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Math Help - Please Bare with Me.. Can you prove this identity?

  1. #1
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    Talking Please Bare with Me.. Can you prove this identity?

    1 - tanx sinx = cos2x + sin²x




    I think i finally created a proper iidentity.. I just need somebody to try and prove it!


    Thanks
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  2. #2
    Senior Member MacstersUndead's Avatar
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    Quote Originally Posted by advancedfunctions2010 View Post
    1 - tanx sinx = cos2x + sin²x
    RHS
    cos2x = cos²x - sin²x

    Therefore, the RHS = cos²x

    LHS
    1 - tanx sinx = cos²x + sin²x - tanx sinx

    LHS =! RHS unless sin²x = tanx sinx
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    Im not sure what you mean?
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by advancedfunctions2010 View Post
    1 - tanx sinx = cos2x + sin²x




    I think i finally created a proper iidentity.. I just need somebody to try and prove it!


    Thanks
    LHS = 1 - (tanx \times sinx)

     = 1 - \frac{sinx}{cosx} \times sinx

     = 1-\frac{sin^2x}{cosx}

    RHS = cos2x + sin^2x = cos^2x-sin^2x+sin^2x =cos^2x

    So the identity you have created cannot be equal.

    however, if you change your LHS to :

    1 - (tanx \times sinx \times cosx)

     = 1 - (\frac{sinx}{cosx} \times sinx \times cosx)

    = 1-(sinx \times sinx) = 1-(sin^2x) = cos^2x

    In this case, your RHS will be equal to LHS giving you a proper identity.
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