Contents 20. Trigonometric Formulas, Identities, and Equations

Transcription

1 Contents 20. Trigonometric Formulas, Identities, and Equations Basic Identities Using Graphs to Help Verify Identities Example Trigonometric Equations Calculator Solutions to Trigonometric Equations Example Example

2 Peterson, Technical Mathematics, 3rd edition Basic Identities Using Graphs to Help Verify Identities A graphing calculator or graphing software can also be used to prove or disprove an identity. With this technique you separately graph each side of the proposed identity over an interval that contains at least one complete period. If both graphs are identical, then you can conclude that the identity is true whenever the functions are defined. A major difficulty arises because neither calculators nor computers define the cot, sec, and csc functions. Example 20.1 Use a graphing calculator to prove the identity sec 2 θ csc 2 θ = sec 2 θ + csc 2 θ. Solution We know that this identity is true because it is the same identity that we proved in Example 20.. The graph of the left-hand side of this identity is shown in Figure 20.2a and the graph of the right-hand side is shown in Figure 20.2b. You can see that these two graphs are identical, hence this is a valid identity. Caution FIGURE 20.2a y = sec 2 θ csc 2 θ [ 1, 10, 1] FIGURE 20.2b y = sec 2 θ + csc 2 θ [ 1, 10, 1] In practice, when you graph the two sides of an identity, they both appear on the same screen. If you graph the second function using the calculator s thick style, as in Figure 20.2c you can easily see if the new graph completely traces over the first function. FIGURE 20.2c

3 Peterson, Technical Mathematics, 3rd edition Trigonometric Equations Calculator Solutions to Trigonometric Equations Earlier we used calculators to help solve equations and inequalities. These techniques can also be applied to trigonometric equations. We will rework Example both graphically and algebraically using a TI-89. Example Use a calculator to solve sin θ tan θ = sin θ for 0 θ < 360. Solution There are two ways in which use graphs to solve this equation. Both methods begin by letting y 1 = sin θ tan θ and y 2 = sin θ. One method uses the intersect feature of the calculator and the other method uses the zero method discussed previously. FIGURE 20.8a FIGURE 20.8b FIGURE 20.8c Using intersect: As you might be able to tell from the title, the intersect method locates where the two curves intersect. Figure 20.8a seems to indicate that the curves intersect in four points in the given interval. (The graph of y 2 = sin θ is drawn using the thick style to make it easier to distinguish between the two graphs.) From the graph window in Figure 20.8a press F5 [Math] 5 [5:Intersection]. Move the cursor close to one of the points of intersection. We will select the second point from the left. Press ENTER twice. The tells the calculator which are the two curves that are intersecting. If there were more than two curves on the calculator screen you could use or to move the cursor to the correct graphs. Next indicate a lower and upper bound for the intersection point. So, move the cursor to the left of the intersection point and press ENTER, then move the cursor to the right of the intersection point and press ENTER. After a short pause you should see something like Figure 20.8b. This indicates that the two curves intersect near the point (5, ). Notice that this gives both the x- and y-values where the curves intersect. Repeat the procedure for each of the other three intersection points. You should get (0, 0), (180, 0), and (225, ). Using zero: We have used the zero technique before. Graph y 1 y 2 and determine where the curve crosses the x-axis. In Figure 20.8c we see that one solution is x = 5. The other solutions are x = 0, x = 180, and x = 225. Notice that the zero method does not give the y-values of the points of intersection. Using solve: We can use the calculator s solve command as shown in Figure 20.8d. The way the answer is displayed needs some explanation. According to the calculator, the solutions are x = 5 3) or x = This is the way that a TI-89 indicates that there are an infinite number of solutions. Let s first look at the second solution, x = This means x = 180n where n is any integer. If n = 1, then 180 ( 1) = 180 is an answer; if n = 0, then 180 (0) = 0 is an answer; if n = 1, then 180 (1) = 180 is an answer, and so on. The first solution, x = 5 3) means that all the solutions are of the form x = 5(n 3), where n is any integer. If n = 1, then 5 (( 1) 3 = 5 ( 7) = 315 is an answer; if n = 0, then 5 ((0) 3 = 5 ( 3) = 135 is an answer; if n = 1, then 5 ((1) 3 = 5 (1) = 5 is an answer, and so on.

4 Peterson, Technical Mathematics, 3rd edition But, we did not want all the solutions. We just wanted those where 0 θ < 360. We can restrict where the calculator looks for solutions. At the end of the solve statement type the with operator be pressing and then add the restrictions that need to be placed on x. The compound inequality 0 x < 360 has to be entered as two simple inequalities: 0 x and x < 360. To do this, use the following key sequence: 0 0 [<] X 2nd alpha ( ) [ ] A N D [ ] alpha X 2nd 0 [<] 360 Pressing 0 [<] creates the symbol abd pressing 2nd alpha puts the calculator in alpha-lock mode so you can type a series of letters without pressing the alpha before each letter. FIGURE 20.8d FIGURE 20.8e The result in Figure 20.8e indicates that the solutions are x = 225, x = 180, x = 5, and x = 0. Example Use a calculator to solve 2 sin 2 x cos x 1 = 0 for 0 x < 2. Solution Since the right-hand side of this equation is zero both the intersection method and the root method will produce the same result. From Figure 20.9a we see that one solution is x (The actual value is 3.) It looks as if the third solution (counting from the left) is approximately We can verify that quickly by pressing 2 X,T,θ,n ENTER. Since the calculator automatically stores the root as x you should get the result x as in Figure 20.9b. The second solution, the one in the middle, appears to be. To verify this press ENTER.

5 Peterson, Technical Mathematics, 3rd edition 5 FIGURE 20.9a [ 3, 3] FIGURE 20.9b [ 3, 3] Thus, the solutions of 2 sin 2 x cos x 1 = 0 for 0 x < 2 are x 1.072, x =, and x Using the solve feature we get the same results. (See Figure 20.9c.) FIGURE 20.9c