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Math Help - trig identity

  1. #1
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    trig identity

    pls solve this
    cot(4/5)-A = cos 2A/(1-sin 2A)
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  2. #2
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    Trigonometry

    Hello vungaralas
    Quote Originally Posted by vungaralas View Post
    pls solve this
    cot(4/5)-A = cos 2A/(1-sin 2A)
    I'm not quite sure what you mean by cot(4/5)-A. But if it's any help, you can express the right-hand side in terms of tan A, using the half-angle formulae, or from the basic double-angle formulae, like this:

    \frac{\cos 2A}{1-\sin 2A} = \frac{\cos^2A - \sin^2A}{1 - 2\sin A \cos A}

    = \frac{1 - \tan^2A}{\sec^2A - 2\tan A}

    = \frac{1 - \tan^2A}{1 + \tan^2A - 2\tan A}

    = \frac{(1 - \tan A)(1+\tan A)}{(1-\tan A)^2}

    = \frac{1 + \tan A}{1 - \tan A}

    Is that any help?

    Grandad
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  3. #3
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    Hello, vungaralas!

    Your equation makes no sense.

    Inspired by Grandad's excellent manipulation,
    . . I took a guess at the expression on the left side.


    \cot(4/5)-A \:= \:\frac{\cos2A}{1-\sin2A}

    If that statement is supposed to be an identity,
    . . I would guess that it should look like this:

    . . \cot\left(\frac{\pi}{4} - A\right) \:=\:\frac{\cos2A}{1-\sin2A}


    We have: . \cot(P - Q) \:=\:\frac{1+\tan P\tan Q}{\tan P - \tan Q}

    So the left side is: . \cot\left(\frac{\pi}{4} - A\right) \;=\;\frac{1 + \tan\frac{\pi}{4}\tan A}{\tan\frac{\pi}{4} - \tan A} \;=\;\frac{1 + \tan A}{1 - \tan A} \;=\;\frac{1 + \frac{\sin A}{\cos A}}{1 - \frac{\sin A}{\cos A}}

    Multiply by \frac{\cos A}{\cos A}\!:\;\;\frac{\cos A + \sin A}{\cos A - \sin A}

    Multiply by \frac{\cos A - \sin A}{\cos A-\sin A}\!:\;\;\frac{\cos A + \sin A}{\cos A - \sin A}\cdot\frac{\cos A - \sin A}{\cos A - \sin A}

    . . = \;\frac{\overbrace{\cos^2\!A - \sin^2\!A}^{\text{This is }\cos2A}}{\underbrace{\cos^2\!A + \sin^2\!A}_{\text{This is 1}} - \underbrace{2\sin A\cos A}_{\text{This is }\sin2A}} \;=\;\frac{\cos2A}{1-\sin2A}

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  4. #4
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    thank u so much
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