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Math Help - [SOLVED] Proof help!

  1. #1
    bsothe1
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    [SOLVED] Proof help!

    cot2x = (cosx+cos3x)/(sin3x+sinx) please list all steps. It's REALLY hard .thx
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  2. #2
    MHF Contributor
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    cot(2x) = (cos(x) +cos(3x)) / (sin(3x) +sin(x)) -------(i)
    Or, rewriting that,
    cot(2x) = [cos(3x) +cos(x)] / [sin(3x) +sin(x)] -------(ii)

    RHS of (ii) =
    = [cos(3x) +cos(x)] / [sin(3x) +sin(x)]

    Using trig identities:
    cosA +cosB = 2cos((A+B)/2)cos((A-B)/2)
    sinA +sinB = 2sin((A+B)/2)cos((A-B)/2),

    = [2cos((3x+x)/2)cos((3x-x)/2)] / [2sin((3x+x)/2)cos((3x-x)/2)]
    = cos(2x) / sin(2x)
    = cot(2x)
    = LHS

    So, proven.
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  3. #3
    Super Member

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    Hello, bsothe1!

    If we are allowed the Sum-to-Product identities, it's easy.

    . . \sin A + \sin B \:=\:2\sin\!\left(\frac{A+B}{2}\right)\cos\!\left(  \frac{A-B}{2}\right)

    . . \cos A + \cos B \:=\:2\cos\!\left(\frac{A+B}{2}\right)\cos\!\left(  \frac{A-B}{2}\right)


    \cot2x \:= \:\frac{\cos3x + \cos x}{\sin3x+\sin x}

    The numerator is: . \cos3x + \cos x \:=\: 2\cos\!\left(\frac{3x+x}{2}\right)\cos\!\left(\fra  c{3x-x}{2}\right) \:=\; 2\cos2x\cos x

    The denominator is: . \sin3x + \sin x \:=\:2\sin\!\left(\frac{3x+x}{2}\right)\cos\!\left  (\frac{3x-x}{2}\right) \:=\:2\sin2x\cos x


    The right side becomes: . \frac{2\cos2x\cos x}{2\sin2x\cos x} \;=\;\frac{\cos2x}{\sin2x} \:=\:\cot2x

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