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Math Help - Trigonomic Proof.

  1. #1
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    Last edited by xMichaelx; July 2nd 2010 at 06:41 PM.
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  2. #2
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    Hello, xMichaelx!

    A Cone has a slant height m and a semi-vertical angle 2a

    Prove that the radius r of the inscribed sphere is given by:

    . . r\:=\: \frac {m\sin 4a}{2(1+\sin 2a)}
    Code:
                          A
                          o
                         /|\
                        / | \
                       /  |2a\
                      /   |   \
                     /    |    \
                    /   * * *   \
                   /*     |     *\  E
                  *       |       o
            m    *        |     *  *    m
                /         |   * r   \
               /*         | *       *\
              / *       P o         * \
             /  *         |         *  \
            /             |             \
           /     *        |r       *     \
          /       *       |       *       \
         /          *     |     *          \
      B o - - - - - - - * o * - - - - - - - o C
                          D

    The side view of the cone is \Delta ABC\!:\;AB = AC = m
    P is the center of the inscribed sphere: . PD = PE = r
    Note that \angle PEA = 90^o

    In right triangle AEP\!:\;\sin2a \:=\:\frac{r}{AP} \quad\Rightarrow\quad AP \:=\:\frac{r}{\sin2a}

    . . Hence: . AD \:=\:AP + PD \:=\:\frac{r}{\sin2a} + r \quad\Rightarrow\quad AD \:=\:r\left(\frac{1+\sin2a}{\sin2a}\right) .[1]


    In right triangle ADC\!:\:\cos2a \:=\:\frac{AD}{m} \quad\Rightarrow\quad AD \:=\:m\cos2a .[2]


    Equate [1] and [2]: . r\left(\frac{1+\sin2a}{\sin2a}\right) \:=\:m\cos2a \quad\Rightarrow\quad r \:=\:\frac{m\sin2a\cos2a}{1+\sin2a}


    Therefore: . r \:=\:\frac{m({\color{red}2}\sin2a\cos2a)}{{\color{  red}2}(1+\sin2a)} \quad\Rightarrow\quad r \;=\;\frac{m\sin4a}{2(1+\sin2a)}

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