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Math Help - Trigo - Showing

  1. #1
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    Exclamation Trigo - Showing

    1. Show that OM = 6 cos θ - 2 sin θ
    2. Show that area of triangle OAM is 10 sin 2(θ + α)
    [OM = R cos (θ + α)]
    [AM = R sin (θ + α)]
    3. Given that θ can vary, find the maximum value of the area of triangle OAM.

    Refer to: http://i49.tinypic.com/1pueqq.png

    ABO & BNO = right angled triangles
    OB = 6 cm
    AB = 2 cm
    line AM perpendicular to ON
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  2. #2
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    Re: Trigo - Showing

    Hey fActor.

    Can you show us what you have tried? (Also show us all the identities that you can derive from the information given). This approach is typically the best one when helping with these kinds of problems.
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  3. #3
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    Re: Trigo - Showing

    Hello, fActor!


    Code:
                    A
                    o
                   *:*
                  * : * 2
                 *  :θ *
                *   :   *
               *    :    * B
              *    P* - - o
             *    6 : *   *
            * α   * :     *
           *  * θ   :     *
        O o * * * * o * * o N
                    M
    \angle ABO = \angle BNO = 90^o,\;OB = 6,\;AB = 2,\;AM \perp ON.

    Note that: \angle BON = \angle BAM = \theta.

    Let \alpha = \angle AOB.

    Draw BP \parallel ON.


    1. Show that: OM \:=\: 6\cos\theta - 2\sin\theta

    We see that: . OM \:=\:ON - MN

    In \Delta BNO\!:\:ON = 6\cos\theta

    In \Delta BPA\!:\:BP = 2\sin\theta = MN

    Therefore: . OM \:=\:6\cos\theta - 2\sin\theta




    2. Show that area of triangle OAM is: 10\sin2(\theta + \alpha)

    The area of \Delta OAM\:=\:\tfrac{1}{2}(OM)(AM)

    We note that: . OA \:=\:\sqrt{2^2+6^2} \:=\:\sqrt{40}

    In \Delta OAM\!:\;\begin{Bmatrix}OM \:=\: 2\sqrt{10}\cos(\theta + \alpha) \\ AM \:=\:2\sqrt{10}\sin(\theta+\alpha) \end{Bmatrix}


    Therefore:

    . . \Delta OAM \;=\;\tfrac{1}{2}\cdot \sqrt{40}\cos(\theta+\alpha)\cdot \sqrt{40}\sin(\theta+\alpha) \;=\;20\sin(\theta+\alpha)\cos( \theta+\alpha)

    . . . . . . . .  =\;10\cdot\underbrace{2\sin(\theta+\alpha)\cos(\th  eta+\alpha)}_{\text{Double-angle identity}} \;=\;10\sin2(\theta+\alpha)




    3. Given that \theta can vary, find the maximum value of the area of \Delta OAM.

    We have: . A \;=\;10\sin2(\theta +\alpha)


    Set the derivative equal to zero and solve.

    . . \frac{dA}{d\theta} \:=\:10\cos2(\theta+\alpha)\cdot 2 \;=\;0 \quad\Rightarrow\quad \cos2(\theta+\alpha) \:=\:0

    . . 2(\theta + \alpha) \:=\:\tfrac{\pi}{2} \quad\Rightarrow\quad \theta + \alpha \:=\:\tfrac{\pi}{4} \quad\Rightarrow\quad \theta \;=\;\tfrac{\pi}{4} - \alpha


    Therefore:

    . . A \;=\;10\sin2(\tfrac{\pi}{4} - \alpha + \alpha) \;=\;10\sin2(\tfrac{\pi}{4}) \;=\;10\sin\tfrac{\pi}{2} \;=\;10\cdot1 \;=\;10
    Last edited by Soroban; January 13th 2013 at 04:15 PM.
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