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AmTran_Bus

Prove this trig identity csc^4-csc^2=cot^4+cot^2

  • one year ago
  • one year ago

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  1. AmTran_Bus
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    |dw:1353025115000:dw|

    • one year ago
  2. lgbasallote
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    look at the LHS...try factoring out csc^2 what do you get?

    • one year ago
  3. AmTran_Bus
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    |dw:1353025247991:dw|

    • one year ago
  4. lgbasallote
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    not exactly....if you factor out csc^2 then csc^4 becomes csc^2 but csc^2 doesn't become csc^2

    • one year ago
  5. lgbasallote
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    |dw:1353025341525:dw|

    • one year ago
  6. AmTran_Bus
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    Would it even exist, that is, would it cancel? Or just be csc?

    • one year ago
  7. lgbasallote
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    what is csc^2/csc^2?

    • one year ago
  8. AmTran_Bus
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    ohhhh. I feel dumb. Got ya.

    • one year ago
  9. lgbasallote
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    so what should the factored form be?

    • one year ago
  10. AmTran_Bus
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    |dw:1353025535036:dw|

    • one year ago
  11. lgbasallote
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    right

    • one year ago
  12. lgbasallote
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    now what is csc^2 - 1?

    • one year ago
  13. AmTran_Bus
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    cot?

    • one year ago
  14. lgbasallote
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    cot^2

    • one year ago
  15. lgbasallote
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    so uou have \[\huge \csc^2 (\cot^2)\] now here comes a tricky part...review your identities and give me the equation for csc^2

    • one year ago
  16. AmTran_Bus
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    csc2 = 1/sin...it also equals cot+1

    • one year ago
  17. lgbasallote
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    cot^2 + 1 not cot + 1

    • one year ago
  18. AmTran_Bus
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    Sorry, your exactly right.

    • one year ago
  19. lgbasallote
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    you should check your notes and make sure you wrote them right....you might get them wrong... \[\huge \sin^2 + \cos^2 = 1\] \[\huge \tan^2 + 1 = \sec^2\] \[\huge \cot^2 + 1 = \csc^2\]

    • one year ago
  20. lgbasallote
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    anyway...back to what i was saying... csc^2 is cot^2 + 1 so \[\huge \csc^2 (\cot^2) \implies (\cot^2 + 1)(\cot^2)\] now expand that

    • one year ago
  21. AmTran_Bus
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    would the lhs say csc cot ^4? The RHS is cot^4 +cot^2

    • one year ago
  22. lgbasallote
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    i think you're confused...let me flip it... \[\huge \cot^2 \theta (\cot^2\theta + 1)\] do you know how to distribute that now?

    • one year ago
  23. AmTran_Bus
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    Its not that I'm confused as much as tired my friend. |dw:1353026333309:dw|

    • one year ago
  24. lgbasallote
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    well anyway... cot^2(cot^2 + 1) becomes cot^4 + cot^2 so you have just proven the identity

    • one year ago
  25. AmTran_Bus
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    Thanks!!!!!!!

    • one year ago
  26. lgbasallote
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    welcome

    • one year ago
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