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Double-Angle and Half-Angle Identities in Trigonometry

Learn how to use double-angle and half-angle identities to simplify trigonometric expressions and solve real-world problems. Practice finding exact values of sine, cosine, and tangent using these identities. Review concepts and examples in this lesson.

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Double-Angle and Half-Angle Identities in Trigonometry

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  1. Splash Screen

  2. Five-Minute Check (over Lesson 13–3) CCSS Then/Now Key Concept: Double-Angle Identities Example 1: Double-Angle Identities Example 2: Double-Angle Identities Key Concept: Half-Angle Identities Example 3: Half-Angle Identities Example 4: Real-World Example: Simplify Using Double-Angle Identities Example 5: Verify Identities Lesson Menu

  3. A. B. C. D. Find the exact value of cos 15°. 5-Minute Check 1

  4. A. B. C. D. Find the exact value of cos 15°. 5-Minute Check 1

  5. A. B. C. D. Find the exact value of sin 105°. 5-Minute Check 2

  6. A. B. C. D. Find the exact value of sin 105°. 5-Minute Check 2

  7. What is the missing step in the identity? ? C. D. A. B. 5-Minute Check 3

  8. What is the missing step in the identity? ? C. D. A. B. 5-Minute Check 3

  9. A. B. C. D. Find the exact value of cos . 5-Minute Check 4

  10. A. B. C. D. Find the exact value of cos . 5-Minute Check 4

  11. Which of the following is equivalent to sin 330°? A. sin 360° cos 30° – cos 360° sin 30° B. cos 360° sin 30° – sin 360° cos 30° C. sin 360° cos 30° + cos 360° sin 30° D. cos 360° sin 30° + sin 360° cos 30° 5-Minute Check 5

  12. Which of the following is equivalent to sin 330°? A. sin 360° cos 30° – cos 360° sin 30° B. cos 360° sin 30° – sin 360° cos 30° C. sin 360° cos 30° + cos 360° sin 30° D. cos 360° sin 30° + sin 360° cos 30° 5-Minute Check 5

  13. Content Standards F.TF.8 Prove the Pythagorean identity sin2 (θ) + cos2 (θ) = 1 and use it to find sin (θ), cos (θ), or tan (θ) given sin (θ), cos (θ), or tan (θ) and the quadrant of the angle. Mathematical Practices 3 Construct viable arguments and critique the reasoning of others. 6 Attend to precision. CCSS

  14. You found values of sine and cosine by using sum and difference identities. • Find values of sine and cosine by using double-angle identities. • Find values of sine and cosine by using half-angle identities. Then/Now

  15. Concept

  16. Find the value of cos 2 if sin = and  is between 0° and 90°. Double-Angle Identities cos 2 = 1 – 2 sin2 Double-angle identity Simplify. Example 1

  17. Find the value of cos 2 if sin = and  is between 0° and 90°. Double-Angle Identities cos 2 = 1 – 2 sin2 Double-angle identity Simplify. Example 1

  18. Find the value of cos 2 if sin = and  is between 0° and 90°. A. B. C. D. Example 1

  19. Find the value of cos 2 if sin = and  is between 0° and 90°. A. B. C. D. Example 1

  20. A. Find the exact value of tan 2 if cos = and  is between 0° and 90°. Double-Angle Identities Step 1 Use the identity sin2 = 1 – cos2 to find the value of cos. cos2 + sin2 = 1 Subtract. Example 2

  21. Double-Angle Identities Take the square root of each side. Step 2 Find tan to use the double-angle identity for tan 2. Definition of tangent Example 2

  22. Double-Angle Identities Simplify. Step 3 Find tan 2. Double-angle identity Example 2

  23. Double-Angle Identities Square the denominator and simplify. Simplify. Answer: Example 2

  24. Answer: Double-Angle Identities Square the denominator and simplify. Simplify. Example 2

  25. B. Find the exact value of sin 2 if cos = and  is between 0° and 90°. Double-Angle Identities Double-angle identity Simplify. Answer: Example 2

  26. B. Find the exact value of sin 2 if cos = and  is between 0° and 90°. Answer: Double-Angle Identities Double-angle identity Simplify. Example 2

  27. A. Find the exact value of cos2 if sin = and  is between 0° and 90°. A. B. C. D. Example 2

  28. A. Find the exact value of cos2 if sin = and  is between 0° and 90°. A. B. C. D. Example 2

  29. B. Find the exact value of tan2 if sin = and  is between 0° and 90°. A. B. C. D. Example 2

  30. B. Find the exact value of tan2 if sin = and  is between 0° and 90°. A. B. C. D. Example 2

  31. Concept

  32. A. Find and  is the second quadrant. Since we must find cos first. Half-Angle Identities cos2 = 1 – sin2 sin2+ cos2 = 1 Simplify. Example 3A

  33. Since  is in the second quadrant, Half-Angle Identities Take the square root of each side. Half-angle identity Example 3

  34. Half-Angle Identities Simplify the radicand. Rationalize the denominator. Multiply. Example 3

  35. Half-Angle Identities Answer: Example 3

  36. Answer: Half-Angle Identities Example 3

  37. Half-Angle Identities B. Find the exact value of sin165. 165 is in Quadrant II; the value is positive. Example 3B

  38. Half-Angle Identities Simplify. Simplify. Answer: Example 3

  39. Answer: Half-Angle Identities Simplify. Simplify. Example 3

  40. A. Find and  is in the fourth quadrant. A. B. C. D. Example 3A

  41. A. Find and  is in the fourth quadrant. A. B. C. D. Example 3A

  42. A. B. C. D. B. Find the exact value of cos157.5. Example 3B

  43. A. B. C. D. B. Find the exact value of cos157.5. Example 3B

  44. FOUNTAINChicago’s Buckingham Fountain contains jets placed at specific angles that shoot water into the air to create arcs. When a stream of water shoots into the air with velocity v at an angle of with the horizontal, the model predicts that the water will travel a horizontal distance of D = sin 2 and reach a maximum height of H = sin2. The ratio of H to D helps determine the total height and width of the fountain. Find Simplify Using Double-Angle Identities Example 4

  45. Simplify Using Double-Angle Identities Original equation Simplify the numerator and the denominator. Example 4

  46. Simplify Using Double-Angle Identities Simplify. sin2 = 2sin cos Simplify. Example 4

  47. Simplify Using Double-Angle Identities Answer: Example 4

  48. Answer: Simplify Using Double-Angle Identities Example 4

  49. Use the identity cos2 = 1 – 2sin2 to help simplify. A. B. C. D. Example 4

  50. Use the identity cos2 = 1 – 2sin2 to help simplify. A. B. C. D. Example 4

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