrant and assign the rest as an eercise. Module
34 Section 1, Lesson 5 Principles of Mathematics 1 A ll M O N ll S Since M is the midpoint of the arc it follows 1 o o MON = d90 i = 45. opposite sin MON = = = hpotenuse OM 1 since the radius of the unit circle is one unit. Furthermore, ou know from Lesson of this module that o 1 1 sin 45 =. Therefore, =. Additionall, MON is isosceles, so = 1. 1 1 Hence, P, = or coincidentall 4 o o P = ( cos 45, sin 45 ). 4 O A 30 30 ll 30 H ll K ll S If points H and K are the trisection points of arc AS, it follows that KOS = HOK = AOH = 30. Therefore, HOS = 60. Module
Principles of Mathematics 1 Section 1, Lesson 5 35 Let s go after the coordinates of K and H! A O 30 K S sin KOS = = since the radius of the unit circle is 1. OK 1 1 And, since KOS = 30 it follows = sin30 = (from Lesson ). 3 Similarl, cos 30 =. Therefore, = cos30 =. OK 3 1 Hence, P =, or P = (cos30, sin30 ). 6 6 You can use the figure below to conclude that: 1 3 P =, or P = (cos60, sin60 ) 3 3 A H O 60 S At this point, b using the smmetr of a circle, ou are capable of completing the coordinates of ever special point on the unit circle. Module
36 Section 1, Lesson 5 Principles of Mathematics 1 Eercise 1 Use the stated coordinates to complete the coordinates of ever other labeled special circular point. A C B D 3 H 1, F I H G K J F H G I K J F I H G K J 1 1 M 1, K 3, E S(1,0) F Q G J R N L Solution F D HG F HG I KJ B 1 3, F CHG 1 1 I, K J 3 1I, KJ A(0, 1) F HG I KJ H 1 3, F M 1 1, HG F HG I K J I KJ 3 1 K, E( 1, 0) S(1, 0) F HG F HG I KJ F 3 1, Q 3 1, I K J G 1 1, F I J 1 3, HG KJ L(0, 1) F HG F HG R 1 1, F N 1 3I, HG KJ I K J I KJ Note: You should know the coordinates of these points b heart, or at least those in Quadrant I then ou can use that information along with the CAST diagram to quickl determine the other points. Module
Principles of Mathematics 1 Section 1, Lesson 5 37 As mentioned previousl, we have a function which matches a real number θ with an ordered pair of numbers (, ). Mathe-maticians prefer to match each real number with onl one real number. This is accomplished b taking a different approach to the trigonometric ratios, an approach which was suggested b realizing that the coordinates of the points on the unit circle are generated b trigonometric function values. Alternate definitions of trigonometric ratios (These are also called circular functions.) If θ is an arc in standard position (this means on a circle of radius 1) and P(θ) = (, ) is the terminal point for the real number θ, then: sin θ = = = r 1 cos θ = = = r 1 and tan θ =, where 0 θ = Since tan sinθ cosθ Eample 1 Use the coordinates on the unit circle and the definitions of the circular functions to find: a) sin b) cos c) tan 6 3 d) sin e) cos f) sin + cos 4 6 4 6 Solution a) 1 This is a good sequence of steps to reason through: 1. Locate where the point terminates on the unit circle. (K in this case). Think of the coordinates of this point. 3 1, 3. Since sin θ =, select the second coordinate as our answer: 1
38 Section 1, Lesson 5 Principles of Mathematics 1 b) 1 This is a good sequence of steps to reason through: 1. Locate where the point terminates on the unit circle: E. Think of the coordinates of this point: ( 1,0) 3. Since cos θ =, select the first coordinate as our answer. c) 3 This is a good sequence of steps to reason through: 1. Locate where the point terminates on the unit circle.. Think of the coordinates of this point. 3. tan θ = d) sin 4 1. arc terminates at M. coordinates of M 3. Since 3 sinθ = = cosθ 1 = 3 sinθ = = 1 sin = 4 1 1 1, e) cos : 6 1. arc terminates at K. coordinates of K 3 1, 3 3. Since cosθ = = 3 cos = 6 1 3 1 3 3+ 1 f) sin + cos = 4 6 + = + = Module
Principles of Mathematics 1 Section 1, Lesson 5 39 Now we ll go backwards. Eample Solve the following over the interval 0 θ <. Give solutions as eact values. 3 a) cosθ = b) sin θ sinθ = 1 Solution 3 a) cosθ = 3 cos θ =. The -coordinate is equal to 3 1 3 1 in two locations: D, and F, 3 1 5 3 1 7 D, P and F, P = 6 = 6 5 7 Solutions :, 6 6 b) If sin θ sinθ = 1 then sin θ sin θ 1 = 0 ( sinθ + 1)( sin 1) = 0 sinθ + 1= 0 or sin θ 1= 0 1 sinθ = or sinθ = 1 1 sin θ =. The -coordinate is equal to or 1 7 3 1 11 3 1 P = F, and P = Q, 6 6 P = A(0,1) 7 11 Solutions :,, 6 6 Module
40 Section 1, Lesson 5 Principles of Mathematics 1 There are three more circular functions defined as reciprocals of the basic three. The are called the secant (written as sec), cosecant (written as csc), and cotangent (written as cot) functions. Definition: If θ is an arc length in standard position, with terminal circular point P(θ), then: 1 sec θ = cos θ 1 csc θ = sin θ cos θ 1 cot θ = or sin θ tan θ provided the denominators are not equal to zero. Eample 3 Determine the eact value of sec 3cot 6 4 Solution = = = sec = = 1 1 4 sec 6 cos 6 3 3 3 6 3 1 1 cot = = = 1 3cot = 3() 1 = 3 4 tan 1 4 4 4 4 9 5 sec 3cot = 3 = = 6 4 3 3 3 3 Module
Principles of Mathematics 1 Section 1, Lesson 5 41 Eample 4 Solve for θ (eactl) for 0 θ< : csc θ + cscθ = 0 Solution If csc θ + cscθ = 0 cscθ( cscθ + 1) = 0 cscθ = 0 or cscθ + 1 = 0 1 sin θ = 0 csc θ = 1 (no solution) 1 sin θ = = 1 = 1 3 P = L(0, 1) 3 θ = The Link Between Angles in S.P. and Coordinates on Unit Circle The following diagram shows the link between the angle in standard position (or the length) in the unit circle and the coordinates of the corresponding point on the unit circle. Given θ in standard position, sin θ =, cos θ =, tanθ = r r and because r = 1 in the unit circle, it is simple to use this diagram to find eact trig. function values of an angle which terminates on an of the points shown. Eamples 3 1 sin = = 4 r 11 3 cos = = 6 r 4 3 1 tan = = 3 = 3 3 Also sec cot 3 1 1 = = 3 cos 3 tan r = = 0 = = = 0 1 Module
4 Section 1, Lesson 5 Principles of Mathematics 1 F HG F HG I K J I KJ 1 3, F 1 1, HG 3 3 3 1I 4, KJ 5 6 (0, 1) F HG I KJ 1 3, F 1 1 I, HG K J 3 4 F 3 1, HG 6 I KJ ( 1, 0) (1, 0) 0 ( ) F HG F HG I KJ 3 1 7, 6 5 1 1 I, K J 4 4 3 F 1 3I, HG KJ 3 (0, 1) F HG 3 7 1 1 5 4, HG 3 F1 3I, HG 11 6 KJ F I K J I KJ 1, For angles outside the interval (0, ), we can still locate the corresponding point on the unit circle. Eample: Find the eact value of 5 csc. 6 Solution 5 is the arc of the unit circle obtained b starting at the 6 positive -ais and travelling CLOCKWISE round the unit 5 circle for a distance of. 6 This is the same point on the unit circle we would have obtained b travelling COUNTER CLOCKWISE from the positve -ais for the rest of the circumference of the circle, i.e., 5 5 5 6 Thus csc 6 6 1 5 = 5 6 6 = csc 6 7 = 7 6 = csc 6
Principles of Mathematics 1 Section 1, Lesson 5 43 Guided Practice This activit is almost identical to the assignment at the end of the previous two lessons ecept ou will be using arc lengths instead of angle measures. 1. Find sin θ, cos θ, and tan θ for each of the following values of θ. In each question, take the etra step to write the coordinates of P(θ). a) 7 b) 3 6 c) 7 d) 5 e) 4 6 f) 14 g) 5 11 h) 53 i) 6 j) 17 k) 3 l) m) 4 3 5 n) o) 3 6 p) 15 7 q) 4 3 r) 4. Determine the eact value of each epression. Note: Although each epression involves functions and should contain notation such as sin (), mathematicians alwas like shortcuts and as long as there is no ambiguit, the write these circular function statements as sin, etc. This convention is used throughout this course. a) cos + tan 3 4 15 13 b) sin cos 0 tan 6 c) tan 6 ( ) d) cos sin 5 3 5 e) tan cos + sin tan 3 4 6 f) cos 6 Module
44 Section 1, Lesson 5 Principles of Mathematics 1 7 3 g) sin cos tan 3 6 4 47 h) sin cos 47 3. Solve the following equations over the interval 0 θ <. Do these questions without the use of a calculator. All answers must be eact special values. 1 a) sin θ = b) tan θ = 3 3 1 c) cos θ = d) cos θ = and tan θ > 0 3 e) sin θ = and cos θ < 0 f) cosθ = θ + θ = θ = g) sin sin 0 h) 4 cos 1 i) tan 3 0 j) sin sin 1 θ + = θ θ = k) cos cos 0 θ θ = ( ) 4. Determine the eact value of each epression. a) sec sin cot b) sec + csc 3 6 4 6 6 5 6 3 3 c) cot d) tan cot 7 4 4 6 e) cos sec f) 4 csc 5. Find the eact solution of the following equations over the interval 0 θ <. a) sec θ = b) csc θ = c) cot θ = 1, and sin θ < 0 d) cot θ cot θsin θ = 0 e) tan θ 1 = 0 f) csc θ sin θ = 0. Check our answers in the Module Answer Ke. Module
Principles of Mathematics 1 Section 1, Lesson 6 45 Lesson 6 Radian Measures Outcomes Upon completing this lesson, ou will be able to: convert an angle from degrees to radians and vice versa convert an angle from degrees to revolutions and vice versa convert an angle from revolutions to radians and vice versa solve angle measure problems, using an of the three units of measure relate arc length and central angles using the formula s = θr Overview You should have noticed in the last lesson that a relationship seems to eist between the arcs on the unit circle and the trigonometric values of the central angles (angles with vertices at the centre). You will develop this relationship in this lesson. You are familiar with degree measure where: one revolution = 360 (once around the circle) A second method of measuring angles is in terms of radians. Definition: When the arc of a circle has the same length as the radius of the circle, then the measure of the angle, with verte at the centre, subtended b this arc is 1 radian. Circle centre O AOB = 1 radian + (1, 0) Module
46 Section 1, Lesson 6 Principles of Mathematics 1 Thus, for an arc of length s units, the radian measure of the corresponding central angle is given b the number of radii needed to make up the length of the arc s. ll ll ll ll s O = radians r For eample, an arc of length 10 cm in a circle of radius 4 cm has a central angle whose measure is given b s O = r 10 = 4 =.5 radians Eercise 1 Use the definition of radian measure to fill in the following table. Mathematical agreement: If no smbol appears after an angle measure, the measure of the angle is in radians. To remind people that the measure is in radians ou can also use a superscript r to denote radian measure. Hence, the following are all equivalent: = radians = r Arc Length Radius Central Angle (Radians) 6 3 0 5 4 16 4 Module
Principles of Mathematics 1 Section 1, Lesson 6 47 Solution Arc Length Radius Central Angle (Radians) 1 6 3 0 5 4 4 0.5 16 4 4 The relationship among these three quantities is apparent. The central angle is equal to the ratio of the arc length to the radius. Stated in a product form: arc length = central angle in radian measure radius Man phsics books use the smbol s for distance and θ for a angle, so the following formula for this relationship has endured: s = θr, where s is the arc length θ is the measure of the central angle in radians r is radius Eample 1 a) Find the length of the arc subtending b a central angle of 3 if the radius of the circle is 0 cm. b) Find the length of the radius of a circle if a two-radian central angle is subtended b a 15 m arc. Solution a) Since s = θr, it follows that s = 3(0) = 60 cm. b) Since s = θr, it follows that 15 = (r) so that r = 7.5 m. Furthermore, since the circumference of a circle is r, it follows that: Using the formula s = θr r = θr = θ So one revolution is radians. Module
48 Section 1, Lesson 6 Principles of Mathematics 1 One revolution = r and since one revolution = 360 it follows that 360 = radians. Eercise a) Convert 1 into radians. b) Convert 1 radian into degrees. This eercise provides ou with a method to easil convert degrees to radians and radians to degrees. Method Since 360 = radians, b dividing b 36 0, 360 = = 360 360 180 = 180 = 180 multipl b. 180 o 1 radians (about 0.0175 radians if ou do the division) o Since 1 radians, to convert degrees to r adians o Since 360 = radians, b dividing b, 360 = 1 radian 180 1 radian, so 1 radian is about 57.3 degr ees. = 180 Since 1 =, to convert radians to degrees 180 multipl b. Module
Principles of Mathematics 1 Section 1, Lesson 6 49 Eample a) Convert 45 into radians. b) Convert radians into degrees. Solution o a) 45 = 45 = radians 180 4 180 o b) = = 180 You are now in a position to link trigonometric ratios and circular function values. Let s look at the circular function values for central angles given in radians. Eample 3 Determine the value (eact if possible, otherwise to three decimal places) of: a) sin (Small superscript "r" means " radians.") 4 r b) cos3 Solution r r Eample 4 Find the eact value, if possible; otherwise use a calculator to find the approimate value. a) cos b) sin 0 3 c) tan d) tan 1 4 r 180 1 (45 is one of our a) sin sin sin45 4 = 4 = = "special" angles) r 180 540 b) cos3 = cos3 = cos 0.990 (calculator) Solution a) is a special value so the memorized answer is cos = 1. o Module
50 Section 1, Lesson 6 Principles of Mathematics 1 b) Since 0 is not a special value, use a calculator, in degree mode, to find sin 0 = 0.340. 3 c) Special value, hence tan = 1. 4 d) Not a special value. Use the calculator, in radian mode, to find tan 1 = 1.55741. Note: it was not 1º so it must mean 1 radian. Unless stated otherwise, ou should alwas attempt to find the trigonometric, or circular function, value without a calculator. If the measure is not a special value, or special value multiple, use a calculator use radian mode for arcs or angles in radian measure, and degree mode for angles stated in degree measure. To check if our graphing calculator is in radian mode or in degree mode, press MODE. To change the mode, move the cursor to the preferred mode (Radian or Degree) and press ENTER. Eample 5 Convert each angle measure into revolutions. r 5 o a) b) 10 6 Solution a) If 1 revolution = r r 5 and revolutions = 6 1 5 5 1 5 then = = = = revolutions r 5 6 6 1 6 b) If 1 revolution = 360º and revolutions = 10º 1 360 1 7 then = = 1 = 7 = revolutions 10 7 1 3 When we solve an equation such as cosθ = for 0 θ <, we get not two but four solutions. Here s wh. Module
Principles of Mathematics 1 Section 1, Lesson 6 51 Eample 6 Solve cosθ = for 0 θ < 6 Using the CAST diagram, θ is positive in Quadrants I and IV. θ = or θ = 6 6 11 θ = or θ = 6 6 11 θ = or θ = 1 1 When we divide b, we reduce the domain of the solutions from [0,p) to [0,p). In order to obtain all solutions for 0 q < p, we need to travel around the unit circle twice. 11 Instead of θ = or 6 6 11 11 we need θ = or or + or + 6 6 6 6 11 13 3 =,,, or 6 6 6 6 11 13 3 Dividing b, θ =,, or 1 1 1 1 We now have four solutions in the domain [0,). In summar: 3 1. The equation sin θ = would also have four solutions over the domain [0,).. The equation sin3θ = 3 would have si solutions over the domain [0,). 3. If we solve trig equations in degrees instead of radians, the equation sin θ = domain [0,360 ). 3 would have four solutions over the Module
5 Section 1, Lesson 6 Principles of Mathematics 1 Guided Practice 1. Convert each degree measure into radian measure. a) 5 b) 15 c) 460 d) 330. Convert each radian measure into degree measure. Round off to one decimal place where necessar. 7 11 a) b) 6 1 c). 634 d) 0. 985 5 3. Determine the complement of. What is the supplement 1 of this angle? (Recall: Two angles are complementar if the sum of their measures is 90. Two angles are supplementar if the sum of their measures is 180.) 4. Epress the supplement of 130 in radian measure, eactl. 5. An isosceles triangle has a base angle with measure. 7 What are the measures of the other two angles? Epress our answer in radians. 6. Two parallel lines have interior alternate angles with measures 4 and Find the value of.. 4º 7. In what quadrant(s) are the following conditions satisfied? a) sin θ < 0 and cos θ > 0 b) tan θ > 0 c) sin θ > 0 and cos θ < 0 d) cos θ < 0 Module
Principles of Mathematics 1 Section 1, Lesson 6 53 Remember the CAST diagram. S T A C 8. An alien being, walking along the equator of her planet, travelled a distance of 1.3 clicks. If the diameter of this planet is 16.4 clicks, through what central angle did this alien travel? Epress our answer to the nearest tenth of a degree. 9. In the diagram AC = 6 cm, BC = 8, and AB is a diameter. Find the circumference of this circle. 10. If a wheel having a circumference of 30 cm rolls 5 cm, how man radians has it turned? How man degrees has it turned? 11. Convert each measure into revolutions: a) 3 radians b) 75 1. Use a calculator, using five digits after the decimal, in order to find each value: a) tan 0 b) cos 0 c) sin 4. d) sec 15. e) cot3 f) sin 5 g) 7 cos h) tan i) csc5 j) sin k) cot 05. l) sinb 3g m) cosb 1g n) csc 4 o) tan b g B A L K M C S N Module
54 Section 1, Lesson 6 Principles of Mathematics 1 13. Use a calculator to solve the following simple equations for, 0 <. Round our answer to two digits after the decimal. (Hint: Use sin 1 function on our calculator. Since is in radians, have our calculator set to radian mode. Be sure to find all angles, not just those in Quadrant I.) a) sin = 013. b) cos = 013. c) tan = d) sin = 05. e) cos = 0. 675 f) sec =. 5 g) cot = 4 h) csc = 6 The following questions are ver important. Attempt to do as much as ou can without using a calculator. If ou must use a calculator, be sure our setting is to the correct mode. 14. Solve each equation for, where 0 < 360. a) sin = sin b) sin = cos c) sin = 1 d) cos = 1 e) csc = 4 f) tan 1 = tan 15. Solve each equation for, where 0 <. a) cos cos = 0 b) tan = 1 ( ) c) csc = sec d) sin 1 = 0 e) cos = 0 f) sin tan + sin = 0 Check our answers in the Module Answer Ke. Review Section 1 before attempting the review questions following the summar on the net page. These questions should help ou consolidate our knowledge as ou prepare Section Assignment.1. Module
Principles of Mathematics 1 Section 1, Summar 55 Summar of Section 1 The Basics The initial arm is the positive direction of the -ais. A positive angle is made b moving in a counter-clockwise direction from the initial arm. A reference angle is alwas made with the -ais. reference angle θ reference angle CAST Diagram S A T C This diagram indicates the quadrants where the basic trigonometric ratios are positive. Eact Values Special trigonometric values can be calculated for Quadrant I and from there the values can be determined for the other quadrants. You should learn this table or learn how to derive these values from triangles (see Lesson 3, Facts 1 and ). The following table summarizes the trigonometric values for the Quadrantal angles: Degrees 0 30 45 60 90 Arc Length 0 6 4 3 sinθ 0 1 1 3 1 cosθ 1 3 1 1 0 tanθ 0 1 3 1 3 undefined Module
56 Section 1, Summar Principles of Mathematics 1 Unit Circle Definitions 1 P (,) s For an point P(,) on the unit circle (radius = 1) with an arc length of s in standard position. θ sinθ = 1 1 cscθ = = sinθ cosθ = 1 1 secθ = = cosθ 1 tanθ = cotθ = = tanθ Solving Equations The equations = sin θ, = cos θ, = tan θ have two solutions for the domain 0 < θ <, or the domain 0 θ< 360. The equations = sin nθ, = cos nθ, = tan nθ, where n is an integer, have n solutions for the domain 0 θ< 360 or the domain 0 θ<. To solve the equations = sec θ, = csc θ, = cot θ, use the reciprocals: 1 1 1 cos θ =, sin θ =, tanθ = Some more information can be found in Michelson s book, Theor and Problems for Senior High Math, on pages 47 49, #1 7, and further problems #1 5 on pages 5 53. Module
Principles of Mathematics 1 Section 1, Review 57 Module, Section 1 Review 1. Find the eact value for each epression. tan o o a) 1 sin30 cos 60 b) 4 5 sin 6 o 3 3 c) csc 10 cot d) sin cos 4 5 o e) sec cot 30 f) cos 135 + sin 55 4. Solve the following equations over the interval 0 θ <. Do these questions without the use of a calculator. All answers must be eact special values. 1 a) sin θ = b) tan θ = 1 1 1 c) cos θ = d) sin θ = and tan θ > 0 3 e) cos θ = and csc θ < 0 f) cosθ = 3 θ + θ = θ = g) cos cos 0 h) 4 cos 1 θ + = θ + θ = i) cot 3 0 j) cos cos 1 θ θ = k) tan tan 0 3. Convert each degree measure into radian measure. a) 45 b) 150 c) 500 d) 600 Module
58 Section 1, Review Principles of Mathematics 1 4. Convert each radian measure into degree measure. Round off to one decimal place where necessar. 5 11 a) b) 6 1 c) 3. 8 d) 0. 543 5. Solve each equation for, where 0 < 360, or for θ, where 0 θ<. θ = θ = a) sin sin b) sin cos c) cos θ = 1 d) sin = 1 θ = = e) csc 4 f) tan 1 tan g) cos θ 1 = 0 h) cos ( θ+ ) = 0 i) cos tan + cos = 0 Check our answers in the Module Answer Ke. Now do the Section Assignment which follows this section. When it is complete, send it in for marking. Module
Principles of Mathematics 1 Section Assignment.1 59 PRINCIPLES OF MATHEMATICS 1 Section Assignment.1 Version 05 Module
60 Section Assignment.1 Principles of Mathematics 1 General Instructions for Section Assignments These instructions appl to all the section assignments, but will not be reprinted each time. Remember them for future sections. (1) Treat this assignment as a test, so do not refer to our module or notes or other materials. A graphing calculator is required. () Where questions require computations or have several steps, showing these can result in part marks for some eercises. Steps must be neat and well-organized, however, or the instructor will onl consider the answer. (3) Alwas read the question carefull to ensure ou answer what is asked. Often unnecessar work is done because a question has not been read correctl. (4) Alwas clearl underline our final answer so that it is not confused with our work. Module
Principles of Mathematics 1 Section Assignment.1 61 Section Assignment.1 Trigonometric Functions Total Value: 48 marks (Mark values in margins) (11 marks) 1. Fill in the blanks with the correct response. a) If sin θ < 0 and sec θ > 0, then θ terminates in Quadrant: 7 b) The eact value of sec is: 6 c) A cm wire is stretched along a circle of radius 4 cm. The degree measure of the central angle subtended b this wire arc is: d) Find the smallest possible angle in standard position of 4400. e) What is the reference angle for 4400? 7 f) What is the reference angle for? 5 g) If 0 < θ < and cos θ = 0.5, then the value of cos (θ) is: F h) The value of cos I 1 3 K J, in degree measure, is: HG cos i) The value of 16 3 is: F I HG K J j) The eact value of cot 5 is: 6 k) What is a coterminal angle with 170? Module
6 Section Assignment.1 Principles of Mathematics 1 F. Find the value of sin 3 cos. 4 HG I K J + (4) 3. a) Evaluate tan 7.5. (1) b) Evaluate eactl 45 cos. 4 (1) 4. As the time changes from 4:10 p.m. to 6:45 p.m. on the analog clock, determine the change in radian measure of the minute hand. () Module
Principles of Mathematics 1 Section Assignment.1 63 5. Solve each equation for, where 0 <. (4) a) sin + 1 = 0 (5) b) cos ( sin 1) = 0 (4) c) cos = 1 (4) d) tan = 1 Module
64 Section Assignment.1 Principles of Mathematics 1 6. Solve each equation or inequalit for : 1 a) sin + = where 0 < (4) b) Solve for 0 < 360 3 tan = 1 (4) c) sec = 3, where 0 < 360 (4) Send in this work as soon as ou complete this section. Total: 48 marks Module