Example Use reference angle and appropriate sign to find the exact value of each expression.

Transcription

1 Example..4. Use reference angle and appropriate sign to find the exact value of each expression. (1) sin 11π and cos 11π () sin () cos ( ) 7π (4) tan 8π 6 Solution. (1) The reference angle of 11π is π, and it lies in Quadrant IV wherein 6 6 sine and cosine are negative and positive, respectively. sin 11π 6 = sin π 6 = 1 cos 11π 6 = cos π 6 = () The angle 7π lies in Quadrant II wherein cosine is negative, and its reference angle is π. 6 ( 6 cos 7π ) = cos π 6 6 = () sin150 = sin0 = 1 (4) tan 8π = tan π = sin π cos π = 1 = Seatwork/Homework.. Use reference angle and appropriate sign to find the exact value of each expression. (1) sin510 Answer: 1 () tan( 5 ) Answer: 1 () sec 1π Answer: (4) cot ( ) 10π Answer: Exercises. 1. Find the exact value. (a) sin600 Answer: (b) tan( 810 ) Answer: Undefined (c) sec585 Answer: 145

2 (d) cos( 40 ) Answer: 1 (e) sin 7π 6 Answer: 1 (f) cos 5π Answer: 1 (g) tan π 4 Answer: 1 (h) sec π Answer: (i) csc 11π 6 Answer: (j) cot 5π Answer: 6 (k) cos ( ) 4π Answer: 1 (l) tan 17π Answer: (m) cos 7π 4 Answer: (n) sec 19π Answer: 4 (o) sin ( ) 4π Answer: (p) sec π Answer: 6 (q) csc 1π Answer: (r) tan 5π 6 Answer:. Find the exact value of each expression. Teaching Notes (sinx) is denoted by sin x. (a) sin 150 +cos 150 Answer: 1 (b) cos( 0 )+sin40 Answer: (c) tan( 5 )+tan405 Answer: 0 (d) sec750 csc( 00 ) Answer: 0 (e) cos π +sin π Answer: 1 (f) sin 11π +cos 5π 6 Answer: 0 (g) cos 5π sin 5π Answer: 0 (h) tan π 4 +cos 8π (i) (j) tan π tan 5π 6 1+tan π tan 5π 6 sin 11π 6 cos π 6 sin( π 6)+cos 5π 6 sin 1π 6 Answer: 1. Compute P(θ), and find the exact values of the six circular functions. (a) θ = 19π 6 Answer: P(θ) = csc 19π 6 =, sec 19π ( ), 1, sin 19π = 1 19π, cos 6 = 6, cot 19π 6 = 146 = 6 Answer: Answer: 1, tan 19π 6 =, Similarly, this notation is used with the other trigonometric functions. In general, for a positive integer n, sin n x = (sinx) n.

3 (b) θ = π ( Answer: P(θ) = csc π = 1, ), sin π = π, sec =, cot π = π, cos = 1 π, tan =, 4. Given the value of a particular circular function and an information about the angle θ, find the values of the other circular functions. (a) cosθ = 1 and π < θ < π Answer: secθ =, sinθ =, tanθ =, cscθ =, cotθ = (b) sinθ = 8 and 0 < θ < π 17 Answer: cscθ = 17 15, cosθ =, tanθ = 8 17, secθ =, cotθ = (c) cosθ = 1 and π < θ < π 1 Answer: secθ = 1, sinθ = 1, tanθ = 1, cscθ = 1, cotθ = 4 Lesson.. Graphs of Circular Functions and Situational Problems Time Frame: 6 one-hour sessions Learning Outcomes of the Lesson At the end of the lesson, the student is able to: (1) determine the domain and range of the different circular functions; () graph the six circular functions with its amplitude, period, and phase shift; and () solve situational problems involving circular functions. Lesson Outline (1) Domain and range of circular functions () Graphs of circular functions () Amplitude, period, and phase shift Introduction There are many things that occur periodically. Phenomena like rotation of the planets and comets, high and low tides, and yearly change of the seasons 147

4 follow a definite pattern. In this lesson, we will graph the six circular functions and we will see that they are periodic in nature...1. Graphs of y = sinx and y = cosx Recall that, for a real number x, sinx = sinθ for an angle θ with measure x radians, and that sinθ is the second coordinate of the point P(θ) on the unit circle. Since each x corresponds to an angle θ, we can conclude that (1) sinx is defined for any real number x or the domain of the sine function is R, and () the range of sine is the set of all real numbers between 1 and 1 (inclusive). From thedefinition, italsofollowsthatsin(x+π) = sinxforanyreal number x. This means that the values of the sine function repeat every π units. In this case, we say that the sine function is a periodic function with period π. Table.11belowshowsthevaluesofy = sinx,wherexistheequivalentradian measure of the special angles and their multiples from 0 to π. As commented above, these values determine the behavior of the function on R. x x 0 π 6 y 0 1 7π 6 π 4 π π 1 π π 4 5π 6 π π 4 4π π 5π 7π 4 11π 6 π y Table.11 From the table, we can observe that as x increases from 0 to π, sinx also increases from 0 to 1. Similarly, as x increases from π to π, sinx also increases from 1 to 0. On the other hand, notice that as x increases from π to π, sinx decreases from 1 to 0. Similarly, as x increases from π to π, sinx decreases from 0 to 1. To sketch the graph of y = sinx, we plot the points presented in Table.11, and join them with a smooth curve. See Figure.1. Since the graph repeats every π units, Figure.1 shows periodic graph over a longer interval. 148

5 Teaching Notes It is a good exercise to construct the graph of the sine function using the height of P(θ). Put the unit circle side-by-side with the coordinate plane for the graph, and trace the height for each value of x onto the graph of y = sinx. Figure.1 Figure.1 We can make observations about the cosine function that are similar to the sine function. y = cosx has domain R and range [ 1,1]. y = cosx is periodic with period π. The graph of y = cosx is shown in Figure.14. Figure.14 From the graphs of y = sinx and y = cosx in Figures.1 and.14, respectively, we observe that sin( x) = sinx and cos( x) = cosx for any real number x. In other words, the graphs of y = cos( x) and y = cosx are the same, while the graph of y = sin( x) is the same as that of y = sinx. In general, if a function f satisfies the property that f( x) = f(x) for all x in its domain, we say that such function is even. On the other hand, we say that a function f is odd if f( x) = f(x) for all x in its domain. For example, the functions x and cosx are even, while the functions x x and sinx are odd. 149

6 ... Graphs of y = asinbx and y = acosbx Using a table of values from 0 to π, we can sketch the graph of y = sinx, and compare it to the graph of y = sinx. See Figure.15 wherein the solid curve belongs to y = sinx, while the dashed curve to y = sinx. For instance, if x = π, then y = 1 when y = sinx, and y = when y = sinx. The period, x-intercepts, and domains are the same for both graphs, while they differ in the range. The range of y = sinx is [,]. Figure.15 In general, the graphs of y = asinx and y = acosx with a > 0 have the same shape as the graphs of y = sinx and y = cosx, respectively. If a < 0, there is a reflection across the x-axis. The range of both y = asinx and y = acosx is [ a, a ]. In the graphs of y = asinx and y = acosx, the number a is called its amplitude. It dictates the height of the curve. When a < 1, the graphs are shrunk vertically, and when a > 1, the graphs are stretched vertically. Teaching Notes Review or teach the reflection across the x-axis when the sign of the function is changed. Now, in Table.16, we consider the values of y = sinx on [0,π]. x 0 y 0 π π π π π π 4 5π 6 π x y 7π 5π π π 5π 7π 4 11π 6 π Table

7 Figure.17 Figure.17 shows the graphs of y = sinx (solid curve) and y = sinx (dashed curve) over the interval [0, π]. Notice that, for sin x to generate periodic values similar to [0,π] for y = sinx, we just need values of x from 0 to π. We then expect the values of sinx to repeat every π units thereafter. The period of y = sinx is π. If b 0, then both y = sinbx and y = cosbx have period given by π b. If 0 < b < 1, the graphs are stretched horizontally, and if b > 1, the graphs are shrunk horizontally. To sketch the graphs of y = asinbx and y = acosbx, a,b 0, we may proceed with the following steps: (1) Determine the amplitude a, and find the period π. To draw one cycle b of the graph (that is, one complete graph for one period), we just need to complete the graph from 0 to π. b () Dividetheintervalintofourequalparts,andgetfivedivisionpoints: x 1 = 0, x, x, x 4, and x 5 = π b, where x is the midpoint between x 1 and x 5 (that is, 1 (x 1 + x 5 ) = x ), x is the midpoint between x 1 and x, and x 4 is the midpoint between x and x 5. () Evaluate the function at each of the five x-values identified in Step. The points will correspond to the highest point, lowest point, and x-intercepts of the graph. (4) Plot the points found in Step, and join them with a smooth curve similar to the graph of the basic sine curve. (5) Extend the graph to the right and to the left, as needed. 151

8 Example..1. Sketch the graph of one cycle of y = sin4x. Solution. (1) The period is π = π, and the amplitude is. 4 () Dividing the interval [0, π ] into 4 equal parts, we get the following x- coordinates: 0, π, π, π, and π () When x = 0, π 4, and π, we get y = 0. On the other hand, when x = π 8, we have y = (the amplitude), and y = when x = π 8. (4) Draw a smooth curve by connecting the points. There is no need to proceed to Step 5 because the problem only asks for one cycle. Example... Sketch the graph of y = cos x. Solution. (1) The amplitude is =, and the period is π 1 = 4π. () We divide the interval [0,4π] into four equal parts, and we get the following x-values: 0, π, π, π, and 4π. () We have y = 0 when x = π and π, y = when x = 0 and 4π, and y = when x = π. (4) We trace the points in Step by a smooth curve. (5) We extend the pattern in Step 4 to the left and to the right. 15

9 Example... Sketch the graph of two cycles of y = 1 sin( ) x. Solution. Since the sine function is odd, the graph of y = 1 sin( ) x is the same as that of y = 1 sin x. (1) The amplitude is 1, and the period is π = π. () Dividing the interval [0, π] into four equal parts, we get the x-coordinates of the five important points: 0+π = π, 0+ π = π 4, π +π = 9π 4. () We get y = 0 when x = 0, π, and π, y = 1 when π 4, and y = 1 when 9π 4. (4) We trace the points in Step by a smooth curve. (5) We extend the pattern in Step 4 by one more period to the right. Seatwork/Homework.. (1) Sketch the graph of one cycle of y = 1 sinx. Answer: () Sketch the graph of two cycles of y = cos ( x ). Answer: 15

10 () Sketch the graph of y = cos4x. Answer: (4) Sketch the graph of one cycle of y = sin ( x ). Answer:... Graphs of y = asinb(x c) + d and y = acosb(x c) + d We first compare the graphs of y = sinx and y = sin ( x ) π using a table of values and the 5-step procedure discussed earlier. Teaching Notes As x runs from π to 7π, the value of the expression x π runs from 0 to π. So for one cycle of the graph of y = sin ( x ) π, we then expect to have the graph of y = sinx starting from x = π. This is confirmed by the values in Table.18. We then apply a similar procedure to complete one cycle of the graph; that is, divide the interval [ π, 7π ] into four equal parts, and then determine the key values of x in sketching the graphs as discussed earlier. The one-cycle graph of y = sinx (dashed curve) and the corresponding one-cycle graph of y = sin ( x ) π (solid curve) are shown in Figure.19. Review or teach the horizontal translation rule: if x is replaced by x h in the equation, the graph is translated h units to the right if h > 0 and to the left if h < 0. x x π sin ( x π ) π 5π 6 4π 11π 6 7π 0 π π π π Table

11 Figure.19 Observe that the graph of y = sin ( x ) π shifts π units to the right of y = sinx. Thus, they have the same period, amplitude, domain, and range. The graphs of y = asinb(x c) and y = acosb(x c) have the same shape as y = asinbx and y = acosbx, respectively, but shifted c units to the right when c > 0 and shifted c units to the left if c < 0. The number c is called the phase shift of the sine or cosine graph. Example..4. In the same Cartesian plane, sketch one cycle of the graphs of y = sinx and y = sin ( x+ π 4). Solution. We have sketched the graph of y = sinx earlier at the start of the lesson. We consider y = sin ( x+ π 4). We expect that it has the same shape as that of y = sinx, but shifted some units. Here, we have a =, b = 1, and c = π. From these constants, we get 4 the amplitude, the period, and the phase shift, and these are, π, and π, 4 respectively. One cycle starts at x = π and ends at x = 4 π +π = 7π. We now compute 4 4 the important values of x. π 4 + 7π 4 = π 4, π 4 + π 4 = π 4, π 4 + 7π 4 = 5π 4 x π π π 5π 7π y = sin ( ) x+ π

12 While the effect of c in y = asinb(x c) and y = acosb(x c) is a horizontal shift of their graphs from the corresponding graphs of y = asinbx and y = acosbx, the effect of d in the equations y = asinb(x c)+d and y = acosb(x c)+d is a vertical shift. That is, the graph of y = asinb(x c)+d has the same amplitude, period, and phase shift as that of y = asinb(x c), but shifted d units upward when d > 0 and d units downward when d < 0. Example..5. Sketch the graph of y = cos ( x π ). 6 Teaching Notes Review or teach the vertical translation rule: if the equation y = f(x) is changed to y = f(x)+k, the graph is translated k units upward if k > 0 and downward if k < 0. Solution. Here, a =, b =, c = π, and d =. We first sketch one cycle of 6 the graph of y = cos ( x 6) π, and then extend this graph to the left and to the right, and then move the resulting graph units downward. The graph of y = cos ( x 6) π has amplitude, period π, and phase shift π. 6 Start of one cycle: π 6 End of the cycle: π 6 +π = 7π 6 π 6 + 7π 6 = π, π 6 + π = 5π π 1, + 7π 6 = 11π 1 x y = cos ( x π 6 ) π 6 5π 1 π 11π 1 7π y = cos ( x π 6)

13 Before we end this sub-lesson, we make the following observation, which will be used in the discussion on simple harmonic motion (Sub-Lesson..6). Different Equations, The Same Graph 1. The graphs of y = sinx and y = sin(x + πk), k any integer, are the same.. The graphs of y = sinx, y = sin(x + π), y = cos(x π ), and y = cos(x+ π ) are the same.. In general, the graphs of and y = asinb(x c)+d, y = asin[b(x c)+π +πk]+d, y = acos[b(x c) π +πk]+d, y = acos[b(x c)+ π +πk]+d, where k is any integer, are all the same. Similar observations are true for cosine. Seatwork/Homework.. (1) In the same Cartesian plane, sketch one cycle of the graphs of y = cosx and y = cos ( x+ π )

14 Answer: () In the same Cartesian plane, sketch one cycle of the graphs of y = 1 4 sinx and y = 1 4 sin( x π 4). Answer: () Sketch the graph of y = sin ( π x). Answer: 158

15 ..4. Graphs of Cosecant and Secant Functions We know that cscx = 1 graph of y = cscx. sinx if sinx 0. Using this relationship, we can sketch the First, we observe that the domain of the cosecant function is {x R : sinx 0} = {x R : x kπ,k Z}. Table.0 shows the key numbers (that is, numbers where y = sinx crosses the x-axis, attain its maximum and minimum values) and some neighboring points, where und stands for undefined, while Figure.1 shows one cycle of the graphs of y = sinx (dashed curve) and y = cscx (solid curve). Notice the asymptotes of the graph y = cscx. x 0 π 6 π 5π 7π π 6 6 π 11π 6 π y = sinx y = cscx und 1 und 1 und Table.0 Figure.1 We could also sketch the graph of cscx directly from the graph of y = sinx by observing the following facts: (1) If sinx = 1 (or 1), then cscx = 1 (or 1). () At each x-intercept of y = sinx, y = cscx is undefined; but a vertical asymptote is formed because, when sinx is close to 0, the value of cscx will have a big magnitude with the same sign as sinx. 159

16 Refer to Figure. for the graphs of y = sinx (dashed curve) and y = cscx (solid curve) over a larger interval. Figure. Like the sine and cosecant functions, the cosine and secant functions are also reciprocals of each other. Therefore, y = secx has domain {x R : cosx 0} = {x R : x kπ, k odd integer}. Similarly, the graph of y = secx can be obtained from the graph of y = cosx. These graphs are shown in Figure.. Figure. Example..6. Sketch the graph of y = csc x. Solution. First, we sketch the graph of y = sin x, and use the technique discussed above to sketch the graph of y = csc x. 160

17 The vertical asymptotes of y = csc x are the x-intercepts of y = sin x: x = 0,±π,±4π,... After setting up the asymptotes, we now sketch the graph of y = csc x as shown below. Example..7. Sketch the graph of y = secx. Solution. Sketchthegraphofy = cosx(notethatithasperiodπ),thensketch the graph of y = secx (as illustrated above), and then move the resulting graph units upward to obtain the graph of y = secx. 161

18 Seatwork/Homework..4 (1) Sketch the graph of y = secx on the interval [0,π]. Answer: () Sketch the graph of y = csc4x 1 on the interval [ π, π ]. Answer: 16

19 ..5. Graphs of Tangent and Cotangent Functions We know that tanx = sinx, where cosx 0. From this definition of the tangent cosx function, it follows that its domain is the same as that of the secant function, which is {x R : cosx 0} = {x R : x kπ, k odd integer}. We note that tanx = 0 when sinx = 0 (that is, when x = kπ, k any integer), and that the graph of y = tanx has asymptotes x = kπ, k odd integer. Furthermore, by recalling the signs of tangent from Quadrant I to Quadrant IV and its values, we observe that the tangent function is periodic with period π. To sketch the graph of y = tanx, it will be enough to know its one-cycle graph on the open interval ( Teaching Notes π, π ). See Table.4 and Figure.5. There is also a way of sketching the x π π π π graph of y = tanx based on the tangent segment to the unit circle, similar to the construction described in sketching the graph of y = sinx. But we do not go anymore into the details of this approach. y = tanx und 1 0 x y = tanx π π Table.4 π π und Figure.5 In the same manner, the domain of y = cotx = cosx sinx is {x R : sinx 0} = {x R : x kπ,k Z}, and its period is also π. The graph of y = cotx is shown in Figure.6. 16

20 Figure.6 In general, to sketch the graphs of y = atanbx and y = acotbx, a 0 and b > 0, we may proceed with the following steps: (1) Determine the period π. Then we draw one cycle of the graph on ( π, π b b b for y = atanbx, and on ( 0, b) π for y = acotbx. () Determine the two adjacent vertical asymptotes. For y = a tan bx, these vertical asymptotes are given by x = ± π. For y = acotbx, the vertical b asymptotes are given by x = 0 and x = π. b () Divide the interval formed by the vertical asymptotes in Step into four equal parts, and get three division points exclusively between the asymptotes. (4) Evaluate the function at each of these x-values identified in Step. The points will correspond to the signs and x-intercept of the graph. (5) Plot the points found in Step, and join them with a smooth curve approaching to the vertical asymptotes. Extend the graph to the right and to the left, as needed. Example..8. Sketch the graph of y = 1 tanx. Solution. The period of the function is π, and the adjacent asymptotes are x = ± π 4,±π,... Dividingtheinterval( π, π 4 4 4) intofourequalparts, thekeyx-values are π, 0, and π. 8 8 ) x π 8 0 π 8 y = 1 tanx

21 Example..9. Sketch the graph of y = cot x on the interval (0,π). Solution. Theperiodofthefunctionisπ, andtheadjacentasymptotesarex = 0 and x = π. We now divide the interval (0,π) into four equal parts, and the key x-values are π 4, π, and 9π 4. x π 4 π 9π 4 y = cot x 0 165

22 Seatwork/Homework..5 (1) Sketch the graph of y = cot( x) on the interval [ π,π]. Answer: () Sketch the graph of y = tan x 4 Answer: on the interval [ π,π]...6. Simple Harmonic Motion Repetitive or periodic behavior is common in nature. The time-telling device known as sundial is a result of the predictable rising and setting of the sun everyday. It consists of a flat plate and a gnomon. As the sun moves across the sky, the gnomon casts a shadow on the plate, which is calibrated to tell the time of the day. 166

23 r.jpg By liz west (Sundial) [CC BY.0 ( via Wikimedia Commons Some motions are also periodic. When a weight is suspended on a spring, pulled down, and released, the weight oscillates up and down. Neglecting resistance, this oscillatory motion of the weight will continue on and on, and its height is periodic with respect to time. t = 0 sec t =.8 sec 167

24 t = 6.1 sec t = 9 sec Periodic motions are usually modeled by either sine or cosine function, and are called simple harmonic motions. Unimpeded movements of objects like oscillation, vibration, rotation, and motion due to water waves are real-life occurrences that behave in simple harmonic motion. Equations of Simple Harmonic Motion The displacement y (directed height or length) of an object behaving in a simple harmonic motion with respect to time t is given by one of the following equations: or y = asinb(t c)+d y = acosb(t c)+d. In both equations, we have the following information: amplitude = a = 1 (M m) - the maximum displacement above and below the rest position or central position or equilibrium, where M is the maximum height and m is the minimum height; period = π b - the time required to complete one cycle (from one highest or lowest point to the next); frequency = b - the number of cycles per unit of time; π c - responsible for the horizontal shift in time; and d - responsible for the vertical shift in displacement. Example..10. A weight is suspended from a spring and is moving up and downinasimpleharmonicmotion. Atstart,theweightispulleddown5cmbelow the resting position, and then released. After 8 seconds, the weight reaches its 168

25 highest location for the first time. Find the equation of the motion. Solution. We are given that the weight is located at its lowest position at t = 0; that is, y = 5 when t = 0. Therefore, the equation is y = 5cosbt. Because it took the weight 8 seconds from the lowest point to its immediate highest point, half the period is 8 seconds. 1 π b = 8 = b = π = y = 5cos πt 8 8 Example..11. Suppose you ride a Ferris wheel. The lowest point of the wheel is meters off the ground, and its diameter is 0 m. After it started, the Ferris wheel revolves at a constant speed, and it takes seconds to bring you back again to the riding point. After riding for 150 seconds, find your approximate height above the ground. Solution. We ignore first the fixed value of m off the ground, and assume that the central position passes through the center of the wheel and is parallel to the ground. Let t be the time (in seconds) elapsed that you have been riding the Ferris wheel, and y is he directed distance of your location with respect to the assumed central position at time t. Because y = 10 when t = 0, the appropriate model is y = 10cosbt for t 0. Given that the Ferris wheel takes seconds to move from the lowest point to the next, the period is. π b = = b = π 16 = y = 10cos πt 16 When t = 150, we get y = 10cos 150π Bringing back the original condition given in the problem that the riding point is m off the ground, after riding for 150 seconds, you are approximately located.8+1 = 16.8 m off the ground. In the last example, the central position or equilibrium may be vertically shifted from the ground or sea level (the role of the constant d). In the same way, the starting point may also be horizontally shifted (the role of the constant c). Moreover, as observed in Sub-Lesson.. (see page 157), to find the function that describes a particular simple harmonic motion, we can either choose or y = asinb(t c)+d y = acosb(t c)+d, and determine the appropriate values of a, b, c, and d. In fact, we can assume that a and b are positive numbers, and c is the smallest such nonnegative number. 169

26 Example..1. A signal buoy in Laguna Bay bobs up and down with the height h of its transmitter (in feet) above sea level modeled by h(t) = asinbt+d at time t (in seconds). During a small squall, its height varies from 1 ft to 9 ft above sea level, and it takes.5 seconds from one 9-ft height to the next. Find the values of the constants a, b, and d. Solution. We solve the constants step by step. The minimum and maximum values of h(t) are 1 ft and 9 ft, respectively. Thus, the amplitude is a = 1(M m) = 1 (9 1) = 4. Because it takes.5 seconds from one 9-ft height to the next, the period is.5. Thus, we have π b =.5, which gives b = 4π 7. Because the lowest point is 1 ft above the sea level and the amplitude is 4, it follows that d = 5. Example..1. A variable star is a star whose brightness fluctuates as observed from Earth. The magnitude of visual brightness of one variable star ranges from.0 to 10.1, and it takes days to observe one maximum brightness to the next. Assuming that the visual brightness of the star can be modeled by the equation y = asinb(t c)+d, t in days, and putting t = 0 at a time when the star is at its maximum brightness, find the constants a, b, c, and d, where a,b > 0 and c the least nonnegative number possible. Solution. a = M m = = 4.05 π π = = b = b 166 d = a+m = = 6.05 For the (ordinary) sine function to start at the highest point at t = 0, the least possible horizontal movement to the right (positive value) is π units. bc = π = c = π b = π π 166 = 49 Example..14. The path of a fast-moving particle traces a circle with equation (x+7) +(y 5) = 6. It starts at point ( 1,5), moves clockwise, and passes the point ( 7,11) for the first time after traveling 6 microseconds. Where is the particle after traveling 15 microseconds? 170

27 Teaching Notes Here, we need an equation with the same graph as y = asin(bt+ π )+d that will fit in the equation y = asinb(t c)+d, where c is the least nonnegative possible number. Recall the observation made on page 157. Solution. As described above, we may choose sine or cosine function. Here, we choose the sine function to describe both x and y in terms of time t in microseconds; that is, we let x = asinb(t c)+d and y = esinf(t g)+h, where we appropriately choose the positive values for a, b, e, and f, and the least nonnegative values for c and g. The given circle has radius 6 and center ( 7,5). Defining the central position of the values of x as the line x = 7 and that of the values of y as the line y = 5, we get a = e = 6, d = 7, and h = 5. From the point ( 1,5) to the point ( 7,11) (moving clockwise), the particle has traveled three-fourths of the complete cycle; that is, three-fourths of the period must be. 4 π b = 4 π f = 6 = b = f = π 4 As the particle starts at ( 1,5) and moves clockwise, the values of x start at its highest value (x = 1) and move downward toward its central position (x = 7) and continue to its lowest value (x = 1). Therefore, the graph of asinbt+d has to move π = 6 units to the right, and so we get c = 6. b As to the value of g, we observe the values of y start at its central position (y = 5) and go downward to its lowest value (y = 1). Similar to the argument used in determining c, the graph of y = esinft+h has to move π = 4 units to b the right, implying that g = 4. Hence, We have the following equations of x and y in terms of t: When t = 15, we get and x = 6sin π 4 (t 6) 7 and y = 6sin π 4 (t 4)+5. x = 6sin π 4 (15 6) 7 = y = 6sin π 4 (15 4)+5 = That is, after traveling for 15 microseconds, the particle is located near the point (.76, 9.4). Seatwork/Homework A weight is suspended from a spring and is moving up and down in a simple harmonic motion. At start, the weight is pushed up 6 cm above the resting 171

28 position, and then released. After 14 seconds, the weight reaches again to its highest position. Find the equation of the motion, and locate the weight with respect to the resting position after 0 seconds since it was released. Answer: y = 6cos π pi t or y = 6sin (t + 7 ), location of the weight after seconds: about 5.4 cm below the resting position. Suppose the lowest point of a Ferris wheel is 1.5 meters off the ground, and its radius is 15 m. It makes one complete revolution every 0 seconds. Starting at the lowest point, find a cosine function that gives the height above the ground of a riding child in terms of the time t in seconds. ( ) Answer: y = 15cos π 15 t Exercises. 1. Sketch two cycles of the graph (starting from x = 0) of the given function. Indicate the amplitude, period, phase shift, domain, and range for each function. (a) y = 4sinx Answer: amplitude = 4, period = π, phase shift = 0, domain = R, range = [ 4,4] (b) y = cosx Answer: amplitude =, period = π, phase shift = 0, domain = R, range = [,] (c) y = cos x 4 Answer: amplitude = 1, period = 8π, phase shift = 0, domain = R, range = [ 1,1] (d) y = sinx Answer: amplitude = 1, period = π, phase shift = 0, domain = R, range = [ 1,1] 17

29 (e) y = +sin4x Answer: amplitude = 1, period = π, phase shift = 0, domain = R, range = [1,] (f) y = 1+cosx Answer: amplitude = 1, period = π, phase shift = 0, domain = R, range = [,0] (g) y = 1 sinx Answer: amplitude = 1, period = π, phase shift = 0, domain = R, range = [ 1, 1] (h) y = sin( x) Answer: amplitude =, period = π, phase shift = 0, domain = R, range = [,] (i) y = cos x Answer: amplitude =, period = 4π, phase shift = 0, domain = R, range = [1,5] (j) y = sin ( ) x π 4 Answer: amplitude = 1, period = π, phase shift = π, domain = R, 4 range = [ 1,1] 17

30 (k) y = cos ( ) x+ π Answer: amplitude =, period = π, phase shift = π, domain = R, range = [,] (l) y = sin(x 4π) Answer: amplitude =, period = π, phase shift = 4π, domain = R, range = [,] (m) y = cos( ) x π Answer: amplitude =, period = π, phase shift = π, domain = R, range = [ 4, 8] (n) y = 4cos ( x ) π + Answer: amplitude = 4, period = π, phase shift = π, domain = R, range = [,6]. Sketch the graph of the following functions. (a) y = sinx Answer: 174

31 (b) y = 4cosx + Answer: (c) y = sin(x+π) 1 Answer:. Sketch the graph of each function over two periods, starting from x = 0. Indicate the period, phase shift, domain, and range of each function. (a) y = csc( x) Answer: period = π, phase shift = 0, domain = {x x kπ, k Z}, range = (, 1] [1, ) (b) y = cot( x) Answer: period = π, phase shift = 0, domain = {x x kπ, k Z}, range = R (c) y = tanx Answer: period = π, phase shift = 0, domain = {x x (k +1) π, k Z}, range = R 175

32 (d) y = secx Answer: period = π, phase shift = 0, domain = {x x (k +1) π, k Z}, range = (, 1] [1, ) (e) y = secx Answer: period = π, phase shift = 0, domain = {x x (k +1)π 6, k Z}, range = (, 1] [1, ) (f) y = cscx Answer: period = π, phase shift = 0, domain = {x x kπ, k Z}, range = (, ] [, ) (g) y = 4sec x Answer: period = π, phase shift = 0, domain = {x x (k+1) π 4, k Z}, range = (, 4] [4, ) (h) y = tan(x+π) 176

33 Answer: period = π, phase shift = π, domain = {x x (k+1) π, k Z}, range = R (i) y = tan ( ) x π Answer: period = π, phase shift = π, domain = {x x kπ, k Z}, range = R (j) y = cot ( ) x+ π 4 Answer: period = π,phase shift = π,domain = {x x 4 (k 1)π, k even integer}, 4 range = R (k) y = cscx Answer: period = π, phase shift = 0, domain = {x x kπ, k Z}, range = (, 1] [5, ) (l) y = 4+secx Answer: period = π, phase shift = 0, domain = {x x (k +1)π, k 6 Z}, range = (,] [6, ) ) (m) y = sec ( x π Answer: period = π, phase shift = π, domain = {x x (k+1)π 4, k Z}, range = (, ] [, ) (n) y = sec x Answer: period = π, phase shift = 0, domain = {x x (k+1) π 4, k Z}, range = (, ] [5, ) 177

34 (o) y = csc ( ) x π Answer: period = π, phase shift = π, domain = {x x (k+1)π, k Z}, range = (, ] [, ) 4. Assuming that there is no vertical shift, find a function that describes a simple harmonic motion with the following properties. (a) sine function; displacement zero at time t = 0; moving up initially; amplitude = 6 cm; period = 4 sec Answer: y = 6sin πt (b) cosine function; highest point 4 cm above the equilibrium at time t = 0; period = 10 sec Answer: y = 5cos πt 5 178

35 (c) cosine function; lowest point 9 cm below the equilibrium at time t = 0; period = 5 sec Answer: y = 9cos π 5 (t 5 ) 5. A point P moving in a simple harmonic motion makes 10 complete revolutions every 1 second. The amplitude of the motion is m. Assuming that P is at its minimum displacement with respect to the equilibrium when t = 0 and there is a vertical shift of m downward, find a sine function that describes the path traced by P in terms of time t. Answer: y = sin0π(t 1 ) The path of a fast-moving particle (assuming constant speed) traces a circle with equation (x ) +(y 4) = 5. It starts at point (, 1), moves counterclockwise, and passes the point (8, 4) for the first time after traveling 7 microseconds. Where is the particle after traveling for 0 microseconds? Answer: about the point ( 1.87, 5.11) Hint. The coordinates (x,y) of the location of the particle at time t (in microseconds) are given by x = 5cos π 14 (t 7)+ and y = 5sin π 14 (t 7) Awoodenballistiedonastring0cmlong, andisoscillatinglikeapendulum. See figure below. It is initially pulled back at 90 angle with the vertical, and is released with a push so that the ball reaches its maximum height back and forth. If it reaches its maximum height again after seconds, find its height 10 seconds after it was released. Answer: 7 cm Hint. The height h(t) (in cm) of the ball at time t (in seconds) is given by h(t) = 0sin π (t ) For what values of k do y = cotx and y = cot(x kπ) have the same graph? Answer: any integer 9. For what values of k do y = secx and y = sec(x kπ) have the same graph? Answer: any even integer 179

36 10. Find the least positive value of c such that the graph of y = sin(x + c) coincide with that of y = cosx. Answer: π Find the largest positive value of c such that the graph of y = cos(x c) coincide with that of y = cos(x ). Answer: + π 1. For what values of a do the graphs of y = acosb(x c) and y = sec π 6 (x 6) never intersect for any values of b and c? Answer: < a < 4 Lesson.4. Fundamental Trigonometric Identities Time Frame: 4 one-hour sessions Learning Outcomes of the Lesson At the end of the lesson, the student is able to: (1) determine whether an equation is an identity or a conditional equation; () derive the fundamental trigonometric identities; () simplify trigonometric expressions using fundamental trigonometric identities; and (4) prove other trigonometric identities using fundamental trigonometric identities. Lesson Outline (1) Domain of an equation () Identity and conditional equation () Fundamental trigonometric identities (4) Proving trigonometric identities Introduction In previous lessons, we have defined trigonometric functions using the unit circle and also investigated the graphs of the six trigonometric functions. This lesson builds on the understanding of the different trigonometric functions by discovery, deriving, and working with trigonometric identities Domain of an Expression or Equation Consider the following expressions: x+1, x 1, x x x 4, 180 x x 1.

37 What are the real values of the variable x that make the expressions defined in the set of real numbers? In the first expression, every real value of x when substituted to the expression makes it defined in the set of real numbers; that is, the value of the expression is real when x is real. In the second expression, not every real value of x makes the expression defined in R. For example, when x = 0, the expression becomes 1, which is not a real number. x 1 R x 1 0 x 1 or x 1 Here, for x 1 to be defined in R, x must be in (, 1] [1, ). In the third expression, the values of x that make the denominator zero make the entire expression undefined. x x 4 = (x 4)(x+1) = 0 x = 4 or x = 1 Hence, the expression x x x 4 is real when x 4 and x 1. Inthefourthexpression, becausetheexpression x 1isinthedenominator, x must be greater than 1. Although the value of the entire expression is 0 when x = 0, we do not include 0 as allowed value of x because part of the expression is not real when x = 0. In the expressions above, the allowed values of the variable x constitute the domain of the expression. Thedomain ofanexpression(orequation)isthesetofallrealvaluesof the variable for which every term(or part) of the expression(equation) is defined in R. In the expressions above, the domains of the first, second, third, and fourth expressions are R, (, 1] [1, ), R\{ 1,4}, and (1, ), respectively. Example.4.1. Determine the domain of the expression/equation. (a) x 1 x+1 x +x 8x 1 x (b) tanθ sinθ cosθ (c) x 1+x = x 1 (d) z cos z 1+sinz = 4sinz 1 181

38 Solution. (a) x + x 8x = x(x + 4)(x ) = 0 x = 0, x = 4, or x = x+1 R x+1 0 x 1 1 x = 0 x = 1 Domain = [ 1, )\{ 4,0,1,} = [ 1,0) (0,1) (1,) (, ) (b) tanθ sinθ cosθ = sinθ sinθ cosθ cosθ cosθ = 0 θ = kπ, k odd integer Domain = R\{ kπ k odd integer} (c) The expression 1+x is always positive, and so 1+x is defined in R. On the other hand, the expression x 1 is also defined in R, but it cannot be zero because it is in the denominator. Therefore, x should not be 1 and 1. Domain = R\{ 1,1} (d) 1+sinz = 0 z = π +kπ, k Z Domain = R\{ π +kπ k Z} Seatwork/Homework.4.1 Find the domain of the expression/equation. x (1) x+1 x+1 x +x+1 x 1 () sec 1 θ = 1 sinθ sinθ Answer: R\{ 1,1} Answer: R\ { (k +1) π k Z} () tan x = cotx+1 Answer: R\ { kπ k Z} (4) 1 tanx = sinx Answer: 1 x R\{ { 1,1} { (k +1) π k Z}}.4.. Identity and Conditional Equation Consider the following two groups of equations: Group A (A1) x 1 = 0 (A) (x+7) = x +49 (A) x 4 x = x 1 Group B (B1) x 1 = (x 1)(x+1) (B) (x+7) = x +14x+49 (B) x 4 x = x+ 18

39 In each equation in Group A, some values of the variable that are in the domain of the equation do not satisfy the equation (that is, do not make the equation true). On the other hand, in each equation in Group B, every element in the domain of the equation satisfies the given equation. The equations in Group A are called conditional equations, while those in Group B are called identities. An identity is an equation that is true for all values of the variable in the domain of the equation. An equation that is not an identity is called a conditional equation. (In other words, if some values of the variable in the domain of the equation do not satisfy the equation, then the equation is a conditional equation.) Example.4.. Identify whether the given equation is an identity or a conditional equation. For each conditional equation, provide a value of the variable in the domain that does not satisfy the equation. (1) x = ( x )( x + x+ 4 ) () sin θ = cos θ+1 () sinθ = cosθ 1 (4) 1 x 1+ x = 1 x+x 1 x Solution. (1) This is an identity because this is simply factoring of difference of two cubes. () Thisisaconditionalequation. Ifθ = 0,thentheleft-handsideoftheequation is 0, while the right-hand side is. () This is also a conditional equation. If θ = 0, then both sides of the equation are equal to 0. But if θ = π, then the left-hand side of the equation is 0, while the right-hand side is. (4) This is an identity because the right-hand side of the equation is obtained by rationalizing the denominator of the left-hand side. Seatwork/Homework.4. Identify whether the given equation is an identity or a conditional equation. For each conditional equation, provide a value of the variable in the domain that does not satisfy the equation. (1) 1+x+ x 1 x = 1 1 x Answer: identity 18

40 () cos θ sin θ cosθ+sinθ = cosθ sinθ Answer: identity () tanθ = cotθ Answer: conditional equation, θ = π (4) cos x = cosx+ Answer: conditional equation, x = The Fundamental Trigonometric Identities Recall that if P(x,y) is the terminal point on the unit circle corresponding to θ, then we have sinθ = y cscθ = 1 tanθ = y y x cosθ = x secθ = 1 x cotθ = x y. From the definitions, the following reciprocal and quotient identities immediately follow. Note that these identities hold if θ is taken either as a real number or as an angle. cscθ = 1 sinθ Reciprocal Identities secθ = 1 cosθ cotθ = 1 tanθ Quotient Identities tanθ = sinθ cosθ cotθ = cosθ sinθ We can use these identities to simplify trigonometric expressions. Example.4.. Simplify: (1) tanθcosθ sinθ Solution. (1) tanθcosθ sinθ () cosθ cotθ = cosθ cosθ sinθ = sinθ cosθ cosθ sinθ = 1 () cosθ cotθ = sinθ If P(x,y) is the terminal point on the unit circle corresponding to θ, then x +y = 1. Since sinθ = y and cosθ = x, we get sin θ+cos θ =

41 Teaching Notes By dividing both sides of this identity by cos θ and sin θ, respectively, we obtain The assumption in the division is that tan θ+1 = sec θ and 1+cot θ = csc θ. the divisor is nonzero. Example.4.4. Simplify: Pythagorean Identities tan θ+1 = sec θ sin θ+cos θ = 1 1+cot θ = csc θ (1) cos θ+cos θtan θ () 1+tan θ 1+cot θ Solution. (1) cos θ+cos θtan θ = (cos θ)(1+tan θ) = cos θsec θ = 1 () 1+tan θ 1+cot θ = sec θ csc θ = 1 cos θ 1 sin θ = sin θ cos θ = tan θ In addition to the eight identities presented above, we also have the following identities. Even-Odd Identities sin( θ) = sinθ tan( θ) = tanθ cos( θ) = cosθ The first two of the negative identities can be obtained from the graphs of the sine and cosine functions, respectively. (Please review the discussion on page Teaching Notes 149.) The third identity can be derived as follows: The corresponding reciprocal functions follow the same Even-Odd Identities: csc( θ) = cscθ sec( θ) = secθ cot( θ) = cotθ. tan( θ) = sin( θ) cos( θ) = sinθ cosθ = tanθ. The reciprocal, quotient, Pythagorean, and even-odd identities constitute what we call the fundamental trigonometric identities. We now solve Example.. in a different way. Example.4.5. If sinθ = 4 and cosθ > 0. Find cosθ. Solution. Using the identity sin θ+cos θ = 1 with cosθ > 0, we have ( cosθ = 1 sin θ = 1 7 = 4)

42 Example.4.6. If secθ = 5 and tanθ < 0, use the identities to find the values of the remaining trigonometric functions of θ. Solution. Note that θ lies in QIV. cosθ = 1 secθ = 5 sinθ = 1 cos θ = cscθ = 1 1 sinθ = ( ) 1 = 5 5 tanθ = sinθ cosθ = = cotθ = 1 1 tanθ = 1 Seatwork/Homework Use the identities presented in this lesson to simplify each trigonometric expression. (a) 1+tanx 1+cotx (b) (c) Solution. 1+tanx 1+cotx = 1+tanx 1+ 1 tanx sinθ 1+cosθ + 1+cosθ sinθ sinθ Solution. 1+cosθ + 1+cosθ = sinθ tany +coty secycscy Solution. tany +coty secycscy = = tanx siny + cosy cosy siny 1 1 cosy siny Answer: tan x Answer: cscθ sin θ sinθ(1+cosθ) + (1+cosθ)(1+cosθ) sinθ(1+cosθ) = sin θ+(1+cosθ+cos θ) sinθ(1+cosθ) = +cosθ sinθ(1+cosθ) = sinθ = cscθ = sin y+cos y cosysiny 1 cosysiny Answer: 1 = sin y +cos y = 1 186

43 (d) 1 cos θ 1+sinθ Solution. 1 cos θ 1+sinθ = 1+sinθ cos θ 1+sinθ = sin θ+sinθ 1+sinθ = sinθ(1+sinθ) 1+sinθ Answer: sin θ = sinθ. Given some initial values, use the identities to find the values of the remaining trigonometric functions of θ. (a) sinθ = and secθ > 0 5 Answer: θ in QI; cscθ = 5, cosθ = 1 sin θ = 1, secθ = 5 1, 5 1 tanθ = 1, cotθ = 1 1 (b) secθ = 8 and tanθ > 0 Answer: θ in QIII; cosθ =, sinθ = 1 cos 8 θ = 7, cscθ = 4 4 7, tanθ = 7, cotθ = (c) tanθ = and cscθ < 0 Answer: θ in QIII; cotθ = 1, secθ = 5, cosθ = 5, cscθ = 5 5, 4 sinθ = (d) cscθ = and secθ < 0 Answer: θ in QII; sinθ =, cosθ = 1 sin θ = 5, secθ = 5, 5 tanθ = 5, cotθ = Proving Trigonometric Identities We can use the eleven fundamental trigonometric identities to establish other identities. For example, suppose we want to establish the identity cscθ cotθ = sinθ 1+cosθ. To verify that it is an identity, recall that we need to establish the truth of the equation for all values of the variable in the domain of the equation. It is not enough to verify its truth for some selected values of the variable. To prove it, we use the fundamental trigonometric identities and valid algebraic manipulations like performing the fundamental operations, factoring, canceling, and multiplying the numerator and denominator by the same quantity. Start on the expression on one side of the proposed identity (preferably the complicated side), use and apply some of the fundamental trigonometric identities and algebraic manipulations, and arrive at the expression on the other side of the proposed identity. 187

44 Expression cscθ cotθ = 1 sinθ cosθ sinθ = 1 cosθ sinθ = 1 cosθ sinθ 1+cosθ 1+cosθ 1 cos θ = (sinθ)(1+cosθ) sin θ = (sinθ)(1+cosθ) = sinθ 1+cosθ Explanation Start on one side. Apply some reciprocal and quotient identities. Add the quotients. Multiply the numerator and denominator by 1+cosθ. Multiply. Apply a Pythagorean identity. Reduce to lowest terms. Upon arriving at the expression of the other side, the identity has been established. There is no unique technique to prove all identities, but familiarity with the different techniques may help. Example.4.7. Prove: secx cosx = sinxtanx. Solution. Example.4.8. Prove: secx cosx = 1 cosx cosx = 1 cos x cosx = sin x cosx = sinx sinx cosx = sinxtanx 1+sinθ 1 sinθ 1 sinθ 1+sinθ = 4sinθsec θ Solution. 1+sinθ 1 sinθ 1 sinθ 1+sinθ = (1+sinθ) (1 sinθ) (1 sinθ)(1+sinθ) = 1+sinθ+sin θ 1+sinθ sin θ 1 sin θ = 4sinθ cos θ = 4sinθsec θ 188

45 Seatwork/Homework.4.4 Prove each identity. 1. tanx+cotx = cscxsecx Answer: tanx+cotx = sinx cosx + cosx sinx = sin x+cos x sinxcosx 1. secθ+tanθ =. Answer: secy +tany cscy +1 Answer: secθ tanθ 1 secθ tanθ = 1 1 sinθ cosθ cosθ = cosθ 1 sinθ = tany secy +tany cscy +1 = = cosθ 1 sinθ 1+sinθ 1+sinθ = (cosθ)(1+sinθ) 1 sin θ = (cosθ)(1+sinθ) cos θ = 1+sinθ cosθ = 1 + siny cosy cosy 1 +1 = siny 1 sinxcosx = 1 cosx 1 sinx = cscxsecx = 1 cosθ + sinθ cosθ = secθ+tanθ 1+sin y cosy 1+sin y siny = siny cosy = tany 4. csc 1 θ = 1 cosθ cosθ 1 Answer: 1 cosθ cosθ = 1+cosθ+1 cosθ 1 cos θ = sin θ = csc θ Exercises.4 1. Find the domain of the equation. (a) x+ x = x Answer: {x x 0} (b) sin x = sinx+1 (c) tanx+cotx = sinx (d) x+1 +cosx = cscx x Answer: R Answer: R\ { kπ k Z} Answer: R\{{ 1,1} {kπ k Z}}

46 . Simplify each expression using the fundamental identities. (a) (b) sin θ sec θ 1 Solution. 1 1+tan x + 1 Solution. (c) 1 cos x 1+sinx (d) sin θ sec θ 1 = sin θ tan θ = sin θ sin θ cos θ = cos θ 1+cot x 1 1+tan x cot x = 1 sec x + 1 csc x = cos x+sin x = 1 Solution. 1 cos x 1+sinx = 1 1 sin x 1+sinx = 1 (1 sinx)(1+sinx) 1+sinx = 1 1 sinx = sinx sinθ cosθtanθ sinθ Solution. cosθtanθ = tanθ 1 tanθ = 1 Answer: cos θ Answer: 1 Answer: sinx Answer: 1. Given some initial information, use the identities to find the values of the trigonometric functions of θ. (a) cscθ = 5 and tanθ > 0 Answer: θ in QI; sinθ =, cosθ = 1 sin θ = 4, secθ = 5, tanθ =, cotθ = 4 (b) tanθ = 1 5 and cosθ < 0 Answer: θ in QII; cotθ = 5 1, secθ = tan θ+1 = 1 5, cosθ = 5 sinθ = 1 cos θ = 1 1, cscθ = 1 1 (c) cscθ = and π < x < π Answer: θ in QIII; sinθ =, cosθ = 1 sin θ = 5, secθ = 5, tanθ = 5, cotθ = (d) cotθ = 7 and π < θ < π 5 Answer: θ in QIV; tanθ = 5, secθ = tan θ+1 = 74, cosθ = 7 74, cscθ = cot θ+1 = 74, sinθ = , 190

47 (e) sinθ = 1 Answer: θ coterminal with π ; cscθ = 1, cosθ = 0, secθ undefined, tanθ undefined, cotθ = 0 (f) cotθ = 1 Answer: θ either in QII or QIV θ in QII: tanθ = 1, sinθ =, cscθ =, cosθ =, secθ = θ in QIV: tanθ = 1, sinθ =, cscθ =, cosθ =, secθ = 4. Determine whether the given equation is an identity or a conditional equation. If it is an identity, prove it; otherwise, provide a value of the variable in the domain that does not satisfy the equation. (a) sinxcosx = 1 Answer: conditional equation, x = 0 (b) sin x = cosx 1 Answer: conditional equation, x = π (c) (sinx cosx) +(sinx+cosx) = Answer: identity Proof. (sinx cosx) +(sinx+cosx) = (cos x cosxsinx+sin x)+(cos x+cosxsinx+sin x) = 1+1 = (d) tan( x)cotx = 1 Answer: conditional equation, x = π (e) sin x = secx+cosx Answer: conditional equation, x = π 5. Prove the following identities. (a) sin x = sinx sinxcos x Solution. sin x = sin x sinx = (1 cosx)sinx = sinx sinxcos x (b) sin 4 x cos 4 x = sin x cos x Solution. sin 4 x cos 4 x = (sin x cos x)(sin x+cos x) = sin x cos x (c) tan( θ) sin( θ) + cos( θ) = sec( θ) Solution. tan( θ)sin( θ)+cos( θ) = tanθsinθ+cosθ = sin θ cosθ +cosθ = sin θ+cos θ cosθ = 1 = secθ = sec( θ) cosθ 191

48 (d) 1+sinu+cosu 1+sinu cosu = 1+cosu sinu Solution 1+sinu+cosu 1+sinu cosu = 1+sinu+cosu sinu+1 cosu 1+cosu 1+cosu = (1+sinu+cosu)(1+cosu) sinu+sinucosu+1 cos u = (1+sinu+cosu)(1+cosu) sinu+sinucosu+sin u = (1+sinu+cosu)(1+cosu) (sinu)(1+cosu+sinu) = 1+cosu sinu Teaching Notes Since you need 1+cosu to retain in the numerator at the end, do not expand the numerator. 6. Express 1 sec x sec x Solution. 1 sec x sec x in terms of sinx. = 1 sec x 1 = cos x 1 = sin x Answer: sin x 7. Express tanxsecx in terms of cosx. Answer: ± 1 cos x cos x Solution. tanxsecx = sinx 1 = sinx = ± 1 cos x cosx cosx cos x cos x 8. Express all other five trigonometric functions in terms of tanx (allowing ± in the expression). tanx Answer: sinx = ± tan x+1 ; cosx = 1 ± tan x+1 ; cotx = 1 tanx ; secx = ± tan x+1; cscx = ± tan x+1 tanx 9. If secθ tanθ =, what is secθ+tanθ? Answer: 1 Solution. tan θ+1 = sec θ = sec θ tan θ = 1 (secθ tanθ)(secθ+tanθ) = 1 = (secθ+tanθ) = 1 = secθ+tanθ =

49 Lesson.5. Sum and Difference Identities Time Frame: one-hour sessions Learning Outcomes of the Lesson At the end of the lesson, the student is able to: (1) derive trigonometric identities involving sum and difference of two angles; () simplify trigonometric expressions using fundamental trigonometric identities and sum and difference identities; () prove other trigonometric identities using fundamental trigonometric identities and sum and difference identities; and (4) solve situational problems involving trigonometric identities. Lesson Outline (1) The sum and difference identities for cosine, sine, and tangent functions () Cofunction identities () More trigonometric identities Introduction In previous lesson, we introduced the concept of trigonometric identity, presented the fundamental identities, and proved some identities. In this lesson, we derive the sum and difference identities for cosine, sine, and tangent functions, establish the cofunction identities, and prove more trigonometric identities The Cosine Difference and Sum Identities Let u and v be any real numbers with 0 < v u < π. Consider the unit circle with points A = (1,0), P 1, P, P, and u and v with corresponding angles shown in Figure.7. Then P 1 P = AP. Recall that P 1 = P(u) = (cosu,sinu), P = P(v) = (cosv,sinv), and P = P(u v) = (cos(u v),sin(u v)), so that while P 1 P = (cosu cosv) +(sinu sinv), AP = [cos(u v) 1] +[sin(u v) 0]. Equating these two expressions and expanding the squares, we get (cosu cosv) +(sinu sinv) = [cos(u v) 1] +sin (u v) 19