3 Unit Circle Trigonometry

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1 0606_CH0_-78.QXP //0 :6 AM Page Unit Circle Trigonometr In This Chapter. The Circular Functions. Graphs of Sine and Cosine Functions. Graphs of Other Trigonometric Functions. Special Identities.5 Inverse Trigonometric Functions.6 Trigonometric Equations Chapter Review Eercises A Bit of Histor The discussion in Section. leads directl to a more analtical approach to trigonometr where the cosine and sine are defined as the - and -coordinates, respectivel, of a point, ) on a unit circle. It is this interpretation of the sine and cosine that enables us to define the trigonometric functions of a real number instead of an angle. It is this last approach to trigonometr that is used in calculus and in advanced applications of trigonometr. Moreover, a trigonometric function of a real number can then be graphed as we would an ordinar function 5 f ), where the variable represents a real number in the domain of f. From a historical viewpoint, it is not known who made this important leap from sines and cosines of angles to sines and cosines of real numbers. The shape of a plucked guitar string, fied at both ends, can be described b trigonometric functions of a real variable.

2 0606_CH0_-78.QXP //0 :6 AM Page. The Circular Functions + = FIGURE.. Unit circle Arc length t t radians Introduction In Chapter we considered trigonometric functions of angles measured either in degrees or in radians. For calculus and the sciences it is necessar to consider trigonometric functions whose domains consist of real numbers rather than angles. The transition from angles to real numbers is made b recognizing that to each real number t, there corresponds an angle of measure t radians. As we see net, this correspondence can be visualized using a circle with radius centered at the origin in a rectangular coordinate sstem. This circle is called the unit circle. From Section. it follows that the equation of the unit circle is 5. In this section the focus will be on the sine and cosine functions. The other four trigonometric functions will be considered in detail in Section.. We now consider a central angle t in standard position; that is, an angle with its verte at the center of a circle and initial side coinciding with the positive -ais. From the definition of radian measure, ) of Section., the angle t is defined to be t 5 s/r, the ratio of the subtended arc of length s to the radius r of the circle. For the unit circle shown in FIGURE.., r 5, and so t 5 s/ or t 5 s. In other words: On a unit circle, the radian measure of an angle of t radians is equal to the measure t of the subtended arc. It follows that for each real number t, the terminal side of an angle of t radians in standard position has traversed a distance of 0 t 0 units along the circumference of the unit circle counterclockwise if t. 0, clockwise if t, 0. This association of each real number t with an angle of t radians is illustrated in FIGURE... t units t radians, 0) t radians, 0) t units a) t 0 FIGURE.. Angle of t radians subtends an arc of length 0 t 0 b) t < 0 units Trigonometric Functions of Real Numbers We are now in a position to define trigonometric functions of a real number. Before proceeding we need the following important definition. DEFINITION.. Values of the Trigonometric Functions The value of a trigonometric function at a real number t is defined to be its value at an angle of t radians, provided that value eists. CHAPTER UNIT CIRCLE TRIGONOMETRY

3 0606_CH0_-78.QXP //0 :57 PM Page 5 For eample, the sine of the real number p/ is simpl the sine of the angle p/6 radian that, as we know, is. Thus there is reall nothing new in evaluating the trigonometric functions of a real number. The unit circle is ver helpful in describing the trigonometric functions of real numbers. If Pt) denotes the point of intersection of the terminal side of the angle t with the unit circle 5 and P, ) are the rectangular coordinates of this point, then from ) of Section. we have sin t 5 r 5 5 and cos t 5 r 5 5. These definitions, along with the definitions of the remaining four trigonometric functions, are summarized net. DEFINITION.. Trigonometric Functions Let t be an real number and Pt) 5 P, ) be the point of intersection on the unit circle with the terminal side of the angle of t radians in standard position. Then the si trigonometric functions of the real number t are sin t 5 cos t 5 tan t 5 sec t 5 cot t 5 csc t 5. ) From the first line in ) of Definition.. we see immediatel that For an real number t, the cosine and sine of t are the - and -coordinates, respectivel, of the point P of intersection of the terminal side of the angle of t radians in standard position) with the unit circle. See FIGURE... As we will soon see, a number of important properties of the sine and cosine functions can be obtained from this result. Because of the role plaed b the unit circle in this discussion, the trigonometric functions ) are often referred to as the circular functions. A number of properties of the sine and cosine functions follow from the fact that Pt) 5 cos t, sin t) lies on the unit circle. For instance, the coordinates of Pt) must satisf the equation of the circle: 5. Substituting 5 cos t and 5 sin t into the foregoing equation gives the familiar result cos t sin t 5. This relationship between the sine and cosine functions is the most fundamental of trigonometric identities, the Pthagorean identit. Bear in mind this identit is not just valid for angles as discussed in Sections. and.; we see now that it is valid for all real numbers t. + = P, ) = Pt) = cos t, sin t) t sin t cos t, 0) FIGURE.. Coordinates of Pt) are cos t, sin t) THEOREM.. For all real numbers t, Pthagorean Identit sin t cos t 5. ). The Circular Functions 5

4 0606_CH0_-78.QXP //0 :58 PM Page 6 Bounds on the Values of Sine and Cosine A number of properties of the sine and cosine functions follow from the fact that Pt) 5 P, ) lies on the unit circle. For instance, it follows that # # and # #. Since 5 cos t and 5 sin t, the foregoing inequalities are equivalent to # cos t # and # sin t #. ) The inequalities in ) can also be epressed using absolute values as 0 cos t 0 # and 0 sin t 0 #. Thus, for eample, there is no real number t for which sin t 5. II III sin t > 0 cos t < 0 0, ) sin t > 0 cos t > 0, 0), 0) sin t < 0 sin t < 0 cos t < 0 cos t > 0 0, ) IV FIGURE.. Algebraic signs of sint and cost in the four quadrants I Domain and Range The observations in ) indicate that both cos t and sin t can be an number in the interval [, ]. Thus we have the sine and cosine functions, f t) 5 sin t and gt) 5 cos t, respectivel, each with domain the set R of all real numbers and range the interval [, ]. The domains and ranges of the other four trigonometric functions will be discussed in Section.. Signs of the Circular Functions The signs of the function values sin t and cos t are determined b the quadrant in which the point Pt) lies, and conversel. For eample, if sin t and cos t are both negative, then the point Pt) and terminal side of the corresponding angle of t radians must lie in quadrant III. FIGURE.. displas the signs of the cosine and sine functions in each of the four quadrants. EXAMPLE Sine and Cosine of a Real Number Use a calculator to approimate sin and cos and give a geometric interpretation of these values. P) = cos, sin ) radians, 0) Solution From a calculator set in radian mode, we obtain cos < and sin < These values represent the and coordinates, respectivel, of the point of intersection P) of the terminal side of the angle of radians in standard position, with the unit circle. As shown in FIGURE..5, this point lies in the second quadrant because p/,,p. This would also be epected in view of Figure.. since cos, the -coordinate, is negative and sin, the -coordinate, is positive. Periodicit In Section. we saw that the angles of t radians and t 6 p radians are coterminal. Thus the determine the same point P, ) on the unit circle. Therefore sin t 5 sint 6 p) and cos t 5 cost 6 p). ) FIGURE..5 The point P) in Eample In other words, the sine and cosine functions repeat their values ever p units. It also follows that for an integer n: sint np) 5 sin t cost np) 5 cos t. 5) DEFINITION.. Periodic Functions A nonconstant function f is said to be periodic if there is a positive number p such that f t) 5 f t p) 6) for ever t in the domain of f. If p is the smallest positive number for which 6) is true, then p is called the period of the function f. 6 CHAPTER UNIT CIRCLE TRIGONOMETRY

5 0606_CH0_-78.QXP //0 :6 AM Page 7 The equations in ) impl that the sine and cosine functions are periodic with period p # p. To see that the period of sin t is actuall p, we observe that there is onl one point on the unit circle with -coordinate, namel, Pp/) 5 cosp/), sinp/)) 5 0, ). Therefore, sin t 5 onl for t 5 p, p 6 p, p 6 p, and so on. Thus the smallest possible positive value of p is p. In summar, the sine function f t) 5 sin t and cosine function gt) 5 cos t are periodic with period p; that is, f t) 5 f t p) and gt) 5 gt p), respectivel. For future reference, we have sint p) 5 sin t and cost p) 5 cos t 7) for ever real number t. EXAMPLE Using Periodicit Evaluate a) sin7p/) b) cosp/). Solution a) Because 7p/ is greater than p and can be written 7p, 5 p p it follows from sint p) 5 sin t with t 5p/, that b) Because 7p sin 5 sinap pb 5 sin p 5!. 9p 5 6p p, it follows from cost np) 5 cos t with n 5 and t 5p/, that 9p cos 5 cos ap 6pb 5 cos p 5. d See Table.. Odd Even Properties The smmetr of the unit circle endows the circular functions with several additional properties. For an real number t, the points Pt) and Pt) on the unit circle are located on the terminal side of an angle of t and t radians, respectivel. These two points will alwas be smmetric with respect to the -ais. FIGURE..6 illustrates the situation for a point Pt) ling in the first quadrant: the -coordinates of the two points are identical, but the -coordinates have equal magnitudes but opposite signs. The same smmetries will hold regardless of which quadrant contains Pt). Thus, for f t) 5 sin t and gt) 5 cos t and an real number t, f t) 5f t) and gt) 5 gt), respectivel. Appling the definitions of odd and even functions from Section.6 we have the following result. THEOREM.. Odd and Even Functions The sine function f t) 5 sin t is odd and the cosine function gt) 5 cos t is even; that is, for ever real number t, sint) 5 sin t and cost) 5 cos t. 8) See the first equation in 7). See the second equation in 7). t t Pt) = cos t, sin t) P t) = cos t), sin t)) FIGURE..6 Coordinates of Pt) and P t). The Circular Functions 7

6 0606_CH0_-78.QXP //0 :6 AM Page 8 EXAMPLE Using the Odd Even Properties Find eact values of sin t and cos t for the real number t 5p/6. Solution From 8) we have sine is an odd function sin a p 6 b 5sin p 6 5, cosine is an even function d See Table.. and cos a p 6 b 5 cos p 6 5!. Note that the signs of the answers are consistent with the fact that the terminal side of the angle p/6 radian lies in quadrant IV. The following additional properties of the sine and cosine functions can be verified b considering the smmetries of appropriatel chosen points on the unit circle. We first saw the results in i) and ii) in the net theorem stated for acute angles in 5) of Section.. THEOREM.. For all real numbers t, Additional Properties i) cos ap tb 5 sin t ii) sina p tb 5 cos t P t = cos t, sin t iii) v) cost p) 5cos t cosp t) 5cos t iv) sint p) 5sin t vi) sinpt) 5 sin t = Pt) = cos t, sin t), 0) For eample, to justif properties i) and ii) of Theorem.. for 0, t,p/, consider FIGURE..7. Since the points Pt) and Pp/ t) are smmetric with respect to the line 5, we can obtain the coordinates of Pp/ t) b interchanging the coordinates of Pt). Thus, cos t 5 5 sin a p tb and sin t 5 5 cos ap tb. FIGURE..7 Geometric justification of i) and ii) of Theorem.. In Section. we will use properties i) and ii) to justif two important formulas for the sine function. EXAMPLE Using Theorem.. In Table.. in Section. we saw that cosp/) 5 sinp/6). This result is a special case of propert i) of Theorem..; with t 5p/ we see that using propert i) of Theorem.. p sin 6 5 sin ap p b 5 cos p. 8 CHAPTER UNIT CIRCLE TRIGONOMETRY Reference Angle Revisited As we noted at the beginning of this section, for each real number t there is a unique angle of t radians in standard position that determines the point Pt), with coordinates cos t, sin t), on the unit circle. As shown in FIGURE..8,

7 0606_CH0_-78.QXP //0 :6 AM Page 9 Pt) Pt ) Pt ) Pt ) t t t t t t Pt) FIGURE..8 Reference angle tr is an acute angle Pt) the terminal side of an angle of t radians with Pt) not on an ais) will form an acute angle with the -ais. We can then locate an angle of tr radians in the first quadrant that is congruent to this acute angle. The angle of tr radians is called the reference angle for the real number t. Because of the smmetr of the unit circle, the coordinates of Ptr) will be equal in absolute value to the respective coordinates of Pt). Hence As the following eamples will show, reference angles can be used to find the trigonometric function values of an integer multiples of p/6, p/, and p/. EXAMPLE 5 Using a Reference Angle Find the eact values of sin t and cos t for the given real number: a) t 5 5p/ b) t 5p/. Solution In each part we begin b finding the reference angle corresponding to the given real number t. a) From FIGURE..9 we find that an angle of t 5 5p/ radians determines a point P5p/) in the fourth quadrant and has the reference angle tr5 p/ radians. After adjusting the signs of the coordinates of Pp/) 5 /,!/) to obtain the fourthquadrant point P5p/) 5 /,!/), we find that 5p sin 5sin p 5! reference angle b) The point Pp/) lies in the third quadrant and has reference angle p/ as shown in FIGURE..0. Therefore sin a p b 5sin p 5! sin t 56sin tr and cos t 56cos tr. T and 5p cos 5 cos p 5. and cos a p b 5cos p 5!. Sometimes, in order to find the trigonometric values of multiples of our basic fractions of p we must use periodicit or the even odd function properties in addition to reference angles. T 5 P =, FIGURE..9 Reference angle in part a) of Eample 5 P =, 5 P =, FIGURE..0 Reference angle in part b) of Eample 5 P =, EXAMPLE 6 Using Periodicit and a Reference Angle Find the eact values of the coordinates of P9p/6) on the unit circle. Solution The point P9p/6) has coordinates cos9p/6), sin9p/6)). We begin b observing that 9p/6 is greater than p, and so we must rewrite 9p/6 as an integer multiple of p plus a number less than p. B division we have 9p 6 5 p 5p 6 5 p) 5p 6.. The Circular Functions 9

8 0606_CH0_-78.QXP //0 :58 PM Page 0 5 P =, P 6 FIGURE.. Reference angle in Eample 6 6 =, Net, from the periodicit equations in 5) with n 5 we know that sina 9p 6 b 5 sina5p 6 b and cosa9p 6 b 5 cosa5p 6 b. Net we see from FIGURE.. that the reference angle for 5p/6 is p/6. Since P5p/6) is a second-quadrant point its -coordinate cos 5p/6) is negative and its -coordinate sin 5p/6) is positive. Finall, using the reference angle as shown in Figure.. we simpl adjust the algebraic signs of the coordinates of the Pp/6) 5 cosp/6), sinp/6)): 9p cos 6 5 cos 5p 6 5cos p 6 5! 9p and sin 6 5 sin 5p 6 5 sin p 6 5. Thus, P9 p/6) 5!/, /).. Eercises Answers to selected odd-numbered problems begin on page ANS-8. In Problems 8, for the given real number t, a) locate the point Pt) 5 cos t, sin t) on the unit circle and b) find the eact values of the coordinates of Pt). Do not use a calculator. 7p.. p. p. p 6 5p 5p p 7. p 8. 6 In Problems 9 6, for the given real number t, a) locate the point Pt) 5 cos t, sin t) on the unit circle and b) use a calculator to approimate the coordinates of Pt) In Problems 7, use periodicit of sin t and cos t to find the eact value of the given trigonometric function. Do not use a calculator. p 6p cos 9p sin cos 0. sin a 5p 6 b 7p. cos 9p. sin 0p. sin. 7p cos In Problems 5 0, justif the given statement b one of the properties of sin t and cos t given in this section. 5. sin p5sin p 6. cosp/) 5 sinp/) 7. sin p) 5sin p) 8. cos 6.8p 5cos.8p 9. cos 0. 5 cos0.) 0. cos.5 p) 5cos.5 0 CHAPTER UNIT CIRCLE TRIGONOMETRY

9 0606_CH0_-78.QXP //0 :6 AM Page. Given that cos t 5 5 and that Pt) is a point on the unit circle in the second quadrant, find sin t.. Given that sin t 5 and that Pt) is a point on the unit circle in the second quadrant, find cos t.. Given that sin t 5 and that Pt) is a point on the unit circle in the third quadrant, find cos t.. Given that cos t 5 and that Pt) is a point on the unit circle in the fourth quadrant, find sin t. In Problems 5 8, the -coordinate of the point P5p/8) on the unit circle is "!. Find the eact value of the given trigonometric function. Do not use a calculator. 5. cos 5p 6. sin a 5p 8 8 pb 7. sin a5p 8. 8 b cos a 5p 8 b In Problems 9, use the unit circle to determine all real numbers t for which the given equalit is true. 9. sin t 5!/ 0. cos t 5. cos t 5. sin t 5 For Discussion. Suppose f is a periodic function with period p. Show that F ) 5 f a), a. 0, is periodic with period p/a.. Graphs of Sine and Cosine Functions Introduction One wa to further our understanding of the trigonometric functions is to eamine their graphs. In this section we consider the graphs of the sine and cosine functions. Graphs of Sine and Cosine In Section. we saw that the domain of the sine function f t) 5 sin t is the set of real numbers `, `) and the interval [, ] is its range. Since the sine function has period p, we begin b sketching its graph on the interval [0, p]. We obtain a rough sketch of the graph given in FIGURE..b) b considering various positions of the point Pt) on the unit circle, as shown in Figure..a). As t varies from 0 to p/, the value sin t increases from 0 to its maimum value. But as t varies from p/ to p/, the value sin t decreases from to its minimum value. We note that sin t changes from positive to negative at t 5p. For t between p/ and p, we see that the corresponding values of sin t increase from to 0. The graph of an periodic function over an interval of length equal to its period is said to be one ccle of its graph. In the case of the sine function, the graph over the interval [0, p] in Figure..b) is one ccle of the graph of f t) 5 sin t.. Graphs of Sine and Cosine Functions

10 0606_CH0_-78.QXP //0 :6 AM Page P P ft) = sin t, 0 t 5 7 t 5 P P a) Unit circle b) One ccle of sine graph FIGURE.. Points Pt) on the unit circle corresponding to points on the graph Note: Change of smbols From this point on we will switch to the traditional smbols and when graphing trigonometric functions. Thus, f t) 5 sin t will either be written f ) 5 sin or simpl 5 sin. The graph of a periodic function is easil obtained b repeatedl drawing one ccle of its graph. In other words, the graph of 5 sin on, sa, the intervals [p, 0] and [p, p] is the same as that given in Figure..b). Recall from Section. that the sine function is an odd function since f ) 5 sin) 5sin 5f ). Thus, as can be seen in FIGURE.., the graph of 5 sin is smmetric with respect to the origin. = sin 5 7 FIGURE.. Graph of 5 sin One ccle B working again with the unit circle we can obtain one ccle of the graph of the cosine function g) 5 cos on the interval [0, p]. In contrast to the graph of f ) 5 sin where f 0) 5 f p) 5 0, for the cosine function we have g0) 5 gp) 5. FIGURE.. shows one ccle in red) of 5 cos on [0, p] along with the etension of that ccle in blue) to the adjacent intervals [p, 0] and [p, p]. We see from this figure that the graph of the cosine function is smmetric with respect to the -ais. This is a consequence of the fact that g is an even function: g) 5 cos) 5 cos 5 g). = cos 5 7 FIGURE.. Graph of 5 cos One ccle CHAPTER UNIT CIRCLE TRIGONOMETRY

11 0606_CH0_-78.QXP //0 :6 AM Page Properties of the Sine and Cosine Functions In this and subsequent courses in mathematics it is important that ou know the -coordinates of the -intercepts of the sine and cosine graphs, in other words, the zeros of f ) 5 sin and g) 5 cos. From the sine graph in Figure.. we see that the zeros of the sine function, or the numbers for which sin 5 0, are 5 0, 6p, 6p, 6p,.... These numbers are integer multiples of p. From the cosine graph in Figure.. we see that cos 5 0 when 56p/, 6p/, 65p/,.... These numbers are odd-integer multiples of p/, that is, 5 n )p/, where n is an integer. Using the distributive law, the zeros of g) 5 cos are often written as 5p/ np. The following list summarizes some of the important properties of the sine and cosine functions that are apparent from their graphs. An odd integer can be written as n, where n is an integer. PROPERTIES OF THE SINE AND COSINE FUNCTIONS The domain of f ) 5 sin and the domain of g) 5 cos is the set of real numbers, that is, `, `). The range of f ) 5 sin and the range of g) 5 cos is the interval [, ] on the -ais. The zeros of f ) 5 sin are 5 np, n an integer. The zeros of g) 5 cos are 5 n )p/, n an integer. The graph of f ) 5 sin is smmetric with respect to the origin. The graph of g) 5 cos is smmetric with respect to the -ais. The functions f ) 5 sin and g) 5 cos are continuous on the interval `, `). As we did in Chapter we can obtain variations of the basic sine and cosine graphs through rigid and nonrigid transformations. For the remainder of the discussion we will consider graphs of functions of the form 5 A sinb C) D or 5 A cosb C) D, where A, B, C, and D are real constants. ) Graphs of 5 A sin D and 5 A cos D We begin b considering the special cases of ): 5 A sin and 5 A cos. For A. 0 graphs of these functions are either a vertical stretch or a vertical compression of the graphs of 5 sin or 5 cos. For A, 0 the graphs are also reflected in the -ais. For eample, as FIGURE.. shows we obtain the graph of 5 sin b stretching the graph of 5 sin verticall b a factor of. Note that the maimum and minimum values of 5 sin occur at the same -values as the maimum and minimum values of 5 sin. In general, the maimum distance from an point on the graph of 5 A sin or 5 A cos to the -ais is 0 A 0. The number 0 A 0 is called the amplitude of the functions or of their graphs. The amplitude of the basic functions 5 sin and 5 cos is 0 A 0 5. In general, if a periodic function f is continuous, then over a closed interval of length equal to its period, f has both a maimum value M and a minimum value m. The amplitude is defined b = sin = sin FIGURE.. Vertical stretch of 5 sin amplitude 5 [M m]. ). Graphs of Sine and Cosine Functions

12 0606_CH0_-78.QXP //0 :7 AM Page = cos = cos FIGURE..5 Graph of function in Eample FIGURE..6 Graph of shifted up unit = + sin 5 sin EXAMPLE Verticall Compressed Cosine Graph Graph 5 cos. Solution The graph of 5 cos is the graph of 5 cos compressed verticall b a factor of and then reflected in the -ais. With the identification A 5 we see that the amplitude of the function is 0 A The graph of cos on the interval [0, p] is shown in red in FIGURE..5. The graphs of 5 A sin D and 5 A cos D are the respective graphs of 5 A sin and 5 A cos shifted verticall, up for D. 0 and down for D, 0. For eample, the graph of 5 sin is the graph of 5 sin Figure..) shifted up unit. The amplitude of the graph of either 5 A sin D or 5 A cos D is still 0 A 0. Observe in FIGURE..6, the maimum of 5 sin is 5 at 5p/ and the minimum is 5 at 5 p/. From ), the amplitude of 5 sin is then [ )] 5. B interpreting as a placeholder we can find the -coordinates of the -intercepts of the graphs of sine and cosine functions of the form 5 A sin B and 5 A cos B considered net). For eample, to solve sin 5 0, we use the fact that the zeros of f ) 5 sin are 5 np, where n is an integer. We simpl replace b to obtain 5 np so that 5 np, n 5 0, 6, 6,... ; that is, sin 5 0 for 5 0, 6 p, 6 p5p, 6 p, 6 p5p, and so on. See FIGURE..7. = sin = sin 5 7 One ccle of = sin One ccle of = sin FIGURE..7 Comparison of the graphs of and 5 sin 5 sin Careful here; sin sin Graphs of 5 A sin B and 5 A cos B We now consider the graph of 5 sin B, for B. 0. The function has amplitude since A 5. Since the period of 5 sin is p, a ccle of the graph of 5 sin B begins at 5 0 and will start to repeat its values when B 5 p. In other words, a ccle of the function 5 sin B is completed on the interval defined b 0 # B # p. Dividing the last inequalit b B shows that the period of the function 5 sin B is p/b and that the graph over the interval [0, p/b] is one ccle of its graph. For eample, the period of 5 sin is p/ 5p, and therefore one ccle of the graph is completed on the interval [0, p]. Figure..7 shows that two ccles of the graph of 5 sin in red and blue) are completed on the interval [0, p], whereas the graph of 5 sin in green) has completed onl one ccle. In terms of transformations, we can characterize the ccle of 5 sin on [0, p] as a horizontal compression of the ccle of 5 sin on [0, p]. CHAPTER UNIT CIRCLE TRIGONOMETRY

13 0606_CH0_-78.QXP //0 :7 AM Page 5 In summar, the graphs of for B. 0, each have amplitude 0 A 0 and period p/b. EXAMPLE Horizontall Compressed Cosine Graph Find the period of 5 cos and graph the function. Solution Since B 5, we see that the period of 5 cos is p/ 5p/. We conclude that the graph of 5 cos is the graph of 5 cos compressed horizontall. To graph the function, we draw one ccle of the cosine graph with amplitude on the interval [0, p/] and then use periodicit to etend the graph. FIGURE..8 shows four complete ccles of 5 cos the basic ccle in red and the etended graph in blue) and one ccle of 5 cos in green) on [0, p]. Notice that 5 cos attains its minimum at 5p/ since cos p/) 5 cos p5 and its maimum at 5p/ since cos p/) 5 cos p 5. If B, 0 in either 5 A sin B or 5 A cos B, we can use the odd/even properties, 8) of Section., to rewrite the function with positive B. This is illustrated in the net eample. EXAMPLE 5 A sin B and 5 A cos B Horizontall Stretched Sine Graph Find the amplitude and period of 5 sin ). Graph the function. Solution Since we require B. 0, we use sin) 5sin to rewrite the function as 5 sin ) 5sin. With the identification A 5, the amplitude is seen to be 0 A Now with B 5 we find that the period is p/ Hence we can interpret the ccle of 5sin 5 p. on [0, p] as a horizontal stretch and a reflection in the -ais because A, 0) of the ccle of 5 sin on [0, p]. FIGURE..9 shows that on the interval [0, p] the graph of 5sin in blue) completes one ccle, whereas the graph of 5 sin in green) completes two ccles. = cos = cos FIGURE..8 Graph of function in Eample = sin = sin FIGURE..9 Graph of function in Eample Graphs of 5 A sinb C) and 5 A cosb C) We have seen that the basic graphs of 5 sin and 5 cos can, in turn, be stretched or compressed verticall, 5 A sin and 5 A cos, shifted verticall, 5 A sin D and 5 A cos D, and stretched or compressed horizontall, 5 A sin B D and 5 A cos B D. The graphs of 5 A sinb C) D and 5 A cosb C) D, are the graphs of 5 A sin B D and 5 A cos B D shifted horizontall. In the remaining discussion we are going to focus on the graphs of 5 A sinb C) and 5 A cosb C). For eample, we know from Section. that the graph of 5 cos p/) is the basic cosine graph shifted to the right. In. Graphs of Sine and Cosine Functions 5

14 0606_CH0_-78.QXP //0 :7 AM Page 6 FIGURE..0 the graph of 5 cos p/) in red) on the interval [0, p] is one ccle of 5 cos on the interval [p/, p/] in blue) shifted horizontall p/ units to the right. Similarl, the graphs of 5 sin p/) and 5 sin p/) are the basic sine graph shifted p/ units to the left and to the right, respectivel. See FIGURES.. and... = cos = sin = sin = cos FIGURE..0 Horizontall shifted cosine graph = sin + FIGURE.. Horizontall shifted sine graph = sin FIGURE.. Horizontall shifted sine graph B comparing the red graphs in Figures..0.. with the graphs in Figures.. and.. we see that the cosine graph shifted p/ units to the right is the sine graph, the sine graph shifted p/ units to the left is the cosine graph, and the sine graph shifted p/ units to the right is the cosine graph reflected in the -ais. In other words, we have graphicall verified the identities cos a p b 5 sin, sin a p b 5 cos, and sin a p b 5cos. ) We now consider the graph of 5 A sinb C), for B. 0. Since the values of sin B C) range from to, it follows that A sinb C) varies between A and A. That is, the amplitude of 5 A sinb C) is 0 A 0. Also, as B C varies from 0 to p, the graph will complete one ccle. B solving B C 5 0 and B C 5 p, we find that one ccle is completed as varies from C/B to p C)/B. Therefore, the function 5 A sinb C) has the period p C B Moreover, if f ) 5 A sin B, then a C B b 5 p B. f a C B b 5 A sin B a C b 5 A sinb C). B ) The result in ) shows that the graph of 5 A sinb C) can be obtained b shifting the graph of f ) 5 A sin B horizontall a distance 0 C 0 /B. If C, 0 the shift is to the right, whereas if C. 0 the shift is to the left. The number 0 C 0 /B is called the phase shift of the graph of 5 A sinb C). EXAMPLE Equation of a Shifted Cosine Graph The graph of 5 0 cos is shifted p/ units to the right. Find its equation. Solution B writing f ) 5 0 cos and using ), we find f a p. b 5 0 cos a p b or 5 0 cos a p b In the last equation we would identif C 5p/. The phase shift is p/. 6 CHAPTER UNIT CIRCLE TRIGONOMETRY

15 0606_CH0_-78.QXP //0 :7 AM Page 7 As a practical matter the phase shift of 5 A sin B C) can be obtained b factoring the number B from B C: Note 5 A sinb C) 5 A sin B a C B b. For convenience the foregoing information is summarized net. SHIFTED SINE AND COSINE GRAPHS The graphs of 5 A sinb C) and 5 A cosb C), B. 0, are, respectivel, the graphs of 5 A sin B and 5 A cos B shifted horizontall b 0 C 0 /B. The shift is to the right if C, 0 and to the left if C. 0. The number 0 C 0 /B is called the phase shift. The amplitude of each graph is 0 A 0 and the period of each graph is p/b. EXAMPLE 5 Graph 5 sin p/). Horizontall Shifted Sine Graph Solution For purposes of comparison we will first graph 5 sin. The amplitude of 5 sin is 0 A 0 5 and its period is p/ 5p. Thus one ccle of 5 sin is completed on the interval [0, p]. Then we etend this graph to the adjacent interval [p, p] as shown in blue in FIGURE... Net, we rewrite 5 sin p/) b factoring from p/: From the last form we see that the phase shift is p/6. The graph of the given function, shown in red in Figure.., is obtained b shifting the graph of 5 sin to the right p/6 units. Remember, this means that if, ) is a point on the blue graph, then p/6, ) is the corresponding point on the red graph. For eample, 5 0 and 5p are the -coordinates of two -intercepts of the blue graph. Thus 5 0 p/6 5p/6 and 5pp/6 5 7p/6 are -coordinates of the -intercepts of the red or shifted graph. These numbers are indicated b the black arrows in Figure... EXAMPLE 6 5 sin a p b 5 sin a p 6 b. Horizontall Shifted Graphs Determine the amplitude, the period, the phase shift, and the direction of horizontal shift for each of the following functions. 6 = sin = sin 7 6 FIGURE.. Graph of function in Eample 5 a) 5 5 cos a5 p b b) 58 sin a p b Solution a) We first make the identifications A 5 5, B 5 5, and C 5p/. Thus the amplitude is 0 A and the period is p/b 5 p/5. The phase shift can be computed either b 0 p 0 /)/5 5 p/0 or b rewriting the function as 5 5 cos 5 a p 0 b. The last form indicates that the graph of 5 5 cos5 p/) is the graph of 5 5 cos 5 shifted p/0 units to the right.. Graphs of Sine and Cosine Functions 7

16 0606_CH0_-78.QXP //0 :59 PM Page 8 b) Since A 58 the amplitude is 0 A With B 5 the period is p/ 5p. B factoring from p/, we see from 58 sin a p b 58 sin a p 8 b that the phase shift is p/8. The graph of 58 sin p/) is the graph of 58 sin shifted p/8 units to the left. = cos + ) = cos FIGURE.. Graph of function in Eample 7 EXAMPLE 7 Horizontall Shifted Cosine Graph Graph 5 cosp p). Solution The amplitude of 5 cos p is 0 A 0 5 and the period is p/p 5. Thus one ccle of 5 cos p is completed on the interval [0, ]. In FIGURE.. two ccles of the graph of 5 cos p in blue) are shown. The -coordinates of the -intercepts of this graph are the values of for which cos p 5 0. The last equation implies p 5 n )p/ or 5 n )/, n an integer. In other words, for n 5 0,,,,,,... we get 56, 6, 6 5, and so on. Now b rewriting the given function 5 cosp p) as 5 cos p ) we see the phase shift is. The graph of 5 cosp p) in red) in Figure.., is obtained b shifting the graph of 5 cos p to the left unit. This means that the -intercepts are the same for both graphs. I 0 0 It) = 0 sin 0 t 0 0 FIGURE..5 Graph of current in Eample 8 60 t EXAMPLE 8 Alternating Current The current I in amperes) in a wire of an alternating-current circuit is given b It) 5 0 sin 0pt, where t is time measured in seconds. Sketch one ccle of the graph. What is the maimum value of the current? Solution The graph has amplitude 0 and period p/0p 560. Therefore, we sketch one ccle of the basic sine curve on the interval [0, 60], as shown in FIGURE..5. From the figure it is evident that the maimum value of the current is I 5 0 amperes and occurs at t 5 0 second since I 0) 5 0 sin 0p # p 0) 5 0 sin Eercises Answers to selected odd-numbered problems begin on page ANS-8. In Problems 6, use the techniques of shifting, stretching, compressing, and reflecting to sketch at least one ccle of the graph of the given function.. 5 cos. 5 cos. 5 sin. 5 sin 5. 5 cos 6. 5 sin 8 CHAPTER UNIT CIRCLE TRIGONOMETRY

17 0606_CH0_-78.QXP //0 :7 AM Page 9 In Problems 7 0, the given figure shows one ccle of a sine or cosine graph. From the figure determine A and D and write an equation of the form 5 A sin D or 5 A cos D for the graph FIGURE..7 Graph for Problem 8 FIGURE..6 Graph for Problem FIGURE..8 Graph for Problem 9 FIGURE..9 Graph for Problem 0 In Problems 6, find the -intercepts for the graph of the given function. Do not graph.. 5 sin p. 5cos. 5 0 cos. 5 sin5) 5. 5 sin a p b 6. 5 cos p) In Problems 7 and 8, find the -intercepts of the graph of the given function on the interval [0, p]. Then find all intercepts using periodicit sin 8. 5 cos In Problems 9, the given figure shows one ccle of a sine or cosine graph. From the figure determine A and B and write an equation of the form 5 A sin B or 5 A cos B for the graph FIGURE..0 Graph for Problem 9 FIGURE.. Graph for Problem 0. Graphs of Sine and Cosine Functions 9

18 0606_CH0_-78.QXP //0 :7 AM Page 0.. FIGURE.. Graph for Problem.. FIGURE.. Graph for Problem FIGURE.. Graph for Problem In Problems 5, find the amplitude and period of the given function. Sketch at least one ccle of the graph sin p sin 7. 5 cos p cos 9. 5 sin 0. 5 sin p p. 5 cos. 5 sin In Problems, find the amplitude, period, and phase shift of the given function. Sketch at least one ccle of the graph. FIGURE..5 Graph for Problem. 6 b. 5. b b p b sin ap. p b 5 sin a p b 5 cos a p 6 b 5 sin a p b 5cos a pb 5 cos ap p b 0 CHAPTER UNIT CIRCLE TRIGONOMETRY In Problems and, write an equation of the function whose graph is described in words.. The graph of 5 cos is verticall stretched up b a factor of and shifted down b 5 units. One ccle of 5 cos on [0, p] is compressed to [0, p/] and then the compressed ccle is shifted horizontall p/ units to the left.. One ccle of 5 sin on [0, p] is stretched to [0, 8p] and then the stretched ccle is shifted horizontall p/ units to the right. The graph is also compressed verticall b a factor of and then reflected in the -ais.

19 0606_CH0_-78.QXP //0 :7 AM Page In Problems 5 8, find horizontall shifted sine and cosine functions so that each function satisfies the given conditions. Graph the functions. 5. Amplitude, period p/, shifted b p/ units to the right 6. Amplitude, period p, shifted b p/ units to the left 7. Amplitude 0.7, period 0.5, shifted b units to the right 5 8. Amplitude, period, shifted b /p units to the left In Problems 9 and 50, graphicall verif the given identit. 9. cos p) 5cos 50. sin p) 5sin Miscellaneous Applications 5. Pendulum The angular displacement u of a pendulum from the vertical at time t seconds is given b ut) 5u 0 cos vt, where u 0 is the initial displacement at t 5 0 seconds. See FIGURE..6. For v5 rad/s and u 0 5p/0, sketch two ccles of the resulting function. 5. Current In a certain kind of electrical circuit, the current I measured in amperes at time t seconds is given b It) 5 0 cos a0pt p b. θ θ 0 FIGURE..6 Pendulum in Problem 5 Sketch two ccles of the graph of I as a function of time t. 5. Depth of Water The depth d of water at the entrance to a small harbor at time t is modeled b a function of the form where A is one-half the difference between the high- and low-tide depths, p/b, B. 0, is the tidal period, and C is the average depth. Assume that the tidal period is hours, the depth at high tide is 8 feet, and the depth at low tide is 6 feet. Sketch two ccles of the graph of d. 5. Fahrenheit Temperature Suppose that p Tt) sin t 8), 0 # t #, is a mathematical model of the Fahrenheit temperature at t hours after midnight on a certain da of the week. a) What is the temperature at 8 AM? b) At what times) does Tt) 5 60? c) Sketch the graph of T. d) Find the maimum and minimum temperatures and the times at which the occur. Calculator Problems In Problems 55 58, use a calculator to investigate whether the given function is periodic. 55. f ) 5 sin a 56. f ) 5 b sin 57. f ) 5 cos ) 58. f ) 5 sin For Discussion dt) 5 A sin B at p b C, 59. The function f ) 5 sin sin is periodic. What is the period of f? 60. Discuss and then sketch the graphs of 5 0 sin 0 and 5 0 cos 0.. Graphs of Sine and Cosine Functions

20 0606_CH0_-78.QXP //0 :8 AM Page. Graphs of Other Trigonometric Functions Introduction Four additional trigonometric functions are defined in terms of quotients and reciprocals of the sine and cosine functions. In this section we will consider the properties and graphs of these new functions. We begin with a definition that follows directl from ) of Section.. DEFINITION.. Four More Trigonometric Functions The tangent, cotangent, secant, and cosecant functions are denoted b tan, cot, sec, and csc, respectivel, and are defined as follows: tan 5 sin cos, cot 5 cos sin, sec 5 cos, csc 5 sin. Note that the tangent and cotangent functions are related b cot 5 cos sin 5 5 sin tan. cos In view of the definitions in ) and the foregoing result, cot, sec, and csc are referred to as the reciprocal functions. ) ) II III tan < 0 cot < 0 sec < 0 csc > 0 tan > 0 cot > 0 sec < 0 csc < 0 tan > 0 cot > 0 sec > 0 csc > 0 tan < 0 cot < 0 sec > 0 csc < 0 I IV FIGURE.. Signs of tan, cot, sec, and csc in the four quadrants Domain and Range Because the functions in ) and ) are quotients, the domain of each function consists of the set of real numbers ecept those numbers for which the denominator is zero. We have seen in Section. that cos 5 0 for 5 n )p/, n 5 0, 6, 6,..., and so the domain of tan and of sec is 5 0 n )p/, n 5 0, 6, 6, Similarl, since sin 5 0 for 5 np, n 5 0, 6, 6,..., it follows that the domain of cot and of csc is 5 0 np, n 5 0, 6, 6, We know that the values of the sine and cosine are bounded, that is, 0 sin 0 # and 0 cos 0 #. From these last inequalities we have 0 sec 0 5 ` cos ` 5 $ 0 cos 0 and 0 csc 0 5 ` ) sin ` 5 $. 0 sin 0 Recall, an inequalit such as ) means that sec $ or sec #. Hence the range of the secant function is `, ] [, `). The inequalit in ) implies that the cosecant function has the same range `, ] [, `). When we consider the graphs of the tangent and cotangent functions we will see that the have the same range: `, `). If we interpret as an angle, then FIGURE.. illustrates the algebraic signs of the tangent, cotangent, secant, and cosecant functions in each of the four quadrants. This is easil verified using the signs of the sine and cosine functions displaed in Figure... ) CHAPTER UNIT CIRCLE TRIGONOMETRY

21 0606_CH0_-78.QXP //0 :8 AM Page EXAMPLE Find tan, cot, sec, and csc for 5p/6. Solution In Eample of Section. we saw that sin a p 6 b 5sin p 6 5 Therefore, b the definitions in ) and ): Eample of Section. Revisited tan a p 6 b 5 /!/ 5!, cot ap 6 b 5! / / 5! sec a p 6 b 5!/ 5!, csc ap 6 b 5 / 5. Table.. summarizes some important values of the tangent, cotangent, secant, and cosecant and was constructed using values of the sine and cosine from Section.. A dash in the table indicates the trigonometric function is not defined at that particular value of. Periodicit Because the cosine and sine functions are p periodic necessaril the secant and cosecant function have the same period. But from Theorem.. of Section. we have tan p) 5 iv) of Theorem.. iii) of Theorem.. Thus 5) implies that tan and cot are periodic with a period p #p. In the case of the tangent function, tan 5 0 onl if sin 5 0, that is, onl if 5 0, 6p, 6p, and so on. Therefore, the smallest positive number p for which tan p) 5 tan is p 5p. The cotangent function, since it is the reciprocal of the tangent function, has the same period. In summar, the secant and cosecant functions are periodic with period p: sec p) 5 sec and csc p) 5 csc. 6) The tangent and cotangent function are periodic with period p: tan p) 5 tan and cot p) 5 cot. 7) Of course it is understood that 6) and 7) hold for ever real number for which the functions are defined. Graphs of 5 tan and 5 cot The numbers that make the denominators of tan, cot, sec, and csc equal to zero correspond to vertical asmptotes of their graphs. For eample, we encourage ou to verif using a calculator that tan S ` as S p and cos a p 6 b 5 cos p 6 5!. sin p) cos p) 5 sin cos 5 tan. and tan S ` as S p. d we could also use cot 5 /tan In other words, 5p/ and 5p/ are vertical asmptotes. The graph of 5 tan on the interval p/, p/) given in FIGURE.. is one ccle of the graph of 5 tan. Using periodicit we etend the ccle in Figure.. to adjacent intervals of length p as shown in FIGURE... The -intercepts of the graph of the tangent function are 0, 0), 6p, 0), 6p, 0),... and the vertical asmptotes of the graph are 56p/, 6p/, 65p/,.... 5) TABLE.. 0 tan cot sec csc p 6 p 0!!!! 0!!!! Also, see Problems 9 and 50 in Eercises.. FIGURE.. One ccle of the graph of 5 tan p p. Graphs of Other Trigonometric Functions

22 0606_CH0_-78.QXP //0 :8 AM Page = tan The graph of 5 cot is similar to the graph of the tangent function and is given in FIGURE... In this case, the graph of 5 cot on the interval 0, p) is one ccle of the graph of 5 cot. The -intercepts of the graph of the cotangent function are 6p/, 0), 6p/, 0), 65p/, 0),... and the vertical asmptotes of the graph are the vertical lines 5 0, 6p, 6p, 6p,.... Note that the graphs of 5 tan and 5 cot are smmetric with respect to the origin. FIGURE.. Graph of of 5 tan = cot THEOREM.. ODD FUNCTIONS The tangent function f ) 5 tan and the cotangent function g) 5 cot are odd functions, that tan ) 5tan and cot ) 5cot 8) for ever real number for which the functions are defined. FIGURE.. Graph of 5 cot Graphs of sec and csc For both 5 sec and 5 csc we know that 0 0 $ and so no portion of their graphs can appear in the horizontal strip,, of the Cartesian plane. Hence the graphs of 5 sec and 5 csc have no -intercepts. As we have alread seen, 5 sec and 5 csc have period p. The vertical asmptotes for the graph of 5 sec are the same as 5 tan, namel, 56p/, 6p/, 65p/,.... Because 5 cos is an even function so is 5 sec 5 /cos. The graph of 5 sec is smmetric with respect to the -ais. On the other hand, the vertical asmptotes for the graph of 5 csc are the same as 5 cot, namel, 5 0, 6p, 6p, 6p,.... Because 5 sin is an odd function so is 5 csc 5 /sin. The graph of 5 csc is smmetric with respect to the origin. One ccle of the graph of 5 sec on [0, p] is etended to the interval [p, 0] b periodicit or -ais smmetr) in FIGURE..5. Similarl, in FIGURE..6 we etend one ccle of 5 csc on 0, p) to the interval p, 0) b periodicit or origin smmetr). = sec = csc FIGURE..5 Graph of 5 sec FIGURE..6 Graph of 5 csc Transformations and Graphs Like the sine and cosine graphs, rigid and nonrigid transformations can be applied to the graphs of 5 tan, 5 cot, 5 sec, and 5 csc. For eample, a function such as 5 A tan B C) D can be analzed in the following manner: vertical stretch/compression/reflection T 5 A tanb C) D. horizontal stretch/compression c b changing period vertical shift T c horizontal shift 9) CHAPTER UNIT CIRCLE TRIGONOMETRY

23 0606_CH0_-78.QXP //0 :8 AM Page 5 If B. 0, then the period of 5 A tanb C) and 5 A cotb C) is p/b, 0) whereas the period of 5 A secb C) and 5 A cscb C) is p/b. ) As we see in 9) the number A in each case can be interpreted as either a vertical stretch or compression of a graph. However, ou should be aware of the fact that the functions in 0) and ) have no amplitude, because none of the functions have a maimum and a minimum value. Of the si trigonometric functions, onl the sine and cosine functions have an amplitude. EXAMPLE Comparison of Graphs Find the period, -intercepts, and vertical asmptotes for the graph of 5 tan. Graph the function on [0, p]. Solution With the identification B 5, we see from 0) that the period is p/. Since tan 5 sin /cos, the -intercepts of the graph occur at the zeros of sin. From the properties of the sine function given in Section., we know that sin 5 0 for 5 np so that 5 np, n 5 0, 6, 6,.... That is, 5 0, 6p/, 6p/ 5p, 6p/, 6p/ 5 p, and so on. The -intercepts are 0, 0), 6p/, 0), 6p, 0), 6p/, 0),.... The vertical asmptotes of the graph occur at zeros of cos. Moreover, the numbers for which cos 5 0 are found in the following manner: p 5 n ) so that 5 n ) p, n 5 0, 6, 6,.... That is, the vertical asmptotes are 56p/, 6p/, 65p/,.... On the interval [0, p] the graph of 5 tan has three intercepts, 0, 0), p/, 0), p, 0), and two vertical asmptotes, 5p/ and 5 p/. In FIGURE..7 we have compared the graphs of 5 tan and 5 tan on the interval. The graph of 5 tan is a horizontal compression of the graph of 5 tan. a) = tan on [0, ] b) = tan on [0, ] FIGURE..7 Graphs of functions in Eample. Graphs of Other Trigonometric Functions 5

24 0606_CH0_-78.QXP //0 :8 AM Page 6 EXAMPLE Comparisons of Graphs Compare one ccle of the graphs of 5 tan and 5 tan p/). Solution The graph of 5 tan p/) is the graph of 5 tan shifted horizontall p/ units to the right. The intercept 0, 0) for the graph of 5 tan is shifted to p/, 0) on the graph of 5 tan p/). The vertical asmptotes 5p/ and 5p/ for the graph of 5 tan are shifted to 5p/ and 5 p/ for the graph of 5 tan p/). In FIGURES..8a) and..8b) we see, respectivel, a ccle of the graph of 5 tan on the interval p/, p/) is shifted to the right to ield a ccle of the graph of 5 tan p/) on the interval p/, p/). a) Ccle of = tan b) Ccle of = tan /) on /, /) on /, /) FIGURE..8 Graph of functions in Eample As we did in the analsis of the graphs of 5 A sinb C) and 5 A cosb C) we can determine the amount of horizontal shift for graphs of functions such as 5 A tanb C) and 5 A secb C) b factoring the number B. 0 from B C. EXAMPLE Graph 5 sec p/). 6 CHAPTER UNIT CIRCLE TRIGONOMETRY Two Shifts and Two Compressions Solution Let s break down the analsis of the graph into four parts, namel, b transformations. i) One ccle of the graph of 5 sec occurs on [0, p]. Since the period of 5 sec is p/, one ccle of its graph occurs on the interval [0, p/]. In other words, the graph of 5 sec is a horizontal compression of the graph of 5 sec. Since sec 5 /cos the vertical asmptotes occur at the zeros of cos. Using the zeros of the cosine function given in Section. we find 5 n ) p or 5 n ) p, n 5 0, 6, 6, FIGURE..9a) shows two ccles of the graph 5 sec ; one ccle on [p/, 0] and another on [0, p/]. Within those intervals the vertical asmptotes are 5p/, 5p/6, 5p/6, and 5p/.

25 0606_CH0_-78.QXP //0 :9 AM Page a) Horizontal compression b) Vertical compression and reflection in -ais FIGURE..9 Graph of function in Eample c) Horizontal shift d) Vertical shift ii) The graph of 5 sec is the graph of 5 sec compressed verticall b a factor of and then reflected in the -ais. See Figure..9b). iii) B factoring from p/, we see from 5 sec a p b 5 sec a p 6 b that the graph of 5 is the graph of 5 sec p/) sec shifted p/6 units to the right. B shifting the two intervals [p/, 0] and [0, p/] in Figure..9b) to the right p/6 units, we see in Figure..9c) two ccles of 5 sec p/) on the intervals [ p/, p/6] and [p/6, 5p/6]. The vertical asmptotes 5p/, 5p/6, 5p/6, and 5p/ shown in Figure..9b) are shifted to 5p/, 5 0, 5p/, and 5 p/. Observe that the -intercept 0, ) in Figure..9b) is now moved to p/6, ) in Figure..9c). iv) Finall, we obtain the graph 5 in Figure..9d) b shifting the graph of 5 sec p/) sec p/) in Figure..9c) units upward.. Eercises Answers to selected odd-numbered problems begin on page ANS-9. In Problems and, complete the given table.. tan p p 5p 6 p 7p 6 5p p p 5p 7p p 6 p cot. sec p p 5p 6 p 7p 6 5p p p 5p 7p p 6 p csc. Graphs of Other Trigonometric Functions 7

26 0606_CH0_-78.QXP //0 :9 AM Page 8 In Problems 8, find the indicated value without the use of a calculator.. p 9p. csc a p 5. tan 6 b 6. sec 7p p cot a p b b 0. tan a 5p 6 b. 0p 7p 9p sec. cot. csc 5p. sec 6 5. sec 0 ) 6. tan csc cot 70 ) In Problems 9, use the given information to find the values of the remaining five trigonometric functions. 9. tan 5, p/,,p 0. cot 5, p,, p/. csc 5, 0,,p/. sec 55, p/,,p. If cos 5 sin, find all values of tan, cot, sec, and csc.. If csc 5 sec, find all values of tan, cot, sin, and cos. In Problems 5, find the period, -intercepts, and the vertical asmptotes of the given function. Sketch at least one ccle of the graph tan p 6. 5 tan p 7. 5 cot 8. 5cot 9. 5 tana 0. 5 p b cot a p b. 5 cot p. 5 tana 5p 6 b In Problems 0, find the period and the vertical asmptotes of the given function. Sketch at least one ccle of the graph. p. 5sec. 5 sec 5. 5 csc p 6. 5 csc 7. 5 seca p 8. 5 csc p) b 9. 5 csca p 0. b 5 sec p) 8 CHAPTER UNIT CIRCLE TRIGONOMETRY In Problems and, use the graphs of 5 tan and 5 sec to find numbers A and C for which the given equalit is true.. cot 5 A tan C). csc 5 A sec C) For Discussion. Using a calculator in radian mode, compare the values of tan.57 and tan.58. Eplain the difference in these values.. Using a calculator in radian mode, compare the values of cot. and cot Can 9 csc 5 for an real number? 6. Can 7 0 sec 5 0 for an real number? 7. For which real numbers is a) sin # csc? b) sin, csc?

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