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AmTran_Bus

  • 3 years ago

Trig identities...harder one

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  1. AmTran_Bus
    • 3 years ago
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    |dw:1353627728354:dw|

  2. hba
    • 3 years ago
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    Try this. First, change all tan(x) and cot(x) into sin(x)/cos(x) and cos(x)/sin(x) respectively. Then combine the fractions using common denominator which is cos(x)sin(x). -Hba.

  3. hba
    • 3 years ago
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    @AmTran_Bus Try it and tell me what you get ?

  4. AmTran_Bus
    • 3 years ago
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    I kept getting it to equal sin2. Let me give it another try.

  5. hba
    • 3 years ago
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    No if you try to reduce it,It will come to \[ \sin^2x -\cos^2x \]

  6. AmTran_Bus
    • 3 years ago
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    Agreed. I see you now. I have it on my paper. But now you have your +2cos^2 on there. I know sin+cos=1 (as long as they are both squared)...but what about the rest?

  7. hba
    • 3 years ago
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    Then put it in the actual ques. \[\sin^2x- \cos^2x + 2\cos^2x = 1\]

  8. hba
    • 3 years ago
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    Now you know how it goes ? @AmTran_Bus

  9. AmTran_Bus
    • 3 years ago
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    I understand that, yes. But maybe I just can't see it because I can't carry it on.

  10. hba
    • 3 years ago
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    |dw:1353628903264:dw|

  11. AmTran_Bus
    • 3 years ago
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    |dw:1353628964067:dw|

  12. hba
    • 3 years ago
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    |dw:1353629001275:dw|

  13. AmTran_Bus
    • 3 years ago
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    Oh yes. Thanks and Happy Thanksgiving.

  14. hba
    • 3 years ago
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    |dw:1353629050958:dw|

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