1 Trigonometry -Ideas and Applications

Transcription

1 1 Trigonometry -Ideas and Applications 1.1 A second look at graphs The sine and cosine are basic entities of trigonometry, for the other four functions can be defined in terms of them. The graphs can be stretched and pulled in various ways and for many applications it is important to be able to express the corresponding functions analytically. As we have seen, the behavior of both sine and the cosine graphs on the interval [0, 2π] is repeated on all subsequent and previous intervals of length 2π. What I mean is that on each of the intervals [2π, 4π], [4π, 6π], [6π, 8π] and [ 2π, 0], [ 4π, 2π], [ 6π, 4π], the graphs look identical to the graph over the interval [0, 2π]. The restriction of the sine function to the interval [0, 2π] is defined to be the basic cycle of the sine function. Similarly the basic cycle of the cosine function is its restriction to the interval [0, 2π]. Earlier in studying general properties of functions given an arbitrary function whose graph had been altered in certain ways, we developed methods for analytically expressing a new function whose graph was the altered graph. We need to apply these ideas Stretching and shrinking vertically, reflecting about the x axis Figure 1: Graphs of y = 3 sin x in blue, y = sin x in green and y = 2 2 sin x in magenta Given either the sine function, y = sin x, or the cosine function, y = cos x, and a constant a,we know that : if a > 1, multiplication by a stretches the graph vertically if 0 < a < 1, multiplication by a shrinks the graph vertically 1

2 if a < 0, multiplication by a first reflects the graph about the x axis and then either stretches it if a < 1 or shrinks it if 1 < a < 0. Given an equation y = a sin x or y = a cos x, the number a is said to be the amplitude of the corresponding equation. Figure 2: Graph of y sin x in green and y = 2 sin x in blue Translating To translate the graph of a function vertically up, add a positive constant to the function, and to translate down, subtract a positive constant from the function definition. Thus for instance, the graph of y = sin x + 2, represents the graph of y = sin x translated two units up. Figure 3: The graph of the cosine in green, its vertical translation in magenta y = cos x+2, and its horizontal translate in blue by π 4 to the right - the graph of y = cos(x π 4 ). 2

3 To translate horizontally we have learned that to translate c units to the right, for c > 0, it is necessary to subtract c from the x variable and to translate c units to the left one adds c to the x variable. The same holds true for the sine and cosine graphs. Thus, for example, the graph of y = cos(x π 4 ) is a translation of π 4 units to the right of the graph of y = cos x. See figure Expanding and compressing horizontally, reflecting about the y axis Again from our previous discussions, we know that in general and specifically for y = sin x or y = cos x, given a constant.b Figure 4: Graphs of y = sin x in green, y = sin 2x in blue, and y = sin x 2 in magenta if b > 1, multiplication of the x variable by b compress the graph horizontally if 0 < b < 1, multiplication of the x variable by b expands the graph horizontally if b < 0, multiplication of the x variable by b first reflects the graph about the the y axis and then either compresses it or expands it according to whether b < 1 or 1 < b < 0 Given either y = sin bx or y = cos bx, the number 2π is called the period of the corresponding equation. It represents the length of the basic cycle associated with the equation. In b figure 4 we show, superimposed on the graph of y = sin x in green, the graph of y = sin 2x in blue,whose basic cycle is of length π, and the graph of y = sin x 2 in magenta, whose period is of length 4π 3

4 1.1.4 Harmonic motion Suppose you throw a stone into still pond. The ripples that gradually radiate from where the stone landed have the shape of a moving 3-dimensional sine curve. Figure 5 shows a picture of ripples in water, and figure 5 shows a mathematical model built from the sine function. Figure 5: Ripples in a pond Figure 6: Mathematical surface that models ripples - graph of z = 0.25 sin(x 2 + y 2 ) In a similar way sound creates ripples or vibrations in the air that travel radially out from the source. A tuning fork used by musicians for tuning their instruments creates vibrations in the air much like the ripples in a pond. Such physical wave phenomena are described in terms of their frequency. In these situations we consider the sine or cosine functions to be functions of time - that is, we imagine the graph of a sine or cosine function passing by us at some fixed speed - think of the movement of ripples in water. The frequency of such a curve is the number of basic cycles that pass per time unit. Usually frequency is measured 4

5 in cycles per second. When we spoke of the period of a sine or cosine curve we defined it to be the length of a basic cycle. Now if we choose to let the x axis represent time instead of physical distance and we rewrite our equations as y = a sin bt and y = a cos bt, for some constants a and b greater than zero, then the period, which we said was 2π b, now refers to the amount of time it takes a basic cycle to occur. If we measure time in seconds, then we can say that the period is the seconds required to complete one cycle. That is seconds per cycle, seconds. The reciprocal of the period then measures cycles per second, y = a sin bt and y = a cos bt this is then Example 1 : Concert pitch A frequency = b 2π cycles second cycle. Thus for the functions Most instruments in an orchestra tune to the so called concert pitch A, a sound with frequency 440 cycles per second. This means in our equations b 2π = 440, so that, b = 880π. The sine equation describing the vibration of the air is then, y = a sin 880π. The constant a is used to model the volume of the sound. Louder sounds require models with larger values of a. Example 2 : Vibrating spring Figure 7: Motion of a weight attached to a spring Imagine a weight at rest on a table but attached to a spring which is anchored in the wall somewhat to the left of the spring, see figure 7 Now pull the weight away from the wall 5

6 stretching the spring. Letting the weight go it springs back to a position to the left of its rest position and then rebounds to a position to the right of the rest position and so forth. Label the rest position zero, the compressed position to the left as -1 and the stretched position to the right +1. Once let go the weight oscillates between the -1 position and the +1 position, gradually due to friction and the nature of the spring the period of the oscillation will shorten and eventually stop. Lets assume an ideal situation in which there is no friction and no fatigue in the spring. Lets further assume that the motion has been modeled by the equation y = 7.2 sin 3πt. We then know that the amplitude a = 7.2 the period is: 2π b the frequency is 3 2 = 2π 3π = 2 3 seconds Hz ( Hz stands for cycles per second) Figure 8: Graph of y = 7.2 sin 3πt in red with a damped version in blue The graph looks like In actual fact the motion of the the weight undergoes a dampening effect whereby the distance traveled gradually decreases. In figure 8 we see the graph of y = 7.2 sin 3πt in red along with a damped version in blue. The later has been chosen to be the the graph of y = e.2(t 10) sin 3πt Inverse trigonometric functions Lets recall what we know about inverse funcitons. Given a function f : A B defined on an interval A, where B = f(a) be the set of all numbers f(x) such that x A. Then we have said that if f is injective f, there will be a function f(x) such that x A with the property that ( ) f f 1 (y) = y for every y B 6

7 ) f (f(x) 1 = x for every x A. Figure 9: Tangent on ( π 2, π 2 ) Now the problem with finding inverses of the trigonometric functions is that none of them satisfy the horizontal line test - none of them are injective. However for each of them they are injective when restricted to certain special intervals. For instance if we consider the tangent only on the interval A = ( π 2, π 2 ), then the graph look is as we see it in figure 9. The graph clearly satisfies the horizontal line test - so an inverse exists. The inverse of the tangent is denoted either as arctan or as tan 1. Since the range of the tangent is is the entire real line, in our definition of an inverse function the set B = R. We then have the relationships, arctan(tan x) = x tan(arctan y) = y Said another way, the arctan of a number y is the angle x whose tangent is y. As for the graph of the arctan, remember that in general given a function f with an inverse f 1, the graph of an f 1 is obtained by reflecting the graph of f about the diagonal line y = x. For the arctan the graph is as in figure 10. Note just as tan is asymptotic to the vertical lines x = π 2 and x = π 2 so arctan is asymptotic to the horizontal lines y = π 2 and y = π 2. The definitions other inverse trigonometric functions depends only on finding a convenient interval over which the associated trigonometric function is injective. The definiton of the inverse cotangent, written arccotan or cot 1, relies on restricting the definition of the cotangent to the interval (0, π). The arccotan then has domain the whole real line and range the interval (0, π). See figures 11 and 12. 7

8 Figure 10: Graph of arctan. The definition of the inverse sine, written arcsin or sin 1, relies on restricting the sine function to the interval [ π 2, π 2 ]. Since the sine has range [ 1, 1], the arcsine then has domain [ 1, 1] and range [ π 2, π 2 ]. Figure 13 shows the graph of the sine in blue and the graph of the arcsine in purple. Note that the sine is injective on the interval [ π 2, π 2 ]. For the inverse cosine, written arc-cosine or cos 1, the domain of the cosine is restricted to the interval [0, π]. The arc-cosine then has this as its range. See figures 14 and 15. For the inverse secant, written arcsec or sec 1, the chosen domain for the secant consists of the intervals [0, π 2 ) and on ( π 2, π] on which the secant is injective. The inverse secant then has domain the range of the secant, namely the two intervals [0, π 2 ) and ( π 2, π], and the range of the inverse secant is the domain of the secant namely the intervals[0, π 2 ) and ( π 2, π]. See figures 16 and 17. The arc-cosecant function may be similarly defined by considering the cosecant function restricted to the intervals [ π 2, 0) and (0, π 2 ). Example 3 Write each of: (a) cos(arcsin 4x) and (b) tan(arcsin 4x) as a function not involving a trigonometric function. What is the domain of the function? Solution: Begin by setting θ equal to the angle whose sine is 4x. In other words set θ = arcsin 4x. Next draw a right triangle as in figure 18. For convenience the triangle is chosen so that the hypotenuse has length one. If this is true and of sin θ = 4x, the 8

9 Figure 11: Cotangent is injective on (0, π) 9

10 Figure 12: Arccotan is defined on all of R with range (0, π) Figure 13: Sine in blue and arcsine in purple; sine is injective on [ π 2, π 2 ], arcsine defined on [ 1, 1] with range [ π 2, π 2 ] 10

11 Figure 14: Cosine injective on[0, π] Figure 15: Arc-cosine with domain[ 1, 1] and range [0, π] 11

12 Figure 16: Secant defined on [0, π 2 ) and on ( π 2, π]. Range:[1, ) (, 1] Figure 17: Arcsecant defined on [1, ) (, 1] with range [0, π 2 ) and ( π 2, π]. 12

13 Figure 18: Right triangle with θ = arcsin 4x. opposite side must have length 4x. Using the Pythagorean theorem, the adjacent side is calculated to have length 1 (4x) 2. We can now read off the answers. First, in order that 1 (4x) 2 not be negative, we must have (4x) 2 < 1 or x 2 < 1 16, which implies that 1 4 x 1 4. Then cos(arcsin 4x) = 1 (4x) 2 tan(arcsin 4x) = 4x 1 (4x) 2 Example 4 A sailor sailing single-handed about the globe encountered a ferocious storm in the south Atlantic. All his electronic equipment was damaged with the exception of a scientific calculator, his compass and his watch which had been set to Greenwich mean time. With the help of his compass he was able to determine with reasonable precision the time ( that is the local Noon) when the sun was directly south. To determine his latitude he however need to determine the elevation of the sun at Noon. He was able to determine that a vertical pole 2 meters high cast a shadow at Noon that was precisely 1.8 meters long. With this information he was able to calculate the elevation of the sun. How was this done? See figure 19. Solution: The sailor being schooled in trigonometry realized that if θ is the angle of elevation, it must be true that the tangent of θ must be = 10 9 = Taking out his calculator he entered and then pushed the inverse tangent button to find the angle whose tangent is The result was tan 1 ( ) = o. 13

14 Figure 19: Find the angle θ 1.2 Further Applications Figure 20: y = sin( x)is the equation of the red curve - when moved π to the left it has equation y = sin(π θ and corresponds to the blue curve y = sin θ Trigonometry allows an easy way to calculate the area of any triangle. First we need to observe that starting with the graph of the sine y = sin θ, reflecting about the x axis gives the graph of y = sin θ which we also know from the section on identities is the same as the graph of y = sin( θ). Using this equation, if we were to translate the graph π units to the left, the corresponding equation is then y = sin( θ + π). The graph we see is precisely the graph of sine. That is we got back to where we had started. See figure 20. Thus we can say that sin θ = sin(π θ). 14

15 1.2.1 Area of a triangle Given an arbitrary triangle we have learned that its area can be calculated by the formula A = base height 2. The difficulty with this formula that except for the case of right triangles, the height of the triangle does not correspond to the length of one of the edges. And if the lengths of edges is all the information available, one is stuck - that is, if one does not know a bit of trigonometry. Figure 21: A base angle θ may be acute or obtuse - in which case the complimentary angle π θ is acute Given a triangle lets let θ be one of the base angles. As we see in figure 21 the angle may be acute (less than 90 o ) or obtuse (greater than 90 o ). In the first case we know that sin θ = h a and in the second case we know that sin(π θ) = h a. But as we have seen, sin(π θ) = sin θ so thus, in both cases, we conclude that h = a sin θ. We then have in either case that the area is 1 2ab sin θ. This is expressed more formally as follows Result 5 Area of triangle The area A of a triangle with two sides of length a and b and an included angle θ is A = 1 ab sin θ Law of sines The law of sines says that for a given triangle ABC the ratio of the sine of angle to the length of its opposite side is independent of the chosen angle. In other words, if a, b and c are the lengths of the sides opposite the angle A, B and C, See figure 22. sin A a = sin B b = sin C. c The law of sines is used in situations where given a triangle it is possible to calculate the ratio of the sine of one of the angles to the length of its opposite side. Then as long as one more piece of information is present, the length of another side or the measurement 15

16 Figure 22: Triangle with angles A, B, and C with opposite sides of length a,b, and c of another angle, all the other dimensions of the triangle can be calculated. For instance suppose we know the size of angle A and the length a of its opposite side. Then if we know the length b of one of the other sides, we have sin A a = sin B b so that sin B = b sin A a. Thus the angle at B has measurement arcsin(b sin A a ). Knowing B we then know the measurement of the angle at C and hence sin C and ultimately the length c of the side opposite C. Here are some examples. Example 6 Given a triangle ABC suppose that A = 83 o, B = 55 o and b = 18. Find the lengths of the other sides. Solution: Since C = 180 ( ) = 42 o, we have sin 55 o 18 = sin 83o a and sin 55 o 18 = sin 42o c Thus a = 18 sin 83o sin 55 o 21.8 and c = 18 sin 42o sin 55 o Example 7 A construction company was awarded a contract by the provincial government to build a foot bridge across a gully on the west shore of Vancouver Island. Although the company knew that the bridge was to go from point A on one side of the gully to point B on the other, they did not know how long the bridge should be. In order to determine the length they set a marker C on the same side as the point B and measured the distance between A and B to be 10 meters. From the other side, with surveying equipment they were able to measure the angel at A, BAC, to be 12 o. The also measured the angle at C, BCA, to be 42 o. With this information they were able to find what the length of the bridge should be. What is this length? Solution: c sin 42 = 10 o sin 12 o We know from the law of sines that if c is the length of the bridge, then 10 sin 42o so that c = sin meters. See figure 23. o 16

17 Figure 23: Calculate the length of the bridge from A to B Using our previous result for the area of a general triangle, the law of sines may be easily proved Proof: Considering figure 22 we can calculate the area in three different ways, depending on which angle we wish to consider. Multiplying each term by 2 abc Area = 1 2 bc sin A = 1 2 ac sin B = 1 ab sin C. 2 gives the result Law of Cosines The law of cosines may be considered a generalization of the Pythagorean theorem. Given an arbitrary triangle as in figure 22, if the angle at C were a right angle then by the Pythagorean theorem c 2 = a 2 + b 2. But what if the angle at C is not a right angle? The law of cosines tells us that c 2 = a 2 + b 2 2ab cos C and in the case that C = 90 o we have cos C = 0 and the law of cosines reduces to the Pythagorean theorem. 17

18 Figure 24: Two boats one traveling N40 o E the other travelings60 o E. Figure 25: position of boats after one hour 18

19 The law of cosines also lets tells us that the converse of the Pythagorean theorem is true, for we then know that if C is not a right angle, then cos C 0 and c 2 a 2 + b 2. From our work in logic we know that the contrapositive of this last statement must also be true - namely, if c 2 = a 2 + b 2, then C is a right angle. This last statement is the converse of the Pythagorean theorem. Here is an example to show how the law of cosines may be applied. Example 8 Two boats leave the same harbor at Noon. Boat A travels at speed of 50 km/h in the direction N45 o E and the other, boat B, travels at a speed of 35 km/h in a direction S60 o E. What is the distance between them at 1:00 P.M.? Solution: We see that the angle between the paths of the two boats is 45 o +30 o = 75 o, and after 1 hour boat A has traveled 50 km and boat B has traveled 35 km. Letting C stand for their common starting point we have a diagram as in figure 25. By the law of cosines the distance c between the two boats is c = (50)(35) cos 75 o 53.1km. Example 9 Given a regular pentagon whose 5 sides are each of length 1. Let A, B, and C be three vertices as in figure 26 and let D be the mid-point of the line segment BC. What is the length b of the line segment AD? Solution: Using the fact that the angle at each of the vertices of a regular pentagon is 108 o, a fact that is not difficult to prove using elementary geometry, by the law of cosines b 2 = (1)(1 ) cos Thus b = 1.25 cos Figure 26: For a regular pentagon with sides of length 1, find length b of the segment AD 19