Identitieslong 9005
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Identitieslong 9005 Presentation Transcript

  • 1. Identities Identities Identities Identities Identities
  • 2. 2.2. Prove the followingIdentity using twoIdentity using twodifferentdifferent methods:
  • 3. For this question. You can workwith the more Method # 1:complicated side. In this case it isthe left side. tan ²θ sin ²θ Line of Separation / “Great Wall of China”
  • 4. 1} In step one, we realized that we can multiply by reciprocal of1} the bottom fraction instead of dividing.
  • 5. 2} For step two you can think of tanθ as2} sinθ/cosθ.
  • 6. 3} In step three we can multiply sinθ by3} sinθ/cosθ.
  • 7. 4} Here we can multiply sin²θ/cosθ out by 1/cosθcos³θ. We can also4} factor out cos²θ in the denominator.
  • 8. 5} In this last step we can say that sin²θ/cos²θ is tan²θ and by the Pythagorean5} identity (1- cos²θ) is sin²θ.
  • 9. QED
  • 10. sinθ tan ²θ cosθ − cos³θ sin ²θ tanθ  tanθ  sinθ   cosθcos³θ   sinθ    sinθ cosθ   cosθ − cos³θ     sin²θ  1   cosθ  cosθ − cos³θ  sin²θ cos²θ − cos 4θ sin²θ cos²θ(1 - cos²θ ) tan²θ sin²θ QED
  • 11. Working withboth sidescan alsowork: Method # 2: tan ²θ sin ²θ Line of Separation / “Great Wall of China”
  • 12. [Right Side] tan ²θ 1} Here we sin ²θ can say that tan²θ is equal sin ²θ to 1} cos ²θ sin²θ/cos²θ. ~by saying sin ²θ that, it can be 1 simplified to 1/cos²θ. cos ²θ
  • 13. [Left Side] 2} Here we recognize cosθ-cos³θ as cosθ - (cosθcos²θ) or cosθ - cosθ2} (1-sin²θ). Also, tanθ as sinθ/cosθ.
  • 14. 3} In this step we can multiply3} cosθ by (1-sin²θ).
  • 15. 4} By multiplying by the reciprocal of sinθ/cosθ we can see that when cosθ gets multiplied out, we get cos²θ- cos²θ+sin²θcosθ. The two cos²θ’s4} cancel and you’re left with sin²θcosθ.
  • 16. 5} In this last step we see that sin²θcosθ/sinθ leaves us sinθcosθ. The5} sinθ’s reduce leaving us 1/cosθ.
  • 17. QED
  • 18. sin θ tan ²θ cos θ − cos ³θ sin ²θ tan θ sin ²θ sin θ cos ²θcos θ − cos θ (1 − sin ²θ ) sin ²θ sin θ 1 cos θ cos ²θ sin θ cos θ − cos θ + sin ²θ sin θ cos θ sin θ sin ²θ cos θ sin θ sin θ sin θ cos θ 1 QED cos θ
  • 19. Identities Identities Identities Identities Identities