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Math Help - [SOLVED] Double angle identities

  1. #1
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    [SOLVED] Double angle identities

    Stuck on all of these and can't see to get my answers anywhere close to the answers

    a) Express \frac{\sin2x}{1 - 2\cos2x} in terms of \tan(x)

    b) Prove the identity \frac{1 + \sin2A - \cos2A}{1 + \sin2A + \cos2A} \equiv \tan(A)

    and finally

    c) Given that \cos2A = \tan^{2}x show that \cos2x = \tan^{2}A
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  2. #2
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    Quote Originally Posted by Beard View Post
    Stuck on all of these and can't see to get my answers anywhere close to the answers

    a) Express \frac{\sin2x}{1 - 2\cos2x} in terms of \tan(x)
     sin(2x) =2sin(x)cos(x)

    cos(2x) = cos^2(x) -sin^2(x)

    \frac{\sin2x}{1 - 2\cos2x}

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

    Divide numerator and denominator by cos^2(x)


    \frac{\frac{2sin(x)}{cos(x)}} {\frac{1}{cos^2(x)}-\frac{2cos^2(x)}{cos^2(x)}+\frac{2sin^2(x)}{cos^2(  x)}}

    I wont tell you that tan(x)= \frac{sin(x)}{cos(x)}


    and that tan^2(x)+1 = sec^2(x)

    Neither will I tell you that

    your answer is


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

    = ~\frac{2tan(x)}{-1+3tan^2(x)}
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  3. #3
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    Quote Originally Posted by Beard View Post
    Stuck on all of these and can't see to get my answers anywhere close to the answers


    b) Prove the identity \frac{1 + \sin2A - \cos2A}{1 + \sin2A + \cos2A} \equiv \tan(A)
    Using the formula






    We will get
    numerator = 1 +2sin(A)cos(A) - cos^2(A) + sin^2(A)

    Now we use
    1-cos^2(x) = sin^2(x)

    So numerator = 2sin^2(A) +2sin(A)cos(A) = 2sin(A)[sin(A) +cos(A)]

    Similarly Denominator will become

    1 +2sin(A)cos(A) +cos^2(A)-sin^2(A)

    = 2cos^2(A) +2sin(A)cos(A)

    = 2cos(A) (sin(A) +cos(A))
    So our term is now

    =\frac{2sin(A)(sin(A) +cos(A))}{2cos(A)(sin(A) +cos(A))}


    Now cancel the terms common in numerator and denominator
    And that's the end of the Second Show
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  4. #4
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by Beard View Post
    c) Given that \cos2A = \tan^{2}x show that
    \cos2x = \tan^{2}A
    cos(2A) = cos^2(A) - sin^2(A) = 2cos^2(x) - 1

    Hence equation now is
     2cos^2(A) - 1 = tan^2(x)


    2cos^2(A)= sec^2(x)


    cos^2(A) = sec^2(x)/2

    Hence sec^2(A) = \frac{2}{sec^2(x) }  = 2 cos^2(x)

    Thus
    sec^2(A)=2 cos^2(x)

    Substract one from both sides

    And
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