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Math Help - Trigo Identity

  1. #1
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    Trigo Identity

    Dear experts,

    I am practicing the Trig identity, however met some issue in the changing {\2tanB/(1+tanB)} TO {\1-3tan^2 B}
    Please kindly advise me:

    identify: \tan3B= \dfrac{\3tanB - tan^2 B }{\1-3tan^2 B}LHS := \dfrac{\ tanB + tan2 B }{\1-tanB.tan2B} = \dfrac{\3tanB - tan^2 B }{\2tanB/(1+tanB)}
    Thank you so much!
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  2. #2
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    Re: Trigo Identity

    Quote Originally Posted by janice91 View Post
    Dear experts,

    I am practicing the Trig identity, however met some issue in the changing {\2tanB/(1+tanB)} TO {\1-3tan^2 B}
    Please kindly advise me:

    identify: \tan3B= \dfrac{\3tanB - tan^2 B }{\1-3tan^2 B}LHS := \dfrac{\ tanB + tan2 B }{\1-tanB.tan2B} = \boxed{\dfrac{\3tanB - tan^2 B }{\2tanB/(1+tanB)}}

    The boxed part is incorrect. Apply the double angle formula again for tan(2B) and you would get the answer.

    Thank you so much!
    .
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  3. #3
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    Lexington, MA (USA)
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    Re: Trigo Identity

    Hello, janice91!

    I met some issue in the changing . \frac{2\tan B}{1+\tan B} \,\text{ to }\,1-3\tan^2B

    No wonder! . . . The two expressions are NOT equal.

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  4. #4
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    Re: Trigo Identity

    I have made a careless mistake. Thank you for guidance!
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