1. 1 Copyright © Cengage Learning. All rights reserved. 7. Analytic Trigonometry 7.1 -7.2 Trigonometric Identities and Trigonometric Equations

2. 2 Basic Trigonometric Identities sin(- x ) = - sin x cos(- x ) = cosx tan( -x) = - tanx sec( -x ) = sec x csc( -x ) = - cscx cot( -x) = - cotx

3. 3 Verifying Trigonometric Identities A trigonometric expression contains symbols involving trigonometric functions. Illustration: Trigonometric Expressions x + sin x

4. 4 Example 1 – Verifying an identity Verify the identity sec  – cos  = sin  tan . Solution: We transform the left-hand side into the right-hand side: reciprocal identity add expressions sin 2  + cos 2  = 1

5. 5 Example 1 – Solution = sin  tan  equivalent expression tangent identity cont’d

6. 6 Example Verify the identity: Solution:

7. 7 Example cont. Solution:

8. 8 Guidelines for Verifying Trigonometric Identities 1.Work with each side of the equation independently of the other side. Start with the more complicated side and transform it in a step  by  step fashion until it looks exactly like the other side. 2.Analyze the identity and look for opportunities to apply the fundamental identities. Rewriting the more complicated side of the equation in terms of sines and cosines is often helpful. 3.If sums or differences of fractions appear on one side, use the least common denominator and combine the fractions. 4.Don't be afraid to stop and start over again if you are not getting anywhere. Creative puzzle solvers know that strategies leading to dead ends often provide good problem  solving ideas.

9. 9 Verifying Trigonometric Identities Another technique for showing that an equation p = q is an identity is to begin by transforming the left-hand side p into another expression s, making sure that each step is reversible—that is, making sure it is possible to transform s back into p by reversing the procedure used in each step. In this case, the equation p = s is an identity. Next, as a separate exercise, we show that the right-hand side q can also be transformed into the expression s by means of reversible steps and, therefore, that q = s is an identity. It then follows that p = q is an identity. This method is illustrated in the next example.

10. 10 Example 4 – Verifying an identity Verify the identity (tan  – sec  ) 2 = Solution: We shall verify the identity by showing that each side of the equation can be transformed into the same expression. First we work only with the left-hand side: (tan  – sec  ) 2 = tan 2  – 2 tan  sec  + sec 2  square expression tangent and reciprocal identities

11. 11 Example 4 – Solution At this point it may not be obvious how we can obtain the right-hand side of the given equation from the last expression. Thus, we next work with only the right-hand side and try to obtain the last expression. equivalent expression add fractions cont’d

12. 12 Example 4 – Solution Multiplying numerator and denominator by the conjugate of the denominator gives us the following: multiply numerator and denominator by 1 – sin  property of quotients sin 2  + cos 2  = 1 cont’d

13. 13 Example 4 – Solution The last expression is the same as that obtained from (tan  – sec  ) 2. Since all steps are reversible, the given equation is an identity. cont’d

14. 14 Verifying Trigonometric Identities In calculus it is sometimes convenient to change the form of certain algebraic expressions by making a trigonometric substitution, as illustrated in the next example.

15. 15 Example 6 – Making a trigonometric substitution Express in terms of a trigonometric function of, without radicals, by making the substitution x = a sin  for –  /2     /2 and a > 0. Solution: We proceed as follows: let x = a sin  law of exponents factor out a 2

16. 16 Example 6 – Solution = |a cos  | = |a||cos  | = a cos  The last equality is true because (1) if a > 0, then |a| = a, and (2) if –  /2     /2, then cos   0 and hence |cos  | = cos . see below |cd| = |c||d| c 2 d 2 = (cd) 2 cont’d sin 2  + cos 2  = 1

17. 17 Example 6 – Solution We may also use a geometric solution. If x = a sin , then sin  = x/a, and the triangle in Figure 10 illustrates the problem for 0 <  <  /2. The third side of the triangle, can be found by using the Pythagorean theorem. Figure 10 cont’d

18. 18 Example 6 – Solution From the figure we can see that or, equivalently, cont’d

19. 19 Copyright © Cengage Learning. All rights reserved. 7.2 Trigonometric Equations

20. 20 Trigonometric Equations A trigonometric equation is an equation that contains trigonometric expressions. If a trigonometric equation is not an identity, we often find solutions by using techniques similar to those used for algebraic equations. The main difference is that we first solve the trigonometric equation for sin x, cos  and so on, and then find values of x or  that satisfy the equation.

21. 21 Trigonometric Equations Solutions may be expressed either as real numbers or as angles. Throughout our work we shall use the following rule: If degree measure is not specified, then solutions of a trigonometric equation should be expressed in radian measure (or as real numbers). If solutions in degree measure are desired, an appropriate statement will be included in the example or exercise.

22. 22 Example 1 – Solving a trigonometric equation involving the sine function Find the solutions of the equation sin  = if (a)  is in the interval [0, 2  ) (b)  is any real number Solution: (a) If sin  =, then the reference angle for  is =  /6.

23. 23 Example 1 – Solution If we regard  as an angle in standard position, then, since sin  > 0, the terminal side is in either quadrant I or quadrant II, as illustrated in Figure 1. Thus, there are two solutions for 0    2  : and cont’d Figure 1

24. 24 Example 1 – Solution (b) Since the sine function has period 2 , we may obtain all solutions by adding multiples of 2  to  /6 and 5  /6. This gives us and for every integer n. cont’d

25. 25 Example 1 – Solution An alternative (graphical) solution involves determining where the graph of y = sin  intersects the horizontal line y =, as illustrated in Figure 2. cont’d Figure 2

26. 26 Example 2 – Solving a trigonometric equation involving the tangent function Find the solutions of the equation tan u = –1. Solution: Since the tangent function has period , it is sufficient to find one real number u such that tan u = –1 and then add multiples of .

27. 27 Example 2 – Solution A portion of the graph of y = tan u is sketched in Figure 3. cont’d Figure 3 y = tan u

28. 28 Example 2 – Solution Since tan (3  /4) = –1, one solution is 3  /4; hence, If tan u = –1, then for every integer n. We could also have chosen –  /4 (or some other number u such that tan u = –1) for the initial solution and written for every integer n. cont’d

29. 29 Example 2 – Solution An alternative solution involves a unit circle. Using tan 3  /4 = –1 and the fact that the period of the tangent is , we can see from Figure 4 that the desired solutions are for every integer n. cont’d Figure 4

30. 30 Example 3 – Solving a trigonometric equation involving multiple angles (a) Solve the equation cos 2x = 0, and express the solutions both in radians and in degrees. (b) Find the solutions that are in the interval [0, 2  ) and, equivalently, [0 , 360  ).

31. 31 Example 3 – Solution (a) We proceed as follows, where n denotes any integer: cos 2x = 0 cos  = 0 cont’d given let  = 2x refer to Figure 5 Figure 5

32. 32 Example 3 – Solution In degrees, we have x = 45  + 90  n. cont’d divide by 2  = 2x

33. 33 Example 3 – Solution (b) We may find particular solutions of the equation by substituting integers for n in either of the formulas for x obtained in part (a). Several such solutions are listed in the following table. cont’d

34. 34 Example 3 – Solution Note that the solutions in the interval [0, 2  ) or, equivalently, [0 , 360  ) are given by n = 0, n = 1, n = 2 and n = 3. These solutions are or, equivalently, 45 , 135 , 225 , 315 . cont’d

35. 35 Example 6 – Solving a trigonometric equation by factoring Find the solutions of 4 sin 2 x tan x – tan x = 0 that are in the interval [0, 2  ). Solution: 4 sin 2 x tan x – tan x = 0 tan x (4 sin 2 x – 1) = 0 tan x = 0, 4 sin 2 x – 1 = 0 tan x = 0, sin 2 x = given factor out tan x zero factor theorem solve for tan x, sin 2 x

36. 36 Example 6 – Solution tan x = 0, sin x =  The reference angle  /6 for the third and fourth quadrants is shown in Figure 8. cont’d solve for sin x Figure 8

37. 37 Example 6 – Solution These angles, 7  /6 and 11  /6, are the solutions of the equation sin x = – for 0  x < 2 . The solutions of all three equations are listed in the following table. Thus, the given equation has the six solutions listed in the second column of the table. cont’d

38. 38 Example 8 – Approximating the solutions of a trigonometric equation Approximate, to the nearest degree, the solutions of the following equation in the interval [0 , 360  ): 5 sin  tan  – 10 tan  + 3 sin  – 6 = 0 Solution: 5 sin  tan  – 10 tan  + 3 sin  – 6 = 0 (5 sin  tan  – 10 tan  ) + (3 sin  – 6) = 0 5 tan  (sin  – 2) + 3(sin  – 2) = 0 given group terms factor each group

39. 39 Example 8 – Solution (5 tan  + 3)(sin  – 2) = 0 5 tan  + 3 = 0, sin  – 2 = 0 tan  = sin  = 2 The equation sin  = 2 has no solution, since –1  sin   1 for every . For tan  = we use a calculator in degree mode, obtaining cont’d factor out (sin  – 2) zero factor theorem solve for tan  and sin 

40. 40 Example 8 – Solution Hence, the reference angle is  R  31 . Since  is in either quadrant II or quadrant IV, we obtain the following solutions:  = 180  –  R  180  – 31  = 149   = 360  –  R  360  – 31  = 329  cont’d

41. 41 Example 9 – Investigating the number of hours of daylight In Boston, the number of hours of daylight D(t) at a particular time of the year may be approximated by with t in days and t = 0 corresponding to January 1. How many days of the year have more than 10.5 hours of daylight?

42. 42 Example 9 – Solution The graph of D is shown in Figure 9. As illustrated in the figure, if we can find two numbers a and b with D(a) = 10.5, D(b) = 10.5, and 0 < a < b < 365, then there will be more than 10.5 hours of daylight in the t th day of the year if a < t < b. cont’d Figure 9

43. 43 Example 9 – Solution Let us solve the equation D(t) = 10.5 as follows: cont’d let D(t) = 10.5 subtract 12 divide by 3

44. 44 Example 9 – Solution If sin  = –, then the reference angle is  /6 and the angle  is in either quadrant III or quadrant IV. Thus, we can find the numbers a and b by solving the equations and cont’d

45. 45 Example 9 – Solution From the first of these equations we obtain and hence t  213 + 79, or t  292. cont’d

46. 46 Example 9 – Solution Similarly, the second equation gives us t  414. Since the period of the function D is 365 days (see Figure 9), we obtain t  414 – 365, or t = 49. Thus, there will be at least 10.5 hours of daylight from t = 49 to t = 292—that is, for 243 days of the year. cont’d Figure 9

47. 47 Trigonometric Equations The next example illustrates how a graphing utility can aid in solving a complicated trigonometric equation.

48. 48 Example 11 – Using a graph to determine solutions of a trigonometric equation Find the solutions of the following equation that are in the interval [0, 2  ): sin x + sin 2x + sin 3x = 0 Solution: We assign sin x + sin 2x + sin 3x to Y 1.

49. 49 Example 11 – Solution Since | sin  |  1 for  = x, 2x, and 3x, the left-hand side of the equation is between –3 and 3, and we choose the viewing rectangle [0, 2 ,  /4] by [–3, 3] and obtain a sketch similar to Figure 10. cont’d Figure 10 [0, 2 ,  /4] by [ –3, 3]

50. 50 Example 11 – Solution Using a root feature, we obtain the following approximations for the x-intercepts—that is, the approximate solutions of the given equation in [0, 2  ): 0, 1.57, 2.09, 3.14, 4.19, 4.71 Changing to degree measure and rounding off to the nearest degree, we obtain 0°, 90°, 120°, 180°, 240°, and 270°. cont’d

51. 51 Example 11 – Solution Converting these degree measures to radian measures gives us Checking these values in the given equation, we see that all six are solutions. Figure 10 suggests that the graph has period 2 . cont’d Figure 10 [0, 2 ,  /4] by [ –3, 3]

52. 52 Example 11 – Solution To change the form of Y 1 and prove that the period is 2  and, therefore, that all solutions of the given equation can be obtained by adding integer multiples of 2 . cont’d