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# PC Lesson 10 - Product-to-Sum and vv.pptx

Precalculus Lesson on Trigo Identities

Precalculus Lesson on Trigo Identities

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### PC Lesson 10 - Product-to-Sum and vv.pptx

1. 1. 7.1 Trigonometric Identities
2. 2. 2
3. 3. 3 Product-Sum Formulas It is possible to write the product sin u cos as a sum of trigonometric functions. To see this, consider the Addition and Subtraction Formulas for Sine: sin(u +) = sin u cos + cos u sin sin(u –) = sin u cos – cos u sin Adding the left- and right-hand sides of these formulas gives sin(u +) = sin(u –) = 2 sin u cos
4. 4. 4 Product-Sum Formulas Dividing by 2 gives the formula sin u cos = [sin(u +) + sin(u –)] The other three Product-to-Sum Formulas follow from the Addition Formulas in a similar way.
5. 5. 5 Product-Sum Formulas The Product-to-Sum Formulas can also be used as Sum-to-Product Formulas. This is possible because the right-hand side of each Product-to-Sum Formula is a sum and the left side is a product. For example, if we let in the first Product-to-Sum Formula, we get so
6. 6. 6 Product-Sum Formulas The remaining three of the following Sum-to-Product Formulas are obtained in a similar manner.
7. 7. 7 Example 5 – Proving an Identity Verify the identity . Solution: We apply the second Sum-to-Product Formula to the numerator and the third formula to the denominator. Sum-to-Product Formulas
8. 8. 8 Example 5 – Solution Simplify Cancel cont’d
9. 9. 9
10. 10. 10 7.4 Basic Trigonometric Equations
11. 11. 11 Basic Trigonometric Equations An equation that contains trigonometric functions is called a trigonometric equation. For example, the following are trigonometric equations: sin2  + cos2  = 1 2 sin – 1 = 0 tan 2 – 1 = 0 The first equation is an identity—that is, it is true for every value of the variable . The other two equations are true only for certain values of . To solve a trigonometric equation, we find all the values of the variable that make the equation true.
12. 12. 12 Basic Trigonometric Equations Solving any trigonometric equation always reduces to solving a basic trigonometric equation—an equation of the form T( ) = c, where T is a trigonometric function and c is a constant. In the next examples we solve such basic equations.
13. 13. 13 Example 1 – Solving a Basic Trigonometric Equation Solve the equation Solution: Find the solutions in one period. Because sine has period 2, we first find the solutions in any interval of length 2. To find these solutions, we look at the unit circle in Figure 1. Figure 1
14. 14. 14 Example 1 – Solution We see that sin  = in Quadrants I and II, so the solutions in the interval [0, 2) are Find all solutions. Because the sine function repeats its values every 2 units, we get all solutions of the equation by adding integer multiples of 2 to these solutions: where k is any integer. cont’d
15. 15. 15 Example 1 – Solution Figure 2 gives a graphical representation of the solutions. Figure 2 cont’d
16. 16. 16 Example 2 – Solving a Basic Trigonometric Equation Solve the equation tan  = 2. Solution: Find the solutions in one period. We first find one solution by taking tan–1 of each side of the equation. tan  = 2  = tan–1(2)   1.12 Given equation Take tan–1 of each side Calculator (in radian mode)
17. 17. 17 Example 2 – Solution By the definition of tan–1 the solution that we obtained is the only solution in the interval (–/2, /2) (which is an interval of length ). Find all solutions. Since tangent has period , we get all solutions of the equation by adding integer multiples of  :   1.12 + k where k is any integer. cont’d
18. 18. 18 Example 2 – Solution A graphical representation of the solutions is shown in Figure 6. You can check that the solutions shown in the graph correspond to k = –1, 0, 1, 2, 3. Figure 6 cont’d
19. 19. 19 Basic Trigonometric Equations In the next example we solve trigonometric equations that are algebraically equivalent to basic trigonometric equations.
20. 20. 20 Example 3 – Solving Trigonometric Equations Find all solutions of the equation. (a) 2 sin  – 1 = 0 (b) tan2  – 3 = 0 Solution: (a) We start by isolating sin  . 2 sin  – 1 = 0 2 sin  = 1 sin  = Given equation Add 1 Divide by 2
21. 21. 21 Example 3 – Solution This last equation is the same as that in Example 1. The solutions are  = + 2k  = + 2k where k is any integer. (b) We start by isolating tan . tan2  – 3 = 0 tan2  = 3 tan  = Given equation Add 3 Take the square root cont’d
22. 22. 22 Example 3 – Solution Because tangent has period , we first find the solutions in any interval of length . In the interval (–/2, /2) the solutions are  = /3 and  = –/3. To get all solutions, we add integer multiples of  to these solutions:  = + k  = – + k where k is any integer. cont’d
23. 23. 23 Solving Trigonometric Equations by Factoring
24. 24. 24 Solving Trigonometric Equations by Factoring Factoring is one of the most useful techniques for solving equations, including trigonometric equations. The idea is to move all terms to one side of the equation, factor, and then use the Zero-Product Property.
25. 25. 25 Example 4 – A Trigonometric Equation of Quadratic Type Solve the equation 2 cos2  – 7 cos  + 3 = 0. Solution: We factor the left-hand side of the equation. 2 cos2  – 7 cos  + 3 = 0 (2 cos  – 1)(cos  – 3) = 0 2 cos  – 1 = 0 or cos  – 3 = 0 cos  = or cos  = 3 Factor Given equation Solve for cos  Set each factor equal to 0
26. 26. 26 Example 4 – Solution Because cosine has period 2, we first find the solutions in the interval [0, 2). For the first equation the solutions are  = /3 and  = 5/3 (see Figure 7). Figure 7 cont’d
27. 27. 27 Example 4 – Solution The second equation has no solution because cos  is never greater than 1. Thus the solutions are  = + 2k  = + 2k where k is any integer. cont’d
28. 28. 28 Example 5 – Solving a Trigonometric Equation by Factoring Solve the equation 5 sin  cos  + 4 cos  = 0. Solution: We factor the left-hand side of the equation. 5 sin  cos  + 2 cos  = 0 cos  (5 sin  + 2) = 0 cos  = 0 or 5 sin  + 4 = 0 sin  = –0.8 Given equation Factor Set each factor equal to 0 Solve for sin 
29. 29. 29 Example 5 – Solution Because sine and cosine have period 2, we first find the solutions of these equations in an interval of length 2. For the first equation the solutions in the interval [0, 2) are  = /2 and  = 3/2 . To solve the second equation, we take sin–1 of each side. sin  = –0.80  = sin–1(–0.80) Second equation Take sin–1 of each side cont’d
30. 30. 30 Example 5 – Solution   –0.93 So the solutions in an interval of length 2 are  = –0.93 and  =  + 0.93  4.07 (see Figure 8). Figure 8 Calculator (in radian mode) cont’d
31. 31. 31 Example 5 – Solution We get all the solutions of the equation by adding integer multiples of 2 to these solutions.  = + 2k  = + 2k   –0.93 + 2k   4.07 + 2k where k is any integer. cont’d
32. 32. 32 7.5 More Trigonometric Equations
33. 33. 33 More Trigonometric Equations In this section we solve trigonometric equations by first using identities to simplify the equation. We also solve trigonometric equations in which the terms contain multiples of angles.
34. 34. 34 Solving Trigonometric Equations by Using Identities
35. 35. 35 Solving Trigonometric Equations by Using Identities In the next example we use trigonometric identities to express a trigonometric equation in a form in which it can be factored.
36. 36. 36 Example 1 – Using a Trigonometric Identity Solve the equation 1 + sin = 2 cos2. Solution: We first need to rewrite this equation so that it contains only one trigonometric function. To do this, we use a trigonometric identity: 1 + sin = 2 cos2 1 + sin = 2(1 – sin2) 2 sin2 + sin – 1 = 0 (2 sin – 1)(sin + 1) = 0 Given equation Pythagorean identity Put all terms on one side Factor
37. 37. 37 Example 1 – Solution 2 sin – 1 = 0 or sin + 1 = 0 sin = or sin = –1  = or  = Because sine has period 2, we get all the solutions of the equation by adding integer multiples of 2 to these solutions. cont’d Set each factor equal to 0 Solve for sin Solve for  in the interval [0, 2)
38. 38. 38 Example 1 – Solution Thus the solutions are  = + 2k  = + 2k  = + 2k where k is any integer. cont’d
39. 39. 39 Example 2 – Squaring and Using an Identity Solve the equation cos + 1 = sin in the interval [0, 2). Solution: To get an equation that involves either sine only or cosine only, we square both sides and use a Pythagorean identity. cos + 1 = sin cos2 + 2 cos + 1 = sin2 cos2 + 2 cos + 1 = 1 – cos2 2 cos2 + 2 cos = 0 Given equation Pythagorean identity Square both sides Simplify
40. 40. 40 Example 2 – Solution 2 cos  (cos  + 1) = 0 2 cos = 0 or cos + 1 = 0 cos = 0 or cos = –1  = or  =  Because we squared both sides, we need to check for extraneous solutions. From Check Your Answers we see that the solutions of the given equation are /2 and . Set each factor equal to 0 Solve for cos Solve for  in [0, 2) cont’d Factor
41. 41. 41 Example 2 – Solution Check Your Answers:  =  =  =  cos + 1 = sin cos + 1 = sin cos  + 1 = sin  0 + 1 = 1 0 + 1 ≟ –1 –1 + 1 = 0 cont’d
42. 42. 42 Example 3 – Finding Intersection Points Find the values of x for which the graphs of f(x) = sin x and g(x) = cos x intersect. Solution 1: Graphical The graphs intersect where f(x) = g(x). In Figure 1 we graph y1 = sin x and y2 = cos x on the same screen, for x between 0 and 2. (a) (b) Figure 1
43. 43. 43 Example 3 – Solution Using or the intersect command on the graphing calculator, we see that the two points of intersection in this interval occur where x  0.785 and x  3.927. Since sine and cosine are periodic with period 2, the intersection points occur where x  0.785 + 2k and x  3.927 + 2k where k is any integer. cont’d
44. 44. 44 Example 3 – Solution Solution 2: Algebraic To find the exact solution, we set f(x) = g(x) and solve the resulting equation algebraically: sin x = cos x Since the numbers x for which cos x = 0 are not solutions of the equation, we can divide both sides by cos x: = 1 tan x = 1 Equate functions Divide by cos x Reciprocal identity cont’d
45. 45. 45 Example 3 – Solution The only solution of this equation in the interval (–/2, /2) is x = /4. Since tangent has period , we get all solutions of the equation by adding integer multiples of : x = + k where k is any integer. The graphs intersect for these values of x. You should use your calculator to check that, rounded to three decimals, these are the same values that we obtained in Solution 1. cont’d
46. 46. 46 Equations with Trigonometric Functions of Multiples of Angles When solving trigonometric equations that involve functions of multiples of angles, we first solve for the multiple of the angle, then divide to solve for the angle.
47. 47. 47 Example 4 – A Trigonometric Equation Involving a Multiple of an Angle Consider the equation 2 sin 3 – 1 = 0. (a) Find all solutions of the equation. (b) Find the solutions in the interval [0, 2). Solution: (a) We first isolate sin 3 and then solve for the angle 3. 2 sin 3 – 1 = 0 2 sin 3 = 1 sin 3 = Given equation Add 1 Divide by 2
48. 48. 48 Example 4 – Solution 3 = Solve for 3 in the interval [0, 2) (see Figure 2) Figure 2 cont’d
49. 49. 49 Example 4 – Solution To get all solutions, we add integer multiples of 2 to these solutions. So the solutions are of the form 3 = + 2k 3 = + 2k To solve for , we divide by 3 to get the solutions where k is any integer. cont’d
50. 50. 50 Example 4 – Solution (b) The solutions from part (a) that are in the interval [0, 2) correspond to k = 0, 1, and 2. For all other values of k the corresponding values of  lie outside this interval. So the solutions in the interval [0, 2) are cont’d
51. 51. 51 Example 5 – A Trigonometric Equation Involving a Half Angle Consider the equation (a) Find all solutions of the equation. (b) Find the solutions in the interval [0, 4). Solution: (a) We start by isolating tan . Given equation Add 1
52. 52. 52 Example 5 – Solution Since tangent has period , to get all solutions, we add integer multiples of  to this solution. So the solutions are of the form Divide by Solve for in the interval cont’d
53. 53. 53 Example 5 – Solution Multiplying by 2, we get the solutions  = + 2k where k is any integer. (b) The solutions from part (a) that are in the interval [0, 4) correspond to k = 0 and k = 1. For all other values of k the corresponding values of x lie outside this interval. Thus the solutions in the interval [0, 4) are cont’d