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Math Help - Analytic trig question

  1. #1
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    Analytic trig question

    2tanx/(1-(tan(x)^2) + 1/2(cosx^2)-1 = cosx+sinx/cosx-sinx
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by bilbobaggins View Post
    2tanx/(1-(tan(x)^2) + 1/2(cosx^2)-1 = cosx+sinx/cosx-sinx
    Will you please rewrite this using parenthesis so we can see what the actual problem is?

    -Dan
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  3. #3
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    Hello, bilbobaggins!

    I think I've guessed what you meant . . .


    We need a number of double-angle identities:

    . . \sin2\theta \:=\:2\sin\theta\cos\theta

    . . \cos2\theta \:=\:\cos^2\!\theta - \sin^2\!\theta \:=\:2\cos^2\!\theta -1

    . . \tan2\theta \:=\:\frac{2\tan\theta}{1 - \tan^2\!\theta}


    \frac{2\tan x}{1-\tan^2\!x} + \frac{1}{2\cos^2\!x-1} \;=\; \frac{\cos x+\sin x}{\cos x-\sin x}

    The left side is: . \tan2x + \frac{1}{\cos2x} \;=\;\frac{\sin2x}{\cos2x} + \frac{1}{\cos2x} \;=\;\frac{\sin2x + 1}{\cos2x} \;=\;\frac{2\sin x\cos x + 1}{\cos^2\!x -\sin^2\!x}


    Since 1 \:=\:\sin^2\!x+\cos^2\!x, the numerator becomes:

    . . 2\sin x\cos x + \sin^2\!x+\cos^2\!x \;= \;\cos^2\!x + 2\sin x\cos x + \sin^2\!x \;=\;(\cos x + \sin x)^2


    The fraction becomes: . \frac{(\cos x + \sin x)^2}{(\cos x - \sin x)(\cos x + \sin x)} \;=\;\frac{\cos x + \sin x}{\cos x - \sin x}<br /> <br />

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