1. ## 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.

2. 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]

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