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Math Help - Trig Identities - Double Angles

  1. #1
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    Trig Identities - Double Angles

    Prove the following using double angle formulas:

    \frac{sin(\pi+x)}{cos(2\pi+x)} + \frac{sec(\pi-x)}{csc(2\pi-x)} = 0

    -----

    My Attempt:

    <br />
LS = \frac{sin\pi cosx + cos\pi sinx}{cos2\pi cosx - sin2\pi sinx} + \frac{1}{cos\pi cosx + sin\pi sinx}

     = \frac {-sinx}{cosx} + \frac{1}{-cosx}\times(\frac{-cosx}{1})

     = \frac {-sinx}{cosx} + \frac {-cosx}{-cosx}

    = \frac {(-sinx)(-cosx) + (-cosx)(cosx)}{(cosx)(-cosx)}
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
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    1,631
    Your
    [sec(pi -x) / csc(2pi -x)] = 1 / [cos(pi)cos(x) +sin(pi)sin(x)]
    is not correct.

    [sec(pi -x) / csc(2pi -x)]
    = [1 / cos(pi -x)] / [1 / sin(2pi -x)]
    = sin(2pi -x) / cos(2pi -x)

    So,
    [sin(pi +x) / cos(2pi +x)] +[sec(pi -x) / csc(2pi -x)] = 0

    LHS =
    = [sin(pi +x) / cos(2pi +x)] +[sec(pi -x) / csc(2pi -x)]

    = [sin(pi +x) / cos(2pi +x)] +[sin(2pi -x) / cos(pi -x)]

    = [{sin(pi)cos(x) +cos(pi)sin(x)} / {cos(2pi)cos(x) -sin(2pi)sin(x)}]
    + [{sin(2pi)cos(x) -cos(2pi)sin(x)} / {cos(pi)cos(x) +sin(pi)sin(x)]

    Now,
    sin(pi) = 0
    cos(pi) = -1
    sin(2pi) = 0
    cos(2pi) = 1

    = [{0*cos(x) +(-1)sin(x)} / {(1)cos(x) -0*sin(x)}]
    + [{0*cos(x) -(1)sin(x)} / {(-1)cos(x) +0*sin(x)]

    = [-sin(x) / cos(x)] +[-sin(x) / -cos(x)]

    = -tan(x) +tan(x)

    = 0

    = RHS

    Therefore, proven.
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  3. #3
    Super Member

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    Lexington, MA (USA)
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    Hello, Macleef!

    Prove: . \frac{\sin(\pi+x)}{\cos(2\pi+x)} + \frac{\sec(\pi-x)}{\csc(2\pi-x)} = 0

    The first fraction is: . \frac{\sin(\pi+x)}{\cos(2\pi+x)} \;=\;\frac{\sin(\pi)\cos(x) + \sin(x)\cos(\pi)}{\cos(2\pi)\cos(x) - \sin(2\pi)\sin(x)}

    . . = \;\frac{0\cdot\cos(x) + \sin(x)\cdot(-1)}{1\cdot\cos(x) - 0\cdot\sin(x)} \;=\; \frac{-\sin(x)}{\cos(x)} \;=\;-\tan(x)


    The second fraction is: . \frac{\sin(2\pi-x)}{\cos(\pi-x)} \;=\;\frac{\sin(2\pi)\cos(x) - \sin(x)\cos(2\pi)}{\cos(\pi)\cos(x) + \sin(\pi)\sin(x)}

    . . = \;\frac{0\cdot\cos(x) - \sin(\pi)\cdot1}{(-1)\cos(x) + 0\cdot\sin(x)} \;=\;\frac{-\sin(x)}{-\cos(x)} \;=\;\tan(x)


    The problem becomes: . -\tan(x) + \tan(x) \;=\;0


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