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Soroban · #2 November 21st, 2008, 09:45 PM

Hello, xwrathbringerx!

Quote:

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

Quote:

(b) Find a similar expression for $BP$ in terms of $\theta.$

In right triangle $BPA\!:\;\;\sin\theta \:=\:\frac{BP}{2} \quad\Rightarrow\quad BP \:=\:2\sin\theta$

Quote:

(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 $PQO\!:\;\begin{array}{c}\cos2\theta = \frac{OQ}{OP} \quad\Rightarrow\quad x \:=\:OQ \:=\:\cos2\theta \\ \\[-4mm] \sin2\theta =\frac{PQ}{OP} \quad\Rightarrow\quad y \:=\:PQ\:=\:\sin2\theta \end{array}$

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

Quote:

(d) Prove that: . $\Delta APQ\,\sim\,\Delta ABP$

Both are right triangles that have a common acute angle: . $\theta = \angle A$

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

The following users thank Soroban for this useful post:
xwrathbringerx
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