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Math Help - Another Trigonometry Word Problem (rather difficults)

  1. #1
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    Another Trigonometry Word Problem (rather difficults)

    In the diagram in the left below, AD is a diameter of the circle and is tangent to line m at D. If AD=1 AND BC is perpendicular to m, show that:

    a) AB + BC = 1 +cos(x) - cos^2(x)

    b) What value of x makes the sum AB + BC a maximum?

    Last edited by violetice; January 28th 2009 at 10:59 AM. Reason: picture doesn't show up
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  2. #2
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    Hello, violetice!

    Code:
                    A
                  * * *
              *     |x *  *
            *       |     * *
           *      E + - - - -* B
                    |     *  |
          *         |θ *     |*
          *        O*        |*
          *         |        |*
                    |        |
           *        |        *
            *       |       *|
              *     |     *  |
          - - - - * * * - - -+- - - - m
                    D        C
    AD is a diameter of circle O and is tangent to line m at D.
    If AD=1 and BC \perp m,
    . . a) show that: . AB + BC \:= \:1 +\cos x - \cos^2\!x
    \Delta OAB is isosceles with OA = OB = \tfrac{1}{2}

    Then: . x \,=\, \angle OAB \,=\, \angle OBA \quad\Rightarrow\quad \angle AOB \,=\,\theta \,=\,180^o - 2x


    Law of Cosines: . AB^2 \:=\:OA^2 + OB^2 - 2(OA)(OB)\cos\theta

    We have: . AB^2 \:=\:\left(\tfrac{1}{2}\right)^2 + \left(\tfrac{1}{2}\right)^2 - 2\left(\tfrac{1}{2}\right)\left(\tfrac{1}{2}\right  )\cos(180^o-2x) \:=\:\tfrac{1}{4} + \tfrac{1}{4} - \tfrac{1}{2}[-\cos2x]

    . . AB^2 \:=\:\frac{1}{2}+\frac{1}{2}\cos2x \:=\;\frac{1+\cos2x}{2} \;=\;\cos^2\!x \quad\Rightarrow\quad\boxed{ AB \:=\:\cos x}


    We see that: . BC \:=\:AD - AE

    In right triangle AEB\!:\;\;AE \:=\:AB\cos x \:=\:\cos^2\!x

    Hence: . \boxed{BC \:=\:1 - \cos^2\!x}


    Therefore: . AB + BC \:=\:\cos x + 1 - \cos^2\!x




    b) What value of x makes the sum AB + BC a maximum?

    We have: . f(x) \:=\:1 + \cos x - \cos^2\!x

    Differentiate and equate to zero: . -\sin x + 2\cos x\sin x \:=\:0

    . . Factor: . \sin x(2\cos x - 1) \:=\:0


    We have: . \sin x \:=\:0\quad\Rightarrow\quad x \:=\:0 .
    This gives a minimum sum.

    And: . 2\cos x-1\:=\:0\quad\Rightarrow\quad \cos x \:=\:\tfrac{1}{2} \quad\Rightarrow\quad \boxed{x \:=\:\frac{\pi}{3}}

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  3. #3
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    A very pretty solution! thank you very much!

    I also just noticed in part b, you could you the quadratic formula for 1-cosx-cosx^2 (and it'd basically be just -x^2-cosx+1) and find the maximum that way.

    ...or am i wrong?
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  4. #4
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    No, he took the derivitive of the function and found the zeros, this is different than just finding the zeros...
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  5. #5
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    No, I meant to not find the zeros, but find the maximum point of the parabola.
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  6. #6
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    ...To find a maximum, you take the derivative of your function and find the zeros of that funciton.
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