Basic Trigonometry. Stephen Perencevich Department of Mathematics Montgomery College 3Π 4 Π 7Π 7Π 6 Π 2 0, 1 1,0 1,0 Π 3 5Π 3 Π 2

Transcription

1 Basic Trigonometry Stephen Perencevich Department of Mathematics Montgomery College 4 0,,0, , 5

2 i All firgures generated by Mathematica. R Revised July, 06 Typeset by L A TEX. Perencevich, Stephen Trigonometry Stephen Perencevich Montgomery College Rockville, MD 0850 stephen.perencevich@montgomerycollege.edu c 06 George Canter Institute. No part of this publication may be reproduced, stored in a retreval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without prior written permission of the author.

3 For Missy, the flower of Lackawanna. ii

4 Contents Right-Triangle Trigonometry. Introduction Right-Triangle Trigonometry Trigonometric Functions of any Angle Trigonometric Identities. Basic Trigonometric Identities Midcourse Review 6 Review Practice Test Radian Measure 8. The Definition of Radian Measure Graphs of Trigonometric Functions 4 4. The Unit Circle Graphs of Trigonometric Functions Inverse Trigonometric Functions Inverse Trigonometric Fucntions Trigonometric Equations Final Review 47 Review Practice Test A The Greek Alphabet 49 B Midcourse Review Answers 50 C Final Review Answers 5 D Answers to Odd-Numbered Problems 54 Index 58

5 Chapter Right-Triangle Trigonometry. Introduction In this section we will recall some theorems on triangles from geometry. Theorem... The sum of the angles in any triangle is 80 C b a A c Figure.. B In terms of Figure..: m A + m B + m C = 80. Definition... C b a A c B D e f F d E Two triangles are similar, denoted ABC DEF if the measures of corresponding angles are equal and if the ratios of the lengths of corresponding sides are equal: a d = b e = c f

6 CHAPTER. RIGHT-TRIANGLE TRIGONOMETRY Theorem... (AA Similarity) If two pairs of angles in two triamgles are congruent, then the triangles are similar. b C a e F d A c B D f E Figure..: Similar Triangles In terms of Figure..: if m A = m D and m B = m E, then ABC DEF. Recall that if two triangles are similar, then ratios of pairs of corresponding sides are equal. Again, in terms of Figure..: if ABC DEF, then (among other ratios) b c = e f and b e = a d Conventions: Note that in Figures.. and.. the angles of the two triangles are labeled by upper case letters and the sides opposite the angles are labeled with the corresponding lower case letter. We will follow this convention throughout these notes. In geometry angles, which are geometrical objects, are rigously distinguished frm their measures, which are numbers. In these notes we will allow not-technically-correct identification of an angle with its measure. Thus we will write A = 6 in place of m A = 6. Theorem..4. (Pythagoras) In a right triangle with side lengths a, b, and c, where c is the length of the hypotenuse, c = a + b a c b Figure..

7 CHAPTER. RIGHT-TRIANGLE TRIGONOMETRY 4 Section. Problems b C a e F d A c B D f E Figure..4 For Problems 4 we are given that ABC DEF.. If a = and b =, find d e.. If a =, b = 4, and d = 5, find e.. If a = and c = 5, find f d. 4. If b = 4, c = 7, and d = 5, find f. For Problems 5 8 we are given that ABC is a right triangle with C a right angle. B a c C b Figure..5 A 5. If a = and b =, find c. 6. If a = and b = 5, find b. 7. If a = and b =, find c. 8. If a = and c =, find b. 9. In ABC, if A = and B = 55 find C. 0. In DEF, if D = and E = 47 find F.. In XY Z, if X = Y and Z = 00 find X.. In P QR, if P = Q and R = 0 find P.. If ABC with C a right angle and A = 7, find B. 4. If ABC with C a right angle and B = 6, find A.

8 CHAPTER. RIGHT-TRIANGLE TRIGONOMETRY 5. Right-Triangle Trigonometry Consider a right triangle that contains an angle with mearure 7. Opp Hyp Adj 7 Figure.. As usual, the side opposite the right angle is called the hypotenuse of the triangle. The side that forms the 7 angle with the hypotenuse is called the adjecent side to the angle and the last side is called the opposite side of the angle. Any right triangle containing a 7 angle will be similar to the one in Figure... In any right triangle containing an angle of measure 7, the ratios opp hyp, adj opp, and will be the same. These ratios are called the sine, cosine and hyp adj tangent, respectively. The same can be said of any acute angle. This leads to the following definition. Definition... Let θ ba an acute angle in a right triangle. We define the following functions on θ. Opp Adj Hyp sin θ = opp hyp cos θ = adj hyp tan θ = opp adj Figure.. In days gone by, the values of the trigonometric functions were tabulated and books on trigonometry contained tables such as Table... Today, of course, these values are easily computed on a scientific calculator.

9 CHAPTER. RIGHT-TRIANGLE TRIGONOMETRY 6 θ sin θ cos θ tan θ Table..: A Short Trigonometric Table To solve a triangle means to find the measures of unknown angles and lengths of unknown sides of a given triangle. Example... Solve the triangle given in the diagram. a C B c.5 4 A Since the acute angles in a right triangle are complementary, B = 90 4 = 56 tan 4 = a.5 a =.5 tan 4 =.4 Using the Pythagoras theorem: c = a + b c = c = = 4. The values in Table.. are decimal approximations of the values of the trigonometric functions. If the exact value of one of the trigonometric functions happens to be known, then the exact value of the others can be found.

10 CHAPTER. RIGHT-TRIANGLE TRIGONOMETRY 7 Example... Given that for acute angle θ, sin θ = 7, find cos θ and tan θ. Choose a right triangle containing θ with opposite side of length and hypotenuse of length 7. Let x be the length of the remaining side. Using the Pythagoras theorem, 7 x 0 x + = 7 x = 40 x = 40 = 0 cos θ = 0 7 tan θ = 0 One of the earliest uses of trigonometry is in indirect measurement problems. The height of a tall building would be difficult to measure directly. It is easier to measure a horizontal distance along the graound and the angle of a sightline and then use trigonometry to deduce the desired height. Example..4. At a point 05 feet away from a building the angle between the ground the top of the building is 55. Find the height of the building. Let h be the height of the building. Then, h tan 55 = h 05 h = 05 tan 55 = 50.0 There are three remaining trigonometric functions: cosecant, secant, cotangent, which are simply the reciprocals of sine, cosine, and tangent, respectively.

11 CHAPTER. RIGHT-TRIANGLE TRIGONOMETRY 8 Definition..5. For an angle θ we define, csc θ = sin θ sec θ = cos θ cot θ = tan θ Example..6. Given that for acute angle θ, cos θ = 4 5, find the values of the remaining 5 trigonometric functions on θ. Choose a right triangle containing θ with adjacent side of length 4 and hypotenuse of length 5. Let y be the length of the remaining side. y 5 4 Using the Pythagoras theorem, y + 4 = 5 y = 9 y = sin θ = 5 csc θ = 5 cos θ = 4 5 sec θ = 5 4 tan θ = 4 cot θ = 4 The Inverse Trigonometric Functions Given an acute angle θ, the trigonometric functions assign a number to that angle. The inverse trigonometric functions reverse that process. That is given the value of a trigonometric function on some angle, the inverse trigonometric function will give the measure of the angle. The inverse trigonometric functions are denoted: sin, cos, and tan. Scientific calculators have buttons for these functions. For example, using a scientific calculator, sin 5 = sin = 5

12 CHAPTER. RIGHT-TRIANGLE TRIGONOMETRY 9 Example..7. ABC is a right triangle with C a right angle. Given that a = 5. and c = 8.4, solve ABC. 5. C B 8.4 b 6.7 A Using the Pythagoras theorem, b = 6.7. sin A = A = sin ( A = 7.4 ) B = = 5.6 Special Triangles In most cases evaluating the trigonometric functions on a given angle θ is a calculator problem. There are three acute angles, 0, 45, and 60, on which the exact value of the trigonometric angles can be found. Consider an eqilateral triangle with side length. An altitude drawn from one of the angles bisects the angle and the opposite side. 0 h Using the Pythagoras theorem, + h = h = h = 60 We now construct a triangle as follows: choose two perpendicular lines and construct a segment of length on each line, each with the point of intersection for one endpoint.

13 CHAPTER. RIGHT-TRIANGLE TRIGONOMETRY 0 45 r 45 Using the Pythagoras theorem, + = r r = r = We can now use the following triangles to evaluate the trigonometric functions on the angles, 0, 45, and Section. Problems Figure..: The Special Trianges For problems 9 ABC is a right triangle with C a right angle. Solve the triangle with the given information. Give your answer correct to one decimal place. B a C b c A. a = 8.7; A =. 4. a =.; b = b = 7.9; B = 47.. b =.; A = a = 4.; c = c = 9.; A = 7.. c = 8.; B = b = 5.8; c = c = 5.5; B = 55.

14 CHAPTER. RIGHT-TRIANGLE TRIGONOMETRY For problems 0 5 ABC is a right triangle with C a right angle. Given the value of one of the trigonometric function, find the values of the remaining 5. Give exact-value answers. 0. sin θ = 7.. cos θ = 5.. tan θ = 5.. sin θ =. 4. cos θ = 5 5. tan θ =. 6. Using Figure.., evaluate the six trigonometric functions on (a) θ = 0 (b) θ = 45 (c) θ = 60 Angle of Elevation Angle of Depression 7. Suzy s eyes are 5 feet above the ground. She is standing feet away from the base of a flag pole and the angle of elevation from her eyes to the top of the pole is. What is the height of the flag pole? 8. An observer on the ground measures the angle of elevation of an airplane to be 8. A second oberver, standing 4. miles from the first observer, is directly below the plane. What is the altitude of the plane? 9. Find the angle of elevation of the top of a 6-foot-tall building from a point on the ground that is 97 feet from the base of the In application problems, the angle of depression is the angle between the line of sight of an observer looking downward and the horizontal. The angle of elevation is the angle between the line of sight of an upward looking observer and the horizontal. building. 0. In the previous problem find the angle of elevation of the top of the building from the eye level of an observer whose eyes are at a height of 5.4 feet. (The observer stands 97 feet from the base of the buikding.). The angle of elevation to the top of an oak tree from a point on the ground 59 feet from the base of the tree is 4. What is the distance from the point to the top of the tree?

15 CHAPTER. RIGHT-TRIANGLE TRIGONOMETRY Evel Knievel was a daredevil who rode his motorcycle over ramps to jump various objects. He estimates that, in order to be safe, the angle of the ramp with the ground should be no more than 7.. Evil Knievel is going to jump over 5 school buses that have a height of 9 feet. If his ramp has an angle of 7, what is the distance along the ground from the base of the ramp to the first school bus?. In over to jump over a -foothigh pen containing mountain lions, Evel Knievel constructs a ramp that measures 50 feet along the ground. What is the angle that the ramp makes with the ground? 4. Evel Knievel is going to construct a ramp th jump over some locomotives that have a height 4 feet. If the angle of the ramp with the ground is 7, what is the shortest ramp (not the distance along the ground) that our daredevil can construct?

16 CHAPTER. RIGHT-TRIANGLE TRIGONOMETRY. Trigonometric Functions of any Angle The trigonometric functions defined in Section. can only be evaluated on acute angles. In this section we expand the definitions so that the trigonometric functions can be evaluated on any angle. Angles in Standard Position Recall the following definition from geometry. Definition... An angle is the union of two rays with a common endpoint. We will adpot a more dynamic view of angles. An angle in standard position is angle in the coordinate plane with its vertex at the origin. The angle is formed by placing both rays along the positive x-axis, fixing one in place and rotating the other. The fixed ray is called the initial ray and the rotated ray is called the terminal ray. In forming such an angle the terminal ray may be rotated either in the counterclockwise or clockwise directions. They are distinguished by defining their measures to be positive or negative, respectively. y 0 0 x Angles formed by counterclockwise rotation have positive measures. Angles formed by clockwise rotation have negative measures. Figure..: Angles in Standard Position

17 CHAPTER. RIGHT-TRIANGLE TRIGONOMETRY 4 Recall that the four quadrants of the coordinate plane are defined as in Figure... II x 0, y 0 I x 0, y 0 Note the inequalities defining the quadrants are strict, so that the coordinate axes from the boundries of the quadrants, but do not belong to the quadrants. III x 0, y 0 IV x 0, y 0 Figure.. Angles whose terminal rays lie on the coordinate axes are called boundary angles II I II I III IV III IV Figure..: Boundary Angles 0 This is a nonstandard usage. Many authors call these angles quandrantal angles. We have adopted the term boundary angle as more discriptive. The terminal ray of such an angle lies in the boundary of the quadrants, not in the quadrants.

18 CHAPTER. RIGHT-TRIANGLE TRIGONOMETRY 5 Definition... ' y x For an angle, θ, in standard position whose terminal ray is interior to one of the quadrants, the reference angle, θ, of θ is the acute angle formed by the terminal ray and the x-axis. Example... Find the reference angle for the following angles: 50, 0, ' ' ' Angles in standard position are formed by rotation of the terminal ray. There is no reason to restrict the rotation to once-round. This can result in different angles having the same terminal ray.

19 CHAPTER. RIGHT-TRIANGLE TRIGONOMETRY 6 Definition Angles with the same terminal ray are called coterminal angles. 70 Example..5. Find three angles coterminal with θ = 0. (i) = 90 (ii) = 750 (iii) 0 60 = 0

20 CHAPTER. RIGHT-TRIANGLE TRIGONOMETRY 7 The Definition of the Trigonometric Functions We now turn to the definition of the trigonometric functions on angles in standard position. Consider first an angle in standard position whose terminal ray lies in the first quadrant. Let (x, y) be a point on the terminal ray other than the origin. Form a triangle by droping a perpendicular from this point to the x-axis. This is called the reference triangle if θ. Let r be the distance from the point to the origin. 90 x, y By the Pythagoras theorem r = x + y 80 r x y 0 60 Using the definitions of the trigonometric functions of an acute angle, sin θ = y r csc θ = r y cos θ = x r sec θ = r x 70 tan θ = y x cot θ = x y We now make the observation that the values of the trigonometric functions given above do not depend on the terminal ray of the angle lying in the first quadrant, so we take them to be our general definition. Definition..6. (The trigonometric functions.) Let θ be an angle in standard position, P (x, y) be a point on the terminal ray of θ, and let r be the distance from P to the origin. We define the values of the trigonometric functions on θ as follows: 90 x, y sin θ = y r csc θ = r y 80 y x r ' 0 60 cos θ = x r tan θ = y x sec θ = r x cot θ = x y 70

21 CHAPTER. RIGHT-TRIANGLE TRIGONOMETRY 8 In Definition..6, r is a distance, so r > 0. However, (x, y) are the coordinates of a point and so they may be positive, negative, or zero. Regardless of the values of x and y, r still satisfies the Pythagoras relation: r = x + y. Example..7. Let θ be an angle in standard position whose terminal ray passes through the point (, 5). Find the values of the six trigonometric functions on θ. 90 r = x + y 80 ' r 70 5, r = + ( 5) = 9 r = 9 sin θ = 5 9 csc θ = 9 5 cos θ = 9 sec θ = 9 tan θ = 5 cot θ = 5 80 II: sin 0 III: tan I: all 0 IV: cos As Example.. shows, the values of the trigonometric functions can sometimes be negative. The signs of the trigonometric functions depend the signs on the signs of the coordinates (x, y). The diagram to the left is helpful in remembering which functions are positive in which quadrants. Figure..4

22 CHAPTER. RIGHT-TRIANGLE TRIGONOMETRY 9 Example..8. Given that sin θ = 4 and 90 < θ < 80 Find the values of the remaining five trigonometric functions on θ. Since sin θ = 4 > 0, the terminal ray of θ must lie in Quadrants I or II. The fact that 90 < θ < 80 implies that the terminal ray of θ lies in Quadrant II. Form the reference triangle of θ using the point (x, ). 80 x, 4 ' x r = x + y 4 = x + x = 7 x = ± 7 In this case, the solution we are looking for is x = sin θ = 4 7 cos θ = 4 tan θ = 7 csc θ = 4 sec θ = cot θ = Finally we observe that for any angle θ interior to one of the quadrants, θ is an ordinary acute angle and, 80 y 90 x, y r ' x 0 60 sin θ = y r cos θ = x r tan θ = y x csc θ = r y sec θ = r x cot θ = x y Hence the value of a trigonometric function on θ is the absolute value of the value of the function on θ. 70

23 CHAPTER. RIGHT-TRIANGLE TRIGONOMETRY 0 Theorem..9. If θ is an angle in stard position whose terminal ray lies interior to one of the quadrants, then sin θ = sin θ csc θ = csc θ cos θ = cos θ sec θ = sec θ tan θ = tan θ cot θ = cot θ Example..0. Find the values of sin 50, cos 50, and tan 50. The reference angle of θ = 50 is θ = ' sin 0 = cos 0 = tan 0 = Using Theorem..9 and Figure sin 50 = + 70 cos 50 = tan 50 = Because it is easy to find points on the coordinate axes, we can directly use the the definition of the trigonometric functions to evaluate on the boundry angles.

24 CHAPTER. RIGHT-TRIANGLE TRIGONOMETRY Example... Find the values of sin 90, cos 90, and tan 90. The point P (0, ) lies on the terminal ray of the 90 angle and has distance r = to the origin. 90 x,y 0, 80 r sin 90 = y r = = 0 60 cos 90 = x r = 0 = 0 tan 90 = y x = 0 (undefined) 70 Section. Problems For problems 9, sketch the given angle in standard position and find its reference angle.. θ = 0.. θ = 00.. θ = θ = θ = 7 6. θ = θ = θ = θ = 0. For problems 0 5, find three angles coterminal to the given angle. 0. θ = 0.. θ = 00.. θ = 95.. θ = θ = θ = 0. For problems 6, evaluate the six trigonometric functions on the angle θ whose terminal ray passes through the given point. Give exact-value answers. 6. P (, 5). 7. P (, 4). 8. P (5, ). 9. P (, 0). 0. P (, 4). (0, ). For problems 7, evaluate the six trigonometric functions on the given angle Give exact-value answers.. θ = 0.. θ = θ = θ = θ = θ = 70.

25 Chapter Trigonometric Identities. Basic Trigonometric Identities While there are six trigonometric functions, they are not independent. In this section we will explore some relationships among them. We will use the righttriangle definition of the trigonometric functions for geometric motivation, but the identities we find will hold for the generalized definitions as well. The Reciprocal Identities This first set of identities is simply a restatement of Definition..5: Reciprocal Identities sec θ = cos θ csc θ = sin θ cot θ = tan θ cos θ = sec θ sin θ = csc θ tan θ = cot θ The trigonometric ratios depend on the fact that any two right triangles containing the acute angle θ must be similar. It is also true that any individual right triangle that contains the angle acute θ must satisfy the Pythagoras theorem. The next two sets of identities establish relationships among the the trigonometric ratios derived from similarity relations and the Pythagoras theorem. The Quotient Identities This set of identities establishes relationships among ratios of trigonometric functions.

26 CHAPTER. TRIGONOMETRIC IDENTITIES b c a Figure.. Referring to Figure.., sin θ cos θ = b/c a/c = b c c a = b a = tan θ cos θ sin θ = a/c b/c = a c c b = a b = cot θ Quotient Identities sin θ cos θ = tan θ The Pythagoras Identities cos θ sin θ = cot θ The right triangle in Figure.. satisfies the Pythagoras theorem: c = a + b. We can use this theroem to investigate relationships among the squares of the quotients that define the trigonometric functions. First we will compute the sum of the squares of the sine and cosine ratios. ( a ) ( b cos θ + sin θ = + = c c) a c + b c = a + b Putting all this together, we have: c = c c = cos θ + sin θ = ( ) Note that sin θ is the notation for (sin θ) and similarly for the other trigonometric functions.

27 CHAPTER. TRIGONOMETRIC IDENTITIES 4 We can now use identity ( ) together with the quotient and reciprocal identities to generate two more identities. Dividing identity ( ) by cos θ: cos θ cos θ + sin θ cos θ = cos θ + tan θ = sec θ Dividing identity ( ) by sin θ: cos θ sin θ + sin θ sin θ = sin θ cot θ + = csc θ Pythagoras Identities cos θ + sin θ = + tan θ = sec θ + cot θ = csc θ Using the basic identities established so far and some algebra, we can establish other identities. The process of proving a trigonometric identity is one of simplification. Pick one side of the identity and use algebra and the basic identities given above to simplify it. The end of the process should yield the other side o fthe identity. Example... Prove the identity: cos θ + sin θ tan θ = sec θ. cos θ + sin θ tan θ = cos θ + sin θ sin θ cos θ = cos θ cos θ cos θ + sin θ cos θ = cos θ cos θ + sin θ cos θ = cos θ + sin θ cos θ = cos θ = sec θ

28 CHAPTER. TRIGONOMETRIC IDENTITIES 5 Example... Prove the identity: cot θ = cos θ. csc θ cot θ cos θ/ sin θ = csc θ / sin θ = cos θ sin sin θ θ = cos θ Example... Prove the identity: tan θ + cot θ = csc θ sec θ. tan θ + cot θ = sin θ cos θ + cos θ sin θ = sin θ sin θ sin θ cos θ + cos θ sin θ cos θ cos θ = sin θ + cos θ sin θ cos θ = sin θ cos θ = sin θ = csc θ sec θ cos θ Section. Problems For problems 0, prove the trigonometric identity... tan θ sin θ tan θ sec θ = sec θ. = sin θ.. sin θ + sin θ cot θ = csc θ. 4. sin θ cos θ tan θ = = cos θ. 5. tan θ cos θ = sin θ 6. sin θ csc θ = csc θ tan θ tan θ = cos θ sin θ. 8. cos θ sin θ = sin θ. 9. sec θ sec θ + = cos θ + cos θ 0. tan θ sin θ = tan θ sin θ.

29 Midcourse Review Review. Find the values of all six trigonometric functions on angle A in the triangle below. A 6 b. Find the reference angles for each of the following angles. (a) 0 (c) 50 (e) 0 (b) (d) 97 (f) 45. Given that α is an acute angle and sin α = /, find the values of the remaining 5 trigonometric functions on θ. 4. Given that 90 < θ < 80 and sin θ = /, find the values of the remaining 5 trigonometric functions on θ. B C 8. Use the special triangles to evaluate: (a) cos 0 (b) tan 0 (c) sin 0 9. Evaluate: (a) cos 90 (b) tan 80 (c) sin 0 0. Given sin θ =.5. (d) cos 5 (e) tan( 60 ) (f) sin 50 (d) sin 70 (e) tan 70 (f) cos 0 (a) Find θ, the referance angle for θ. (b) Given that θ lies in Quad. IV, evaluate: i. cos θ ii. tan θ. Solve the right triangle in the figure below. B 5. Solve the right triangle in the figure below.. c B a c C A.5 6. Evaluate sin + cos. 7. Given that sin = 0.907, find: (a) sin 57 (d) cos 67 (b) sin 0 (e) cos (c) sin 7 (f) cos( ) C 4.7. Prove the following trigonometric identities: (a) cot θ csc θ = cos θ (b) cot α + tan α = sec α csc α (c) tan β sin β + cos β = sec β (d) sin θ = tan θ sec θ + sin θ A 6

30 CHAPTER. TRIGONOMETRIC IDENTITIES 7 Practice Test. If θ is an acute angle and sin θ =, find tan θ. Give an exactvalue 7 answer. (a) sin 60 (b) cos 90 (c) tan 00 (d) sin 50 (e) cos 40 (f) tan 80. In the triangle below, find the length of side a. Give your answer correct to one decimal place. a C B 4.7 c Problem A 8. In the triangle below, find the measure of angle A. Give your answer correct to two decimal places. 4. C B 6. c Problem 8 A. Given that cos θ = 4 5 and 70 < θ < 60, find the exact value of sin θ. 9. Find the value of the six trigonometric functions on the angle shown in the diagram below 4. Find the reference angle for θ = Given that sin θ = 0.57 and θ lies in Quad IV. For this problem, give your answers correct to three digits after the decimal. (a) Find θ, the reference angle for θ. (b) cos θ = (c) tan θ = 6. Given that tan θ = 5, find the exact value of cot(90 θ). y x 7. Use your knowledge of the special triangles and the definition of the trigonometric functions to give the exact value of the following: Problem 9 0. Prove the following identity: csc θ(csc θ sin θ) = cot θ.

31 Chapter Radian Measure. The Definition of Radian Measure Up to now we have been measuring angles in degrees without much comment. We now wish to step back and consider what is meant by degree mrasure. In order to define an angle measure, divide a circle into n equal parts by choosing n equally spaced points on the top half of the circle and drawing diameters from those points. The result is that n equal central angles of the circle are formed. We then take the measure of one of these central angles to be the standard angle measure. n 8 The measure of an angle is then the number of nonoverlapping standard angles that fit into the angle. Degree measure is the result of choosing to divide the circle into 60 equal parts. The choice of 60 is abritrary. Another choice will result in an equally valid angle measure, rather like switching from feet to inches in linear measure. In contrast to linear measure, it is possible to construct an intrinsic or natural measure for angles, that is a measure that does not involve an arbitrary choice. That is the content of this section. 8

32 CHAPTER. RADIAN MEASURE 9 Definition... (Radian Measure) Let θ be an central angle of a circle or radius r, and let s be the length of the arc subtended by θ. r s The radian measure of θ is given by: θ = s r It may seem that Definition.. depends on the choice of a circle, but since all circles are similar, the defining ratio will remain the same from circle to circle. The definition is also reminiscent of the definition of π. Recall that π is the ratio of the circumference to the diameter of a circle (Again, this does not depend on the circle chosen.): d c π = c d In fact, this observation will be the key to converting angle measures between degrees and radians. Radian-Degree Conversions s c 80 r d If θ = 80, then the radian measure of θ is given by Hence, θ = s r = c/ d/ = c d = π 80 = π

33 CHAPTER. RADIAN MEASURE 0 We note that radian measure is actually unitless. The radius and arc length are measured in linear units, say inches. Then θ = s in r in = s r (no units) Nevertheless, it is a useful mnemonic device to append rad to radian measure as if it were a unit. For the purposes of converting between the two angle measures it is also useful to append deg to an angle rather than the usual. With these conventions, we have two conversion factors: 80 deg = π rad 80 deg π rad = π rad 80 deg = Example... (i) Convert 60 to radian measure; (ii) Convert 5π 6 to degree measure. (i) 60 deg π rad 80 deg = π (ii) 5π 6 rad 80 deg π rad = 50 Radian measure is rarely used when actually measuring angles in applications in engineering, phisics or even more every-day applications such as carpentry. The reason is that it is somewhat clumbsy. An angle of mearure radian has a degree measure of 57.. The real utility of radian measure is that since it is unitless, it converts the trigonometric functions from functions assigning numbers to geometric objects (angles) to ordinary functions on the real numbers. Figure.. shows the boundry angles and the special triangles with angles measured in radians. It would be worth while to become familiar with the radian values of the special angles.

34 CHAPTER. RADIAN MEASURE II III I IV Figure..: Special Angles in Radians Example... Using diagrams in Figure.., (i) sin π 6 = (ii) cos π 4 = (iii) tan π = It is just as easy to evaluate the trigonometric functions on the boundry angles. Example..4. r 0, Using the diagram on the left and the definition of the trigonometric functions,, 0 0,, 0 0 sin π = y r = = cos π = x r = = tan π = y x = 0 (undefined)

35 CHAPTER. RADIAN MEASURE The procedure for evaluating the trigonometric functions is exactly that same as that for angles measured in degrees. Example..5. Evaluate tan π. π lies in Quad II and has reference angle π π = π. ' Using Figure.., 0 tan π = tan π = Section. Problems For problems 9, convert the angle to radian measure.. θ = 0.. θ = 90.. θ = θ = θ = θ = θ = θ = θ = 5. For problems 0 5, find three angles coterminal to the given angle. 0. θ = π 6.. θ = π 4.. θ = π.. θ = 5π θ = π 6 5. θ = 5π. For problems 6 4, sketch the angle in standard position and find the reference angle for the given angle.

36 CHAPTER. RADIAN MEASURE 6. θ = 7π θ = 5π 6.. θ = 5π. 7. θ = 5π θ = π. 0. θ = 5π 4.. θ = π. θ π 6 For problems 0, evaluate the trigonometric function. 4. θ = 5π.. sin π 6.. tan π θ = π. 5. cos 5π. 6. sin 7π 6 7. cos π. 8. cos π. 9. sin π 0. sin 7π 4.

37 Chapter 4 Graphs of Trigonometric Functions 4. The Unit Circle Let us recall the definition of the sine and cosine functions. If θ is an angle in standard position and (x, y) is a point on the terminal ray of θ and r is the distance from (x, y) to the origin, then, sin θ = y r cos θ = x r (4.) (4.) x, y, 0 0, r, 0 0 If (x, y) is the point of intersection of the terminal ray with the unit circle, then, since r =, sin θ = y r = y 0, cos θ = x r = x Figure 4..: The Unit Circle 4

38 CHAPTER 4. GRAPHS OF TRIGONOMETRIC FUNCTIONS 5 This is often taken as the definition for the sine and cosine functions. We will state this observation as a theorem. Theorem 4... If (x, y) is the point of intersection of the terminal ray of θ with the unit circle, then (x, y) = (cos θ, sin θ). r cos, sin 0 Example 4... For θ = 5π 6 : (i) Draw the terminal ray of θ; (ii) Find the reference angle of θ; (iii) Find the coordnates of the point of intersection of θ with the unit circle. Since π < 5π 6 < π, θ lies in the second quadrant. Hence θ = π 5π 6 = π 6. 6 x, y ' cos 5π 6 = sin 5π 6 = (x, y) = ( /, / )

39 CHAPTER 4. GRAPHS OF TRIGONOMETRIC FUNCTIONS 6 Section 4. Problems For each angle given, (i) Draw the terminal ray of θ; (ii) Find the reference angle of θ; (iii) Find the coordnates of the point of intersection of θ with the unit circle.. θ = π 4. θ = 4π. θ = π 6 4. θ = π 5. θ = π 6. θ = π 4. Graphs of Trigonometric Functions We now turn to the graphs of the sine and cosine functions. Let θ increase from 0 to π as in Figure 4... The y-coordinate of the point of intersection of the terminal ray of θ and the unit circle begins at 0, increases to a maximum value of at θ = π/ returns to 0 at θ = π, decreases to a minimum value of at θ = π/, and then returns back to 0 at θ = π. The graph of y = sin θ is than as shown in Figure ,,0, , 5 Figure 4..: y = sin θ

40 CHAPTER 4. GRAPHS OF TRIGONOMETRIC FUNCTIONS 7 Of course increasing θ above would result in the terminal ray of θ tracing over the same points on the unit circle several times, so the shape of the sine curve in Figure 4.. will simply repeat itself. Similarly for negative values of θ. We say that the sine function is periodic with period π. y sin 4 Figure 4.. A similar analysis of the cosine function yields the following cosine curve. y cos 4 Figure 4.. In general, it suffices to sketch a single period of a periodic curve. Figure 5.. shows the sine and cosine curves for 0 θ π. These curves should be memorized. y sin y cos (a) y = sin θ (b) y = cos θ Figure 4..4

41 CHAPTER 4. GRAPHS OF TRIGONOMETRIC FUNCTIONS 8 Consider now the circle of radius A centered at the origin. If θ is an angle in standard position and (x, y) is a point on the terminal ray of θ and r is the distance from (x, y) to the origin, then, by definition, sin θ = y r cos θ = x r x, y A, 0 0, A r A A, 0 0 If (x, y) is the point of intersection of the terminal ray with the unit circle, then, since r = A, sin θ = y A y = A sin θ 0, A cos θ = x A x = A cos θ Figure 4..5 Whereas the minimum and maximum values of y = sin θ and y = cos θ are and, the minimum and maximum values of y = A sin θ and y = A cos θ are A and A. The graphs retain the same basic shapes, but are stretched vertically. Example 4... On the same set of axes, plot: (i) y = sin θ and y = sin θ (ii) y = cos θ and y = cos θ y y (a) y = sin θ (b) y = cos θ Figure 4..6

42 Like with any other function, the graphs of y = sin θ and y = cos θ are the graphs of y = sin θ and y = cos θ reflected across the x-axis. Example 4... On the same set of axes, plot: (i) y = sin θ and y = sin θ (ii) y = cos θ and y = cos θ y y (a) y = sin θ (b) y = cos θ Figure 4..7 Definition 4... For the functions y = A sin θ (4.) y = A cos θ (4.4) A is called the amplidtue of the function. Section 4. Problems For problems -6 graph the given trigonometric function for 0 θ π. Label the θ-axis with multiples of π.. y = sin θ. y = 5 sin θ. y = 4 cos θ 4. y =.5 sin θ 5. y = cos θ 6. y = cos θ 9

43 CHAPTER 4. GRAPHS OF TRIGONOMETRIC FUNCTIONS 40 Chapter 5 Inverse Trigonometric Functions 5. Inverse Trigonometric Fucntions Let us recall than, interms of their graphs, functions assign points on the x-axis to points on the y-axis and that the inverse of a function reverses the process. y y f x y x x f y x Figure 5.. Recall also that a function may not have multiple outputs for a single input. This presents a problem for the sine function: it s inverse is not a function. y sin Figure 5..

44 CHAPTER 4. GRAPHS OF TRIGONOMETRIC FUNCTIONS 4 We do wish to define an inverse function for the sine function. To this end we recall that a function has three components: a domain, a range, and a rule that assigns elements of the domain to elements of the range. When defining a function, we have complete control over the first two components. We now define a new function whose rule agrees with that of the sine function, but we take the domain to be the interval π θ π. We denote this function Sin θ. This new (restricted) function has the same range as the sine function (the interval [, ]), but has the advantage that each element of the range is the image of only one element of the domain. y Sin Figure 5.. We denote the inverse of Sin θ by Sin x. Sin x = θ (i) sin θ = x if and only if (ii) π θ π We make a similar restriction for the cosine function, except that we cannot use the same interval. We define Cos θ = cos θ for 0 θ π. Note this function has a capital S to distinguish it from the ordinary (nonrestricted) sine function.

45 CHAPTER 4. GRAPHS OF TRIGONOMETRIC FUNCTIONS 4 y Cos Figure 5..4 Cos x = θ (i) cos θ = x (ii) 0 θ π if and only if For the tangent function, the restriction is similar to that of the sine function. Tan θ = tan θ for π < θ < π. y Tan Tan x = θ if and only if (i) tan θ = x (ii) π < θ < π Figure 5..5 Example 5... ( Evaluate Cos ). By the definition of Cos, this problem is equivalent to: find θ such that cos θ = and 0 θ π. Since cos θ = < 0, the terminal ray of θ lies in either Quad. II or Quad. III.

46 CHAPTER 4. GRAPHS OF TRIGONOMETRIC FUNCTIONS 4 cos θ = cos θ = 0 θ = π Since we require that 0 θ π, θ must lie in Quad. II and 6 θ = π π = π Figure 5..6 Example 5... Evaluate Sin ( ). By the definition of Sin, this problem is equivalent to: find θ such that sin θ = and π θ π. Since sin θ = < 0, the terminal ray of θ lies in either Quad. III or Quad. IV. sin θ = sin θ = θ = π 4 Since we require that π θ π must lie in Quad. IV and 4 θ = π 4 Figure 5..7 Example 5...

47 CHAPTER 4. GRAPHS OF TRIGONOMETRIC FUNCTIONS 44 ( Evaluate Tan 5 ). Since tan θ = 5, θ is not a special angle, so we 4 4 use a scientific calculator: ( Tan 5 ) = Be sure that your calculator is in radian mode. Section 5. Problems For problems -9 evaluate the inverse trigonometric function. Give exactvalue answers in radians. (. Sin 4. Sin ) 7. Tan ( ) 6 ). Cos (. Tan 5. Cos ( 6. Cos ) 5. Trigonometric Equations 8. sin 9. cos ( ) 4 The simplest trigonometric is one of the form: sin θ = (5.) 4 To solve Equation 5. means to find all angles θ that satisfy the equation. Since < 0, we should expect to find solutions only in Quadrants III and IV, where the sine function is negative. Recall that sin θ = sin θ, where θ is the reference angle of θ. sin θ = sin θ = ) θ = Sin ( = π Two solutions of Equation 5. are: θ = π π 6 = π 6 θ = π + π 6 = 7π 6 6 Figure 5..

48 CHAPTER 4. GRAPHS OF TRIGONOMETRIC FUNCTIONS 45 Any other solution of Equation 5. general solution is are coterminal with these two, so the θ = 7π 6 + kπ θ = π 6 + kπ k = 0, ±, ±, ±, In general, it suffices to find all solutions θ of a trigonometric equation that satisfy 0 θ < π. Any other solution will be coterminal with one of these. In most cases, some algebra is involved to put a trigonometric equation in the form of Equation 5.. Example 5... Find all solutions to cos θ = 0 that satisfy 0 θ < π. We first solve algebraically for cos θ: cos θ = 0 cos θ = cos θ = Since cos θ = > 0, we will find solutions only in Quadrants I and IV, where the cosine function is positive cos θ = cos θ = ( ) θ = Cos = π 4 The two solutions of the equation are: 4 θ = π 4 Figure 5.. θ = π π 4 = 7π 4 4

49 CHAPTER 4. GRAPHS OF TRIGONOMETRIC FUNCTIONS 46 Of course, the solutions to trigonometric equations are not always special angles. Example 5... Find all solutions to tan θ + = 0 that satisfy 0 θ < π. We first solve algebraically for cos θ: tan θ + = 0 tan θ = tan θ = Since tan θ = < 0, we will find solutions only in Quadrants II and IV, where the tangent function is negative. tan θ = ' ' 0 tan θ = ( ) θ = Tan = The two solutions of the equation are: Be sure you calculator is in radian mode. θ = π θ = θ = π θ =.588 Figure 5.. Section 5. Problems For problems -9 find all solutions for the trigonometric equation satisfying 0 θ < π. Where possible, give exact-value answers. (This is when the solutions are special angles.). sin θ = sin θ +. = cos θ = 0 6. cos θ + = 0 5. cos θ = 0 8. sin θ = 0. tan θ = tan θ + 4 = 0 9. tan θ = 0 4 4

50 Final Review Review. Convert the following angles to radian measure. Give exact-value answers. (a) 5 (b) 5 (c) 0 (d) 70 (e) 60 (f) 0. Convert the following angles to degree measure. (a) π 8 (b).5 (c) 5.5 (d) π 6 (e).7 (f) 4π. A 6-foot ladder is resting against a wall. The top of the ladder is 5 feet from the ground. What angle does the ladder make with the wall? 4. From a point 00 feet from the base of a Roman aqueduct in northern Spain the angle of elevation to the top of the aqueduct is 9. Find the height of the aqueduct. 5. Find the values of the six trigonometric functions on an angle whose terminal ray passes throug the point P (7, 4). 6. Find the exact value of sin θ, cos θ, and tan θ for the following angles. (a) θ = π (b) θ = π 4 (c) θ = 5π 6 (d) θ = π (e) θ = 7π 4 (f) θ = 4π 7. Use a calculator to evaluate the following expressions. Give your answer accurate to 4 decimal places. (a) sin 7 (b) cos( 6 ) (c) sec( 4.45) (d) csc Find two values θ with 0 θ < π that satisfy the trigonometric equation. Give exact-value answers. (a) sin θ = (b) cos θ = (c) tan θ = (d) tan θ = (e) sin θ = (f) cos θ = 9. Graph the following functions. State the domain, range, amplitude and period. (a) y = sin θ (b) y = cos θ 0. Solve the following equations in the interval 0 θ < π. (a) sin θ = 0 (b) cos θ = (c) cos θ + = 0 (d) tan θ + 5 = 0. Use the definitions of sin θ = y r, cos θ = x r, and tan θ = y to prove the following x trigonometric identities. (a) sin θ + cos θ = (b) tan θ = sin θ cos θ (c) + tan θ = sec θ (d) + cot θ = sec θ. Multiply and simplify: (a) (sin θ cos θ)(sin θ + cos θ) (b) (sin θ cos θ) (c) ( + tan θ) (d) tan θ(cos θ csc θ) 47

51 CHAPTER 4. GRAPHS OF TRIGONOMETRIC FUNCTIONS 48 Practice Test. Convert 6 to radian measure. Give an exact-value answer.. Convert 5π to degree measure. 8 Give your answer accurate to two decimal places.. At a point 80 feet away from the base of a building, the angle of elevation to the top of the building is 8. What is the height of the building? Give your answer correct to two decimal places. 4. Use a calculator to evaluate the following. Give your answers correct to four decimal places. (Be sure to check your mode.) (a) sin 7 (b) cos(.4) 5. Evaluate the following. Give exact-value answers. (a) sin π (b) cos π (c) tan 5π 4 (d) sin π (e) tan π (f) cos π 6 6. Find the exact value in radians of ( cos ) 7. Find the exact value in radians of tan () 8. Sketch the graph of y = sin θ for 0 θ π. Label the θ-axis with multiples of π and state the amplitude of the function. 9. Multiply and simplify fully: sin θ(cot θ + tan θ) 0. Find the value of the six trigonometric functions on the angle shown in the diagram below y Problem 0 x

52 Appendix A The Greek Alphabet alpha α A nu ν N beta β B xi ξ Ξ gamma γ Γ omicron o O delta δ pi π epsilon ε E rho ρ P zeta ζ Z sigma σ Σ eta η E tau τ T theta θ upsilon υ Υ iota ι I phi ϕ Φ kappa κ K chi χ X lambda λ Λ psi ψ Ψ mu µ M omega ω Ω 49

53 Appendix B Midcourse Review Answers A. Find the values of all six trigonometric functions on angle A in the triangle below. 6 b B C Using Pythagoras, b = 5. sin A = 6 csc A = 6 cos A = 5 6 sec A = 6 5 tan A = 5 cot A = 5 a C B 5. Solve the right triangle in the figure below..5 c A B = 90 = 57 tan = a.5 a =.5 tan =. Using Pythagoras, c =.5 +. = 4.. Find the reference angles for each of the following angles. (a) 60 (c) 0 (e) 0 (b) (d) 8 (f) 45. Given that α is an acute angle and sin α = /, find the values of the remaining 5 trigonometric functions on θ. 5 Using Pythagoras, the remaining side is 5. sin θ = csc θ = cos θ = 5 sec θ = 5 tan θ = 5 cot θ = 5 6. Evaluate sin + cos. sin + cos = 7. Given that sin = 0.907, find: Note that 90 = 67, so cos 67 = sin. After finding the reference angle for each angle below, and considering the quadrant in which the angle lies, we find: (a) sin 57 =.907 (d) cos 67 =.907 (b) sin 0 =.907 (e) cos =.907 (c) sin 7 =.907 (f) cos( ) = Use the special triangles to evaluate: (a) cos 0 = (b) tan 0 = (d) cos 5 = (e) tan( 60 ) = Given that 90 < θ < 80 and sin θ = /, find the values of the remaining 5 trigonometric functions on θ. ' Α sin θ = / θ = α from problem. Since θ lies in Quad IV, sin θ = csc θ = cos θ = 5 sec θ = 5 tan θ = 5 cot θ = 5 (c) sin 0 = 9. Evaluate: (a) cos 90 = 0 (b) tan 80 = 0 (c) sin 0 = 0 (See Figure B.0.) (f) sin 50 = (d) sin 70 = (e) tan 70 : undefined (f) cos 0 = 50

54 APPENDIX B. MIDCOURSE REVIEW ANSWERS Prove the following trigonometric identities: 80,0 0,,0 0, 0 60 (a) cot θ csc θ = cos θ cot θ cos θ/ sin θ = csc θ / sin θ = cos θ sin sin θ θ = cos θ (b) cot α + tan α = sec α csc α Given sin θ =.5. Figure B.0. (a) Find θ, the referance angle for θ. sin.5 = (b) Given that θ lies in Quad. IV, evaluate: i. cos θ = cos = ii. tan θ = tan = 0.. Solve the right triangle in the figure below. C B (d) 4.7 c A Using Pythagoras, c = = 5.6. A = tan 4.7 =.4 B = 90 =.4 sin θ = tan θ sec θ + sin θ = 56.6 sin θ + sin θ = sin θ + sin θ + sin θ + sin θ sin θ sin θ = + sin θ ( sin θ) sin θ cot α + tan α = cos α sin α + sin α cos α = cos α + sin α cos α sin α = (c) tan β sin β + cos β = sec β = sin θ cos θ = sin θ cos θ cos α sin α = cos α = sec α csc α sin α tan β sin β + cos β = sin β sin β + cos β cos β = sin β cos β + cos β = sin β cos β + cos β cos β = sin β + cos β cos β = cos β = sec β = tan θ sec θ cos θ

55 Appendix C Final Review Answers. Convert the following angles to radian measure. Give exact-value answers. (a) π (b) 5π 4 (c) π 6 (d) π (e) π (f) π. Convert the following angles to degree measure. (a) sin π = 0 cos π = tan π = 0 (b) sin π 4 = cos π 4 = tan π 4 = (c) sin 5π 6 = cos 5π 6 = tan 5π 6 = cos π = 0 tan π : undef. (e) sin 7π 4 = cos 7π 4 = tan 7π 4 = (f) sin 4π = cos 4π = (a) 67.5 (b) 85.9 (c) 00.8 (d) 0 (e) 54.7 (f) 40. cos θ = 5 6 θ = ( ) cos 5 6 = tan 9 = h 00 h = 00 tan 9 = 68.9ft 5. x = 7, y = 4. r = 7 + ( 4) = 65 (a) sin θ = 4 65 (b) cos θ = 7 65 (c) tan θ = 4 7 (d) csc θ = (e) sec θ = 7 (f) cos θ = 7 4 (d) sin π = tan 4π = 7. Use a calculator to evaluate the following expressions. Give your answer accurate to 4 decimal places. (a) (b) (c).855 (d) Find two values θ with 0 θ < π that satisfy the trigonometric equation. Give exact-value answers. (a) θ = π 6, 5π 6 (b) θ = 5π 6, 7π 6 (c) θ = 5π 6, π 6 (d) θ = π 4, 5π 4 (e) θ = 7π 6, π 6 (f) θ = π, 5π 6. Find the exact value of sin θ, cos θ, and tan θ for the following angles. 9. Graph the following functions. State the domain, range, amplitude and period. 5

56 APPENDIX C. FINAL REVIEW ANSWERS 5 y sin amplitude: ; period: π. y cos (c) (d). Multiply and simplify: + tan θ = + y x = x + y x = r x = sec θ + cot θ = + x y = y + x y = r y = sec θ amplitude: ; period: π. 0. Solve the following equations in the interval 0 θ < π. (a) θ = π 4, π 4 (b) θ = 0.8, (c) θ = π, 4π (d) θ =.95, Use the definitions of sin θ = y r, cos θ = x r, and tan θ = y to prove the following trigonometric x identities. (a) (b) ( y ) ( x ) sin θ + cos θ = + r r = y + x r = r r = (a) (sin θ cos θ)(sin θ + cos θ) = sin θ cos θ (b) (c) (d) (sin θ cos θ) = (sin θ cos θ) (sin θ cos θ) = sin θ sin θ cos θ + cos θ = sin θ cos θ ( + tan θ) = ( + tan θ) ( + tan θ) = + tan θ + tan θ = sec θ + tan θ tan θ(cos θ csc θ) = sin θ cos θ ( cos θ sin θ ) = sin θ cos θ = sin θ sec θ sin θ cos θ = y/r x/r = y r r x = y x = tan θ

57 Appendix D Answers to Odd-Numbered Problems Section Section.. B = 68 b =.5 c =.. A = 8 a = 5.0 b = A = 4.5 B = 55.5 c = A = 4 a = 7.4 c = A = 7 a =. b = sin θ = csc θ = cos θ = 5 sec θ = 5 tan θ = 5 cot θ = 5 sin θ = csc θ = cos θ = sec θ = tan θ = cot θ = sin θ = csc θ = cos θ = sec θ = tan θ = cot θ = ft ft..5 ft Section

58 APPENDIX D. ANSWERS TO ODD-NUMBERED PROBLEMS , 90, , 500, , 00, sin θ = 4 5 csc θ = 5 4 cos θ = 5 sec θ = 5 tan θ = 4 cot θ = 4 sin θ = 0 cos θ = tan θ = 0 Section... tan θ sin θ csc θ : undef. sec θ = cot θ : undef. sin θ/ cos θ = sin θ = sin θ cos θ sin θ = cos θ = sec θ sin θ + sin θ cot θ = sin θ( + cot θ) = sin θ csc θ = sin θ sin θ = sin θ = csc θ sin θ = cos θ = 0 csc θ = sec θ : undef. tan θ : undefined cot θ = 0 sin 0 = csc 0 = cos 0 = sec 0 = tan 0 = cot 0 = sin 80 = 0 cos 80 = tan 80 = 0 sin 70 = cos 70 = 0 csc 80 : undef. 80 θ = cot : undefined csc 70 = sec 70 : undef. tan 70 : undefined cot 70 = 0 sin θ/ cos θ tan θ cos θ = cos θ = sin θ sin θ + tan θ + tan θ = cos θ sin θ cos θ = cos θ + sin θ cos θ cos θ sin θ cos θ = cos θ + sin θ cos θ = cos θ + sin θ = cos θ sin θ cos θ sin θ cos θ cos θ sin θ

59 APPENDIX D. ANSWERS TO ODD-NUMBERED PROBLEMS sec θ sec θ + = = Section.. π. π π 7. π π 4. 5 cos θ cos θ + cos θ cos θ + cos θ cos θ = cos θ cos θ = cos θ + cos θ cos θ + cos θ θ = π 4 9. θ = π 6. θ = π Section 4. ( ).,. (, 5. (, 0) ) Section 4. y sin y 4cos 4 4 Problem Problem y cos Problem 5

60 APPENDIX D. ANSWERS TO ODD-NUMBERED PROBLEMS 57 Section 5.. π 4. π π 9. π Section 5.. θ = π 6. θ = π 4 θ = 5π 4 5. θ = 0.84 θ = θ = π θ = 5π 9. θ = π θ = 5π 6 θ = 4π

61 Index amplitude, 7 angle, boundary angles, coterminal, 4 initial ray, period, 5 periodic, 5 Pythagoras, radian measure, 7 reference angle, reference triangle, 5 similar, solve a triangle, 6 standard position, terminal ray, 58