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Math Help - Trigonometric Identities

  1. #1
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    Trigonometric Identities

    Help!
    1. Verify the identity: (1/1-cosx)-(cosx/1+cosx)=2csc(squared)x-1
    2.Verify the identity: cos5t-cos3t= -8sin(squared)t(2cos(cubed)t-cost)
    3. Verify the identity: 2cos4xsin2x=2sin3xcos3x-2cosxsinx
    Last edited by kelsey3; April 6th 2008 at 03:02 PM. Reason: added third problem
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  2. #2
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    #1:

    \frac{1}{1-cosx}-\frac{cosx}{1+cosx}=2csc^{2}x-1

    The left side, you can cross multiply into:

    \frac{1+cosx-(cosx-cos^{2}x)}{1-cos^{2}x}

    =\frac{1+cos^{2}x}{1-cos^{2}x}

    =\frac{1+cos^{2}x}{sin^{2}x}

    \frac{1}{sin^{2}x}+\frac{cos^{2}x}{sin^{2}x}

    =csc^{2}x+cot^{2}x

    Use the identity: csc^{2}x-1=cot^{2}x

    csc^{2}x+(csc^{2}x-1)=2csc^{2}x-1
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  3. #3
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    Do you know how to do the other two?....lol
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  4. #4
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    Hello, kelsey3!

    We need these double-angle identities:
    . . \sin2\theta \:=\:2\sin\theta\cos\theta
    . . \cos2\theta \:=\:2\cos^2\!\theta-1

    And this sum-to-product identity: . \cos A - \cos B \:=\:-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)


    2. Verify the identity: . \cos5t-\cos3t\l= \; -8\sin^2\!t\cos t\left(2\cos^2\!t-1\right)

    The left side is: . -2\sin\left(\frac{5t+3t}{2}\right)\sin\left(\frac{5  t-3t}{2}\right)

    . . . . . . . . . . =\;-2\sin4t\sin t

    . . . . . . . . . . = \;-2(2\sin2t\cos2t) \sin t

    . . . . . . . . . . =\;-4\sin2t\cos2t\sin t

    . . . . . . . . . . = \;-4(2\sin t\cos t)(2\cos^2\!t-1)\sin t

    . . . . . . . . . . = \;-8\sin^2\!t\cos t(2\cos^2\!t - 1)

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