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burhan101

  • 2 years ago

Simplify this expression

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  1. burhan101
    • 2 years ago
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    \[ \huge \ln e^{\sin^2x} + \ln e^{\cos^2x} \]

  2. carlos74923
    • 2 years ago
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    hah., a

  3. burhan101
    • 2 years ago
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    the answer is one.. but how ?!

  4. carlos74923
    • 2 years ago
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    its a nice problem

  5. carlos74923
    • 2 years ago
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    its 0

  6. burhan101
    • 2 years ago
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    no my answerkey says 1

  7. carlos74923
    • 2 years ago
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    :'( walks in shame

  8. chrissy401
    • 2 years ago
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    \[\cos^2x \ln e+\sin^2x \ln e\] but \[\ln e\] is 1

  9. chrissy401
    • 2 years ago
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    since \[\ln e = \log_{e} ^e\]

  10. burhan101
    • 2 years ago
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    \[\sin^2x+\cos^2x\]

  11. chrissy401
    • 2 years ago
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    we know that \[\cos^2x+\sin^2x=1\] in trigonometry

  12. chrissy401
    • 2 years ago
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    thast why the answer is one

  13. carlos74923
    • 2 years ago
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    log e is = to lin

  14. burhan101
    • 2 years ago
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    thank you :D

  15. carlos74923
    • 2 years ago
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    so log e to e to 1 is 1

  16. chrissy401
    • 2 years ago
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    you're welcome..

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