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Math Help - Double Angle Identities

  1. #1
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    Double Angle Identities

    I'm stuck on this question...


    I've tried expressing tan^2 (x/2) as a function of tan x and tan x/2, then substituting for cos/sin but that doesn't seem to work (unless I've done the algebra wrong...) can anyone give me some pointers please?

    Identities here:
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  2. #2
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    Re: Double Angle Identities

    $\tan^2(\frac x 2)=$

    $\sec^2(\frac x 2)-1=$

    $\dfrac 1 {\cos^2( \frac x 2)} - 1=$

    $\dfrac 1 {\dfrac {1+\cos(x)}2}-1=$

    $\dfrac 2 {1+\cos(x)} - 1=$

    $\dfrac{1-\cos(x)}{1+\cos(x)}$
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  3. #3
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    Re: Double Angle Identities

    Quote Originally Posted by romsek View Post
    $\tan^2(\frac x 2)=$

    $\sec^2(\frac x 2)-1=$

    $\dfrac 1 {\cos^2( \frac x 2)} - 1=$

    $\dfrac 1 {\dfrac {1+\cos(x)}2}-1=$

    $\dfrac 2 {1+\cos(x)} - 1=$

    $\dfrac{1-\cos(x)}{1+\cos(x)}$
    Thanks, as always!
    How about this question?
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  4. #4
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    Re: Double Angle Identities

    Quote Originally Posted by Bude8 View Post
    Thanks, as always!
    How about this question?
    just apply the double angle formulas as needed. You can figure it out
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