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Math Help - Prove the identity

  1. #1
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    Prove the identity

    Hi again,

    I would appreciate it a lot if anyone could tell me if this is correct:

    Prove the identity:
    cotθ - tanθ = 2cot2θ

    cosθ/sinθ - sinθ/cosθ = 2(cosθ/sinθ)2θ
    - sinθ/cosθ =
    - tanθ/4 = θ

    Thanks.
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  2. #2
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    Quote Originally Posted by laoch View Post
    Hi again,

    I would appreciate it a lot if anyone could tell me if this is correct:

    Prove the identity:
    cotθ - tanθ = 2cot2θ

    cosθ/sinθ - sinθ/cosθ = 2(cosθ/sinθ)2θ
    - sinθ/cosθ =
    - tanθ/4 = θ

    Thanks.
    Sorry, what?

    Anyway here is my proof.
    cot T - tan T = 2 cot 2T
    iff cos T / sin T - sin T / cos T = 2 cos 2T / sin 2 T
    iff (cos^2 T - sin^2 T) / sin T cos T = 2 (cos^2 T - sin^T) / 2 sin T cos T
    QED.
    Identities used: cos 2T = cos^2 T - sin^2 T and sin 2T = 2 sin T cos T
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by laoch View Post
    Hi again,

    I would appreciate it a lot if anyone could tell me if this is correct:

    Prove the identity:
    cotθ - tanθ = 2cot2θ

    cosθ/sinθ - sinθ/cosθ = 2(cosθ/sinθ)2θ
    - sinθ/cosθ =
    - tanθ/4 = θ

    Thanks.
    Here's one approach:

    Use \tan 2\theta=\frac{2 \tan \theta}{1-\tan^2 \theta}

    \cos \theta - \tan \theta = 2\cot 2\theta

    \frac{1}{\tan\theta}-\tan \theta=2 \cot 2\theta

    \frac{1-\tan^2 \theta}{\tan \theta}=\frac{2}{\tan 2\theta}

    \frac{1-\tan^2 \theta}{\tan \theta}=\frac{2}{\frac{2 \tan \theta}{1-\tan^2 \theta}}

    \frac{1-\tan^2 \theta}{\tan \theta}=\frac{1-\tan^2 \theta}{\tan \theta}
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  4. #4
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    Hello, laoch!

    Prove the identity: . \cot\theta - \tan\theta \:= \:2\cot2\theta

    The right side is: . \frac{2}{\tan2\theta} \;=\;\frac{2}{\frac{2\tan\theta}{1-\tan^2\!\theta}} \;=\;\frac{2(1-\tan^2\!\theta)}{2\tan\theta}

    . . =\:\frac{1-\tan^2\!\theta}{\tan\theta} \;= \;\frac{1}{\tan\theta} - \frac{\tan^2\!\theta}{\tan\theta} \;=\; \cot\theta - \tan\theta

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