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    • Thread Starter
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    how can you get sin2x - tanx to

    2cos2x?
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    sin2x - tanx
    =2sinxcosx - sinx/cosx
    =sinx(2cosx - 1/cosx)
    =sinx[(2cos²x-1)/cosx]
    =sinx[(cos2x)/cosx]
    =tanxcos2x

    hmmmm maybe not that way :s
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    i got 2 tanXcos2X...dnt knw how toget furtha??
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    Q must b wrong
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    (Original post by nas7232)
    how can you get sin2x - tanx to

    2cos2x?
    What is the full question? to prove that they are equal?

    If so: sin2x - tanx

    => 2sinxcosx - sinx/cosx

    => [2sinxcos2x - sinx]/cosx

    => [sinx(2cos2x - 1)]/cosx

    => tanxcos2x

    Now tanx =/= cos2x so I guess I have done something wrong or the question is wrong.
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    (Original post by mackin boi)
    i got 2 tanXcos2X...dnt knw how toget furtha??
    hmmm how u get the 2? Can u show ur working?
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    i cant see how its possible either
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    (Original post by samdavyson)
    What is the full question? to prove that they are equal?

    If so: sin2x - tanx

    => 2sinxcosx - sinx/cosx

    => [2sinxcos2x - sinx]/cosx

    => [sinx(2cos2x - 1)]/cosx

    => tanxcos2x

    Now tanx =/= cos2x so I guess I have done something wrong or the question is wrong.
    ops sorry, that was part a) done right

    part b is sin2x - tanx = 2cos2x
    solve that equation
    in the interval of 0<x<180

    Sorry! i typed the question out wrong lol
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    (Original post by Syncman)
    hmmm how u get the 2? Can u show ur working?
    The question must be a mistake.

    I don't think there should be a 2 in that either.
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    ah i can do it now...what a waste of thread
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    (Original post by nas7232)
    how can you get sin2x - tanx to

    2cos2x?
    sin2x - tanx
    = 2sincosx - sinx/cosx
    = (2sinxcos^2x - sinx)/cosx
    = {2sinx[(cos2x + 1)/2] - sinx}/cosx
    = {sinx(cos2x + 1) - sinx}/cosx
    = {sinxcos2x + sinx - sinx}/cosx
    = sinxcos2x/cosx
    = tanx.cos2x

    So your aim was impossible.
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    ops sorry, that was part a) done right

    part b is sin2x - tanx = 2cos2x
    solve that equation
    in the interval of 0<x<180

    Sorry! i typed the question out wrong lol
    .
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    (Original post by nas7232)
    ah i can do it now...what a waste of thread
    I still can't see how to solve it.

    (Original post by nas7232)
    ops sorry, that was part a) done right

    part b is sin2x - tanx = 2cos2x
    solve that equation
    in the interval of 0<x<180

    Sorry! i typed the question out wrong lol
    Ahh. Thank the heavens.

    sin2x - tanx = 2cos2x

    => cos2xtanx = 2cos2x

    => cos2x = 0

    OR => tanx = cos2x

    => sinx/cosx = cos2x - sin2x

    => Then what?
    • Thread Starter
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    part b is sin2x - tanx = 2cos2x
    solve that equation
    in the interval of 0<x<180


    as you worked out:

    sin2x - tanx = tanx.cos2x

    tanx.cos2x = 2cos2x
    tan.cos2x - 2cos2x = 0
    cos2x(tanx - 2)=0
    cos2x = 0
    or tanx = 2

    solutions
    45,63.4,135

    once again, sry for not typing it out properly. Now i know why i couldn't initially do it.
 
 
 
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