In solving trigonometric equations (as in Section 3.4) we typically reduce

Transcription

1 CHAPTER 6 Inverse Trigonometric Functions In solving trigonometric equations as in Section.4 we tpicall reduce a comple trigonometric equation to a simple one, such as sin =, with the variable occurring inside the trig function. Then we have to reason backwards and ask ourselves What value of makes sin =? Reflecting on the unit circle, we might arrive at the solution =. This is eactl the kind of backwards thinking needed to mentall evaluate inverse functions. We are in essence treating sin as if it had an inverse and reasoning as sin = angle for which sin = =. But there is a slight problem with this. There are man values for which sin =, namel = +k for an integer k. This is clear from the unit circle. Start at = radians and take k laps of length around the circle to return to the same point, but at = + k radians. We still have sin + k =. + k This is also clear from the graph of sin, as sin = at each = + k. The graph reveals the cru of our problem: sin is not one-to-one because the horizontal line = meets it at ever = +k. So sin has no inverse. = sin + + We have a dilemma. An inverse sin would be useful, but it doesn t eist. We will overcome this b restricting the domain of sin.

2 The Function sin The Function sin The function sin is not one-to-one, so technicall it has no inverse. But in the diagram below, the part of its graph bold over the interval [, ] is a one-to-one function. This function has an inverse; we will call it sin. For input, sin accepts values of between and the outputs of sin. Given this input, the output of sin is the angle for which sin =, with the additional stipulation. See the diagram below. = sin sin To repeat, sin is the inverse of sin with its domain restricted to [, ]. Notice sin has domain [,], and its range is [, ]. There is a simple wa to visualize sin. The left side of the bo below reminds us that sin is the height of the point P at on the unit circle: For input we get an output sin, the -coordinate of P. The left side of the bo reverses this. The input is the height of a point on the unit circle; the output sin is the angle for which sin =. The functions sin and sin sin = -coordinate of point P sin = angle with and sin = P sin sin This is a picture of sin. It sas that a right triangle with hpotenuse of length and opposite leg of length has an angle measuring sin radians. From this we can mentall work out sin for man.

3 80 Inverse Trigonometric Functions Here are a few eamples. We will start out drawing a picture for each one, though ou will quickl reach the point of bpassing the picture in favor of visualizing the situation in the mind s ee.. Draw a half unit First let s find sin circle representing angles between and, and. This forms a fa- put a point on it at height miliar triangle with the 60 or angle at the origin. Thus angle for sin = which = and sin =. For another eample, let s find sin. First we draw the half-circle and locate the point at height, this time below the -ais. This forms an angle of radian measure 4. Thus sin sin angle for sin = which = and sin = 4. 4 Now let s do sin. The point on the halfcircle at height forms the angle, so angle for sin = which =. and sin = sin From the same picture we can see sin = and sin 0 = 0. Of course we can t do ever such problem mentall. Consider sin. This poses a problem because we can t think of a for which sin =. But using the sin button on our calculator gives sin Use our calculator sparingl. Do not allow it to obstruct the simple meaning of sin Your calculator should allow ou to toggle between radian and degree mode. Test this b working out sin. With degree mode ou should get sin = 90. Radian mode gives sin = which is.

4 The Function sin 8 Finall, let s investigate the graph of sin. Below is the graph of = sin with the part on domain [, ] drawn bold. Reflecting this across the line = gives the graph of = sin, drawn dashed. = sin = sin Erasing the clutter, the graph of sin is shown again as Figure 6.. Notice that the domain of sin is the interval [,]. The range is [, ]. = sin Figure 6.. The graph of sin. It has domain [,] and range [, ]. We remark that sin sin = for an real number in the domain of sin, in accordance with the propert f f =. But we caution that sin sin = is not alwas true. Indeed consider =. Then sin sin = sin 0 = 0, so sin sin in this particular case. The reason for this apparent contradiction is that sin does not reall have an inverse. The function sin is the inverse of sin with a restricted domain. The value = is not in the restricted domain [, ] of sin and thus lies outside the range of sin.

5 8 Inverse Trigonometric Functions Eample 6. Find all solutions of the equation 4sin + sin = that are in the interval [0,]. Because of the second power of sin, it seems reasonable to tr to solve b factoring. For this we need zero on one side, so we rewrite as 4sin + sin = 0. This factors as 4sin sin + = 0. So must be such that it makes one or the other factor zero. We consider these cases separatel. First, if sin+ = 0, then sin =. There is onl one in [0,] that makes this true, and that is = which is immediatel evident from the unit circle. Therefore one solution to the equation is =. Net suppose 4sin = 0, so sin = 4. We know no angle making sin = 4, so we cannot solve this from direct inspection of the unit circle. But we can take sin of both sides of sin = 4, and doing this ields a solution = sin 4, which is the indicated point on the circle at height 4, in the first quadrant. There is another point at height 4, in the second quadrant, and this is at sin 4. Taking sin of either of these angles results in 4 so the are the solutions of sin = 4. sin 4 sin 4 So the equation has three solutions, =, sin 4, and sin 4. Resorting to a calculator, , , Eercises for Section 6. Work these problems mentall, without a calculator.. sin. sin. sin 5. sin 6. sin 7. sin 4. sin 8. sin

6 The Function cos 8 6. The Function cos The cos function is certainl not one-to-one, so it doesn t have an inverse unless we restrict its domain. The diagram below reveals that cos is one-to-one on the domain [0,]. This part of the graph is drawn bold. = cos 0 cos We define cos to be the inverse of the function cos with domain [0,]. The boed diagrams below are visual descriptions of cos and cos. For an angle we know that cos is the -coordinate of the point P at on the unit circle. Reversing this, to find cos, envision as the -coordinate of a point on the upper half circle encompassing all angles 0. Then cos is the angle between 0 and for which cos =. The functions cos and cos cos = -coordinate of point P cos = angle for which 0 and cos = P cos cos 0 To illustrate, let s compute cos. We first draw a point on the upper circle with -coordinate. Our eperience with the unit circle tells us that this forms an angle of 6 radians. Thus cos = 6. cos 0 6

7 84 Inverse Trigonometric Functions Net consider the problem of cos. We first locate the point on the upper circle with -coordinate. Knowledge of the unit circle tells us that this forms an angle of radians, so cos =. cos 0 Now let s do cos 0. The point on the upper circle with -coordinate 0 is at radians. Therefore cos 0 =. You can also see from this drawing how cos = 0 and cos =. cos Now that we can compute cos let s think about its graph. Below we get the graph of = cos dashed b reflecting across the line = the relevant part of the graph of = cos bold. = cos = 0 = cos Erasing the preliminar steps, we get our final graph, in Figure 6.. = cos = cos 0 Figure 6.. The graph of cos. It has domain is [,] and range [0,].

8 The Functions tan and sec The Functions tan and sec Now we will develop the functions tan and sec, beginning with tan. The function tan is definitel not one-to-one, but its graph suggests that we can make it one-to-one b restricting its domain to the interval [, ]. = tan 5 Our visual descriptions of tan and sec will use a right triangle whose adjacent side has length, shown here. The opposite side is OPP = OPP = OPP ADJ = tan, while the hpotenuse is HYP = HYP = HYP ADJ = sec. This is a picture of tan and sec. The number tan is the length of the opposite side of the triangle; sec is its hpotenuse. sec tan Placing this on the unit circle gives a visual description of both tan and tan. The left side of the bo below states that tan is the opposite side of our triangle. The right side reverses this: If the opposite side is, then the angle is tan. The functions tan and tan angle with tan = opposite side of triangle tan = and tan = tan tan

9 86 Inverse Trigonometric Functions As an eample, consider tan. You can probabl alread do this mentall without resorting to a picture: tan is the angle in [, ] for which tan =. As tan 4 =, we get tan = 4. Drawing a below reinforces this. It shows a triangle with adjacent side and opposite side =. This is a triangle, so the angle tan must be 4. tan tan a b Now think about tan. Drawing b shows the relevant triangle. Because is negative, we orient the triangle so that the vertical side is below the -ais. This is a triangle with the angle tan at the 60 corner. Therefore tan =. Notice that as the opposite side of our triangle grows larger, the hpotenuse becomes more vertical, and the angle tan rotates closer to 90, or radians. Thus as increases to the value of tan increases, approaching. This is reflected in the graph of tan below the line = is a horizontal asmptote. This graph is the reflection across the line = of the graph of = tan with domain [, ]. The vertical asmptote = of tan reflects to the horizontal asmptote =. If approaches, the triangle is below the -ais, and the angle tan approaches. tan = tan = tan Figure 6.. The graph of tan. Lines = ± are horizontal asmptotes.

10 The Functions tan and sec 87 Now we eplore the function sec. As usual, because sec is not one-to-one we will have to restrict its domain in order to get an inverse. Its graph suggests that we should restrict it to the interval [0, ]. Actuall, to be absolutel precise, the point = is not in the domain, so we restrict sec to [0,,]. This is indicated b the bold part of the graph below. = sec 5 We therefore define sec to be the inverse of the function sec with restricted domain [0,,]. At the beginning of this section we remarked that sec is the length of the hpotenuse of a right triangle with angle and adjacent side having length. This gives the following interpretation, at least for values of with 0 <. The functions sec and sec sec = hpotenuse of triangle sec = angle for which 0 and sec = sec sec 0 In other words, if the above triangle has hpotenuse, then sec is the radian measure its angle. One slightl unsettling thing about this situation is that the picture changes when <, when is in the second quadrant. Then the triangle flips to the left of the -ais, as illustrated below. Note sec is negative for these, so we have to interpret the hpotenuse as a negative number. Thus the diagram on the right shows the angle sec for negative values of.

11 88 Inverse Trigonometric Functions sec = hpotenuse of triangle sec = angle for which 0 and sec = sec sec 0 Here is regarded as negative, so this is the picture of sec for negative. Regardless of the diagrams, sec is alwas the angle between 0 and with sec =. Thus sec 0, though its input can be negative. Eample 6. Find sec, sec and sec 4. Each of these can be solved without resorting to a diagram. Below we work them first without a diagram, but then add the diagram to highlight the geometric meaning of the inverse secant function. First we find sec. If = sec, then 0 and sec =, which means cos = =. From this we see = 4, that is, sec = 4. The diagram reinforces this. The triangle with hpotenuse is a triangle, with angle 4. Now for sec. If = sec, then 0 and sec =, or cos =. From this see =, so sec =. Again a diagram reinforces this. We interpret the input of as the hpotenuse of a triangle in the second quadrant. Here it happens that the triangle is half of an equiangular triangle with sides of length. We read off sec = 0, or radians. sec 0 sec 0 Finall, consider sec 4. If = sec 4, then sec = 4 and cos = 4. Here there is no familiar angle for which cos = 4. We have to resort to a calculator approimation sec This is in radians. In degree mode our calculator will give sec degrees.

12 The Functions tan and sec 89 Now we investigate the graph of sec. Our strateg is to use the fact that the graph of the inverse of a function is the function s graph reflected across the line =. In this case we start with the graph of = sec restricted to the domain [0, ], shown bold below. The other portions of = sec are drawn in light gra. This bold graph = sec reflects across the line = dotted to the dashed graph, which is the graph of = sec. Notice how the vertical asmptote = of sec reflects to a horizontal asmptote = of sec. = sec = = sec Erasing all the junk, we get the clean graph below. Notice that the domain of sec is, ] [,. The range is [ 0,,. = tan = sec Figure 6.4. The graph of sec. The line = is a horizontal asmptote. Take note that the interval, is not a part of the domain of sec. This is because all outputs of sec are either greater than or equal to, or less than or equal to. As the inputs of sec are the outputs of sec, no numbers in, can be plugged into sec. This follows also from our diagrams. The in sec is identified with the hpotenuse of a right triangle on the unit circle, which never has a length between and.

13 90 Inverse Trigonometric Functions Eercises for Section 6. These problems are for both Sections 6. and 6.. Evaluate the following inverse trig functions. It is important to do them without a calculator this will greatl sharpen our understanding of inverse trig functions.. cos. cos. cos 4. cos 5. tan 0 6. tan 7. tan 8. tan 9. tan 0. sec. sec. sec. sec 4. sec 5. sec 7. sin sin5 8. sin sin 9. tan tan 5 4. sec sec tan tan 4. sec sec 4. cos cos8 4. cos cos 6. sec 5. Solve the equation tan =. Use a calculator to approimate the final answer, if ou wish. 6. Find all solutions of tan + tan = 0 that are in the interval [,]. Use a calculator to approimate the final answer, if ou wish. 7. Find all solutions of 5sin = that are in the interval [,]. Use a calculator to approimate the final answer, if ou wish. 8. Find all solutions of the equation cossin + 6cos sin = 0 that are in the interval [0,]. Use a calculator to approimate the final answer, if ou wish. 9. Eplain wh cos = sin + for all in the domain of cos and sin. 0. Eplain wh cot = tan for all 0.. Eplain wh sec = cos.. Eplain wh cos + cos = for all in the domain of cos.. Eplain wh sin + sin = 0 for all in the domain of sin.

14 The Functions cot and csc The Functions cot and csc These functions are occasionall useful. But we will see that the are epressible in terms of tan and sec, so the need onl a brief mention. = cot 0 First, consider cot. The above graph of cot shows that it is one-to-one on the interval [0,]. We thus define cot as the inverse of cot with its domain restricted to [0,]. Below is the graph of = cot, which is the reflection of the bold part of = cot above across the line =. Notice how the vertical asmptote = of cot reflects across to a horizontal asmptote of cot. Likewise the -ais a vertical asmptote of cot reflects across to the -ais, a horizontal asmptote of cot. 0 = cot Figure 6.5. The function cot has domain R and the range [0,]. Given an input, the corresponding output cot is the angle for which 0 < < and cot =. Compare the graphs of cot in Figure 6.5 and tan in Figure 6. on page 86. It appears that the graph of cot is that of tan reflected across the -ais and then moved up units. That is, we might guess cot = tan. This is indeed the case, as ou are invited to verif. The fact that cot can be epressed in terms of tan is one reason that we do not need to invest much mental energ into the intricacies of cot, as least as long as we have a good grip on tan.

15 9 Inverse Trigonometric Functions Now we ll investigate the last remaining inverse trig function, csc. We start with the graph of csc, below. This function is one-to-one on the domain,0 0,, where the point = 0 is not included because it is not in the domain of csc. This part of the graph is shown bold. = csc 0 The function csc is defined to be the inverse of csc on this restricted domain. Reflecting the bold part of the graph of csc across the line = gives the graph of csc, below. 0 = csc Figure 6.6. The graph of csc. Its domain is, ] [, and its range is,0 0,. The -ais is a horizontal asmptote. Given an input, the corresponding output csc is the angle for which < < and csc =. Comparing the graphs of csc in Figure 6.6 and sec in Figure 6.4 on page 89, it looks as if the graph of csc is the graph of sec reflected across the -ais, then moved up units. That is, we might guess csc = sec. This is indeed the case, as ou can verif. Because csc can be epressed in terms of sec, we can downpla the importance of csc.

16 Simplifications Simplifications On page 8 we noted that sin sin =. Obviousl the right side of this equation is simpler than the left side. Taking sin of sin wipes out the trig functions. What if we did cos sin? This too simplifies dramaticall, but the answer is not. To understand how this epression simplifies, let s eamine the angle sin that we are taking cos of. We saw in Section 6. how to draw a picture of sin. It is the angle of the following triangle whose hpotenuse has length and whose opposite side has length. sin Taking cos of this angle, we get the length of the adjacent side, as follows. sin }{{} cos sin Now, we can solve for this adjacent side using the Pthagorean theorem: cos sin + = cos sin = cos sin =. Therefore we obtain the simplification cos sin =, which holds true for all in the domain [, ] of sin. In essence, taking cos of the angle sin wipes out the trig functions, leaving a simpler algebraic epression. There will be man instance where such simplifications are ver useful. This section eplains how to simplif epressions such as cos sin, or cos tan, or sec tan, etc., that involve the composition of a trig function with an inverse trig function. In ever case the answer can be found b appling the Pthagorean theorem to an appropriate triangle.

17 94 Inverse Trigonometric Functions Success at doing this revolves around understanding the geometric, triangle interpretation of the si trig functions. This is summarized in the si triangles below. In each case a trig function is interpreted as the length of a triangle edge, where one other edge has length. For eample, in the first triangle, OPP = OPP = OPP ADJ = sin, that is, the opposite side has length sin. You should check that the remaining triangles are labeled correctl. sin tan cot sec cos csc Reversing input and output as we have learned to do in this chapter ields corresponding triangles for each of the si inverse trig functions. sin tan cot cos sec csc Now we will use these to solve a few instances of simplifications. Eample 6. Simplif sec tan. Start with the triangle for tan, redrawn here. B the Pthagorean theorem, the hpotenuse is +, as labeled. Now, sec tan = HYP ADJ = HYP = + = +. Therefore we have obtained sec tan = +. + tan

18 Simplifications 95 Eample 6.4 Simplif sin tan. Again we start with the triangle for tan. As before, the Pthagorean theorem sas the hpotenuse has length +. Now, sin tan = OPP HYP = =. + + Therefore we have obtained our answer sin tan = +. + tan Eample 6.5 Simplif tan sec. We start b drawing the triangle for sec. This time things are slightl trickier. Recall from Section 6. that we interpret the hpotenuse length to be positive or negative depending on whether the triangle is in the first or second quadrant. positive negative sec sec Either wa, the Pthagorean theorem sas the opposite side has length, which is positive, regardless of the sign of. positive negative sec sec For the triangle in the first quadrant, tan sec = OPP ADJ = Whereas in the second quadrant, tan sec = OPP ADJ = Therefore our final answer is the piecewise function tan sec { if is positive = if is negative =. =. We will use this formula as well as other simplifications in this section later in the course.

19 96 Inverse Trigonometric Functions Eercises for Section 6.5 Simplif the given compositions.. tan sin =. tan cos =. tan tan = 4. sin cos = 5. sin sec = 6. sin sin = 7. cos sin = 8. cos tan = 9. cos sec = 0. sec sin =. sec cos =. sec tan =