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Math Help - Verify Trig Identities

  1. #1
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    Verify Trig Identities

    (CosθCotθ)/1-Sinθ - 1 = Cscθ

    4(Tan^4)x + (tan^2)x - 3 = (sec^2)x[4(tan^2)x - 3]

    Sin(α/3)Cos(α/3) = .5Sin(2α/3)
    Last edited by RosieLaird; August 5th 2010 at 02:52 PM.
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  2. #2
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    Quote Originally Posted by RosieLaird View Post
    (CosθCotθ)/1 = Cscθ

    4(Tan^4)x + (tan^2)x - 3 = (sec^2)x[4(tan^2)x - 3]

    Sin(α/3)Cos(α/3) = .5Sin(2α/3)
    1. recheck the first equation ... as you typed it, it makes no sense.

    2. factor the left side ... then use the identity 1+\tan^2{x} = \sec^2{x}

    3. use double angle formula for sine on the right side
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  3. #3
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    Hello, RosieLaird!

    \dfrac{\cos\theta \cot\theta}{1-\sin\theta} - 1 \:=\: \csc\theta

    The left side is: . \dfrac{\cos\theta\cdot\frac{\cos\theta}{\sin\theta  }}{1-\sin\theta} - 1  \;\;=\;\;\dfrac{\cos^2\!\theta}{\sin\theta(1-\sin\theta)} - 1

    . . =\;\;\dfrac{\cos^2\theta - \sin\theta(1 - \sin\theta)}{\sin\theta(1-\sin\theta)} \;\;=\;\;\dfrac{\overbrace{\cos^2\theta + \sin^2\theta}^{\text{This is 1}} - \sin\theta}{\sin\theta(1-\sin\theta)}


    . . =\;\;\dfrac{1-\sin\theta}{\sin\theta(1-\sin\theta)} \;\;=\;\;\dfrac{1}{\sin\theta} \;\;=\;\;\csc\theta

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  4. #4
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    4(Tan^4)x + (tan^2)x - 3 = (sec^2)x[4(tan^2)x - 3]

    4tan^2(x)tan^2(x) + tan^2(x) - 3

    4tan^2(x)(sec^2x-1) + tan^2(x) - 3

    4tan^2(x)sex^2(x) - 4tan^2(x) + tan^2(x) -3

    4tan^2(x)sex^2(x) - 3tan^2(x) -3

    4tan^2(x)sex^2(x) - 3(tan^2(x) + 1)

    4tan^2(x)sex^2(x) - 3sec^2(x)

    sec^2(x)[4tan^2(x) - 3]
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