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Math Help - Identities

  1. #1
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    Identities

    For some reason, i have a problem with these, and only these.

    [sin(x)+cos(2x)-1]/[cos(x)-sin(2x)]=tan(x)
    and
    tan(x/2)=(sin(x))/(1+cos(x)), for which I can get it to tan(x/2)=sin(x)tan(x), but don't know the identity to go further.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by dnlstffrd View Post
    For some reason, i have a problem with these, and only these.

    [sin(x)+cos(2x)-1]/[cos(x)-sin(2x)]=tan(x)
    and
    tan(x/2)=(sin(x))/(1+cos(x)), for which I can get it to tan(x/2)=sin(x)tan(x), but don't know the identity to go further.
    I already answered the first one in the other thread.

    I would think of the second one with the replacement y = x/2. (I hate 1/2 angle problems!) That transforms the problem to showing:
    tan(y) = sin(2y) / [1 + cos(2y)]

    The RHS is
    sin(2y) = 2sin(y)cos(y)
    cos(2y) = 2[cos(y)]^2 - 1

    So 1 + cos(2y) = 1 + 2[cos(y)]^2 - 1 = 2[cos(y)]^2

    Dividing:
    sin(2y) / [1 + cos(2y)] = [2sin(y)cos(y)]/[2{cos(y)}^2] = sin(y)/cos(y) = tan(y).

    - Dan
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