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Identitieslong 9005
Identitieslong 9005
Identitieslong 9005
Identitieslong 9005
Identitieslong 9005
Identitieslong 9005
Identitieslong 9005
Identitieslong 9005
Identitieslong 9005
Identitieslong 9005
Identitieslong 9005
Identitieslong 9005
Identitieslong 9005
Identitieslong 9005
Identitieslong 9005
Identitieslong 9005
Identitieslong 9005
Identitieslong 9005
Identitieslong 9005
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Identitieslong 9005

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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

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