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Splash Screen. Five-Minute Check (over Lesson 5-2) Then/Now Example 1: Solve by Isolating Trigonometric Expressions Example 2: Solve by Taking the Square Root of Each Side Example 3: Solve by Factoring Example 4: Real-World Example: Trigonometric Functions of Multiple Angles

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  1. Splash Screen

  2. Five-Minute Check (over Lesson 5-2) Then/Now Example 1: Solve by Isolating Trigonometric Expressions Example 2: Solve by Taking the Square Root of Each Side Example 3: Solve by Factoring Example 4: Real-World Example: Trigonometric Functions of Multiple Angles Example 5: Solve by Rewriting Using a Single Trigonometric Function Example 6: Solve by Squaring Lesson Menu

  3. A. B. C. D. Verify the identity sin x tan x = sec x – cos x. 5–Minute Check 1

  4. A. B. C. D. Verify the identity 1 = tan x cos x csc x. 5–Minute Check 2

  5. A. B. C. D. Verify the identity sec x – sec x sin2x = cos x. 5–Minute Check 3

  6. Which of the following is equivalent to? A.cos θ B.sec θ C.sin θ D.csc θ 5–Minute Check 4

  7. You verified trigonometric identities. (Lesson 5-2) • Solve trigonometric equations using algebraic techniques. • Solve trigonometric equations using basic identities. Then/Now

  8. Solve . Original equation Subtract 3cos x from each side to isolate the trigonometric expression. Solve for cos x. Solve by Isolating Trigonometric Expressions Example 1

  9. The period of cosine is 2π, so you only need to find solutions on the interval [0, 2π). The solutions on this interval are . The solutions on the interval (–∞, ∞) are then found by adding integer multiples 2π. Therefore, the general form of the solutions is x = , where n is an integer. Answer: Solve by Isolating Trigonometric Expressions Example 1

  10. Solve sin x + = – sinx. A. B. C. D. Example 1

  11. Divide each side by 3. Take the square root of each side. Rationalize the denominator. Solve by Taking the Square Root of Each Side Solve 3 tan2x – 4 = –3. 3 tan2x – 4 = –3 Original equation 3 tan2x = 1 Add 4 to each side. Example 2

  12. The period of tangent is π. On the interval [0, π), tan x = when x = and tan x = when x = . The solutions on the interval (–∞, ∞) have the general form , where n is an integer. Answer: Solve by Taking the Square Root of Each Side Example 2

  13. A. B. C. D. Solve 5 tan2x – 15 = 0. Example 2

  14. A. Find all solutions of on the interval [0, 2π). Original equation Isolate the trigonometric terms. Factor. Zero Product Property Solve by Factoring Example 3

  15. Solve for x on [0, 2π). On the interval [0, 2π), the equation has solutions . Answer: Solve for cos x. Solve by Factoring Example 3

  16. Original equation Factor. Zero Product Property Solve for sin x. Solve for x on [0, 2π). Solve by Factoring B. Find all solutions of 2sin2x + sinx – 1 = 0 on the interval [0, 2π). Example 3

  17. On the interval [0, 2π), the equation 2sin2x + sinx – 1 = 0 has solutions . Answer: Solve by Factoring Example 3

  18. A. B. C. D. Find all solutions of 2 tan4x – tan2x – 15 = 0 on the interval [0, π). Example 3

  19. PROJECTILES A projectile is sent off with an initial speed vo of 350 m/s and clears a fence 3000 m away. The height of the fence is the same height as the initial height of the projectile. If the distance the projectile traveled is given by , find the interval of possible launch angles to clear the fence. Trigonometric Functions of Multiple Angles Example 4

  20. Original formula d = 3000 and v0 = 350 Simplify. Multiply each side by 9.8. Divide each side by 122,500. Definition of inverse sine. Trigonometric Functions of Multiple Angles Example 4

  21. Trigonometric Functions of Multiple Angles Recall from Lesson 4-6 that the range of the inverse sine function is restricted to acute angles of θ in the interval [–90°, 90°]. Since we are finding the inverse sine of 2θ instead of θ, we need toconsider angles in the interval [–2(90°), 2(90°)] or [–180°, 180°]. Use your calculator to find theacute angle and the reference angle relationship sin (180° − θ) = sin θ to find the obtuse angle. sin–10.24 = 2Definition of inverse sine 13.9° or 166.1° = 2 sin–1(0.24) ≈13.9° and sin(180° – 13.9°) = 166.1° 7.0° or 83.1° =  Divide by 2. Example 4

  22. Original formula   = 7.0° or  = 83.1°  Use a calculator. Trigonometric Functions of Multiple Angles The interval is [7.0°, 83.1°]. The ball will clear the fence if the angle is between 7.0° and 83.1°. Answer:7.0° ≤  ≤ 83.1° CHECK Substitute the angle measures into the original equation to confirm the solution. Example 4

  23. GOLF A golf ball is sent off with an initial speed vo of 36 m/s and clears a small barricade 70 m away. The height of the barricade is the same height as the initial height of the ball. If the distance the ball traveled is given by , find the interval of possible launch angles to clear the barricade. A. 1.6° ≤ ≤ 88.5° B. 3.1° ≤ ≤ 176.9° C. 16.0° ≤ ≤ 74.0° D. 32° ≤ ≤ 148.0° Example 4

  24. Solve by Rewriting Using a Single Trigonometric Function Find all solutions of sin2x – sin x + 1 = cos2x on the interval [0, 2π). sin2x – sin x + 1 = cos2x Original equation –cos2x + sin2x – sin x + 1 = 0 Subtract cos2 x from each side. –(1 – sin2x) + sin2x – sin x + 1 = 0 Pythagorean Identity 2sin2x – sin x = 0 Simplify. sinx (2sin x – 1) = 0 Factor. Example 5

  25. Solve for x on [0, 2π). x = 0, π Solve by Rewriting Using a Single Trigonometric Function sin x = 0 2sin x – 1 = 0 Zero Product Property 2sin x = 1 Solve for sin x. Example 5

  26. CHECKThe graphs of Y1 = sin2x – sin x + 1 and Y2 = cos2x intersect at on the interval [0, 2π) as shown.  Answer: Solve by Rewriting Using a Single Trigonometric Function Example 5

  27. A. B. C. D. Find all solutions of 2sin2x = cosx + 1 on the interval [0, 2). Example 5

  28. Solve by Squaring Find all solutions of sin x – cos x = 1 on the interval [0, 2π). sin x – cos x = 1 Original equation sin x = cos x + 1 Add cosx to each side. sin2x = cos2x + 2cos x + 1 Square each side. 1 – cos2x = cos2x + 2cos x + 1 Pythagorean Identity 0 = 2cos2x + 2cos x Subtract 1 – cos2x from each side. 0 = cos2x + cos x Divide each side by 2. 0 = cos x(cos x + 1) Factor. Example 6

  29. , x = π Solve for x on [0, 2). Original formula sin π – cos π = 1 Substitute Simplify.   Solve by Squaring cos x = 0 cos x + 1 = 0 Zero Product Property cos x = –1 Solve for cos x. Example 6

  30. Therefore, the only valid solutions are on the interval . Answer: Solve by Squaring Example 6

  31. A. B. C. D. Find all solutions of 1 + cos x = sin x on the interval [0, 2π). Example 6

  32. End of the Lesson

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