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Math Help - Trig Problem

  1. #1
    Senior Member
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    Trig Problem



    (a) Explain why AP = 2cosθ.

    (b) Find a similar expression for BP in terms of θ.

    (c) Explain why the coordinates of pt P is (cos2θ, sin2θ).

    (d) Prove that triangle APQ is similar to triangle ABP.

    (e) Hence deduce that
    (i) sin2θ = 2sin θcosθ
    (ii) cos2θ = 2cos^2 θ - 1
    (iii) tan2θ = (2tanθ)/(1-tan^ 2 θ)

    Could someone please help me on this problem?? I've worked out parts a and b but added them in just in case it is required for the other parts.
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  2. #2
    Super Member

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    Hello, xwrathbringerx!

    Code:
                  * * *
              *           *  P
            *               o
           *           *  * |*
                   *    *   |
          *   * θ     * 2θ  | *
        A o - - - - o - - - * o B
          *         O       Q *
     
           *                 *
            *               *
              *           *
                  * * *

    (a) Explain why AP \:=\:2\cos\theta
    Draw chord PB.

    \angle APB is inscribed in a semicircle. .Hence: . \angle APB = 90^o

    In right triangle BPA\!:\;\;\cos\theta \:=\:\frac{AP}{AB}\quad\Rightarrow\quad AP \:=\:AB\!\cdot\cos\theta

    Since AB = 2\!:\;\;AP \:=\:2\cos\theta




    (b) Find a similar expression for BP in terms of \theta.
    In right triangle BPA\!:\;\;\sin\theta \:=\:\frac{BP}{2} \quad\Rightarrow\quad BP \:=\:2\sin\theta



    (c) Explain why the coordinates of pt P are: (\cos2\theta,\:\sin2\theta)
    Note that the radius is: OP = 1

    Inscribed \angle PAB = \theta is measured by \tfrac{1}{2}\text{arc}\:\!PB

    Central \angle POB is measured by \text{arc}\:\!PB

    Hence: . \angle POB = 2\theta

    In right triangle PQO\!:\;\begin{array}{c}\cos2\theta = \frac{OQ}{OP} \quad\Rightarrow\quad x \:=\:OQ \:=\:\cos2\theta \\ \\[-4mm]<br />
\sin2\theta =\frac{PQ}{OP} \quad\Rightarrow\quad y \:=\:PQ\:=\:\sin2\theta \end{array}

    Therefore, the coordinates of P are: . (\cos2\theta,\:\sin2\theta)




    (d) Prove that: . \Delta APQ\,\sim\,\Delta ABP
    Both are right triangles that have a common acute angle: . \theta = \angle A

    Therefore: . \Delta APQ \,\sim\,\Delta ABP

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