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Math Help - Stupid trig identities

  1. #1
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    Stupid trig identities

    the problem is tan(x)+cot(x)=2csc(2x). I can get to the point where I have (2sin(x)cos(x))/(sin(x)cos(x)). I am not quite sure what to do beyond this point.
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  2. #2
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    Quote Originally Posted by dnlstffrd View Post
    the problem is tan(x)+cot(x)=2csc(2x). I can get to the point where I have (2sin(x)cos(x))/(sin(x)cos(x)). I am not quite sure what to do beyond this point.
    tan(x)+cot(x)=sin(x)/cos(x)+cos(x)/sin(x)=[sin^2(x)+cos^2(x)]/[cos(x)sin(x)]
    ............... =1/[cos(x)sin(x)]

    but cos(x)sin(x)=(1/2)sin(2x), so:

    tan(x)+cot(x)=2csc(2x).

    RonL
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  3. #3
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    wait a sec. It should be 2/sin(2x), not 1/2sin(2x) because that would equal (1/2)csc(2x), not 2csc(2x).
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  4. #4
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    Quote Originally Posted by dnlstffrd View Post
    wait a sec. It should be 2/sin(2x), not 1/2sin(2x) because that would equal (1/2)csc(2x), not 2csc(2x).
    1/[cos(x)sin(x)]=1/[(1/2)sin(2x)]=2/sin(2x)=2csc(2x).

    RonL
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  5. #5
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    oops, I missed that part

    I actually have another that I also cannot figure out for some reason:

    [sin(x)+cos(2x)-1]/[cos(x)-sin(2x)]=tan(x)
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  6. #6
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    Quote Originally Posted by dnlstffrd View Post
    oops, I missed that part

    I actually have another that I also cannot figure out for some reason:

    [sin(x)+cos(2x)-1]/[cos(x)-sin(2x)]=tan(x)
    You should post this in a new thread.

    sin(2x) = 2sin(x)cos(x)
    cos(2x) = 1 - 2[sin(x)]^2 = 2[cos(x)]^2 - 1 = [cos(x)]^2 - [sin(x)]^2

    so
    sin(x)+cos(2x)-1 = sin(x) + 1 - 2[sin(x)]^2 - 1 = -2[sin(x)]^2 + sin(x) = sin(x) {-2sin(x) + 1}

    cos(x)-sin(2x) = cos(x) - 2sin(x)cos(x) = cos(x) {1 - 2sin(x)}

    Dividing the two cancels the common 1 - 2sin(x) leaving sin(x)/cos(x) = tan(x).

    -Dan
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  7. #7
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    tan(x) + cot(x) = 2 csc(2x)


    LHS = sin(x)/cos(x) + cos(x)/sin(x)

    = (sin^2(x) + cos^2(x)) / (sin(x) cos(x))

    = 1 / (sin(x) cos(x))


    RHS = 2 / sin(2x)

    = 2 / (2 sin(x) cos(x))

    = 1 / (sin(x) cos(x))

    = LHS
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