1. ## Trigo - Showing

1. Show that OM = 6 cos θ - 2 sin θ
2. Show that area of triangle OAM is 10 sin 2(θ + α)
[OM = R cos (θ + α)]
[AM = R sin (θ + α)]
3. Given that θ can vary, find the maximum value of the area of triangle OAM.

Refer to: http://i49.tinypic.com/1pueqq.png

ABO & BNO = right angled triangles
OB = 6 cm
AB = 2 cm
line AM perpendicular to ON

2. ## Re: Trigo - Showing

Hey fActor.

Can you show us what you have tried? (Also show us all the identities that you can derive from the information given). This approach is typically the best one when helping with these kinds of problems.

3. ## Re: Trigo - Showing

Hello, fActor!

Code:
                A
o
*:*
* : * 2
*  :θ *
*   :   *
*    :    * B
*    P* - - o
*    6 : *   *
* α   * :     *
*  * θ   :     *
O o * * * * o * * o N
M
$\angle ABO = \angle BNO = 90^o,\;OB = 6,\;AB = 2,\;AM \perp ON.$

Note that: $\angle BON = \angle BAM = \theta.$

Let $\alpha = \angle AOB.$

Draw $BP \parallel ON.$

1. Show that: $OM \:=\: 6\cos\theta - 2\sin\theta$

We see that: . $OM \:=\:ON - MN$

In $\Delta BNO\!:\:ON = 6\cos\theta$

In $\Delta BPA\!:\:BP = 2\sin\theta = MN$

Therefore: . $OM \:=\:6\cos\theta - 2\sin\theta$

2. Show that area of triangle $OAM$ is: $10\sin2(\theta + \alpha)$

The area of $\Delta OAM\:=\:\tfrac{1}{2}(OM)(AM)$

We note that: . $OA \:=\:\sqrt{2^2+6^2} \:=\:\sqrt{40}$

In $\Delta OAM\!:\;\begin{Bmatrix}OM \:=\: 2\sqrt{10}\cos(\theta + \alpha) \\ AM \:=\:2\sqrt{10}\sin(\theta+\alpha) \end{Bmatrix}$

Therefore:

. . $\Delta OAM \;=\;\tfrac{1}{2}\cdot \sqrt{40}\cos(\theta+\alpha)\cdot \sqrt{40}\sin(\theta+\alpha) \;=\;20\sin(\theta+\alpha)\cos( \theta+\alpha)$

. . . . . . . . $=\;10\cdot\underbrace{2\sin(\theta+\alpha)\cos(\th eta+\alpha)}_{\text{Double-angle identity}} \;=\;10\sin2(\theta+\alpha)$

3. Given that $\theta$ can vary, find the maximum value of the area of $\Delta OAM.$

We have: . $A \;=\;10\sin2(\theta +\alpha)$

Set the derivative equal to zero and solve.

. . $\frac{dA}{d\theta} \:=\:10\cos2(\theta+\alpha)\cdot 2 \;=\;0 \quad\Rightarrow\quad \cos2(\theta+\alpha) \:=\:0$

. . $2(\theta + \alpha) \:=\:\tfrac{\pi}{2} \quad\Rightarrow\quad \theta + \alpha \:=\:\tfrac{\pi}{4} \quad\Rightarrow\quad \theta \;=\;\tfrac{\pi}{4} - \alpha$

Therefore:

. . $A \;=\;10\sin2(\tfrac{\pi}{4} - \alpha + \alpha) \;=\;10\sin2(\tfrac{\pi}{4}) \;=\;10\sin\tfrac{\pi}{2} \;=\;10\cdot1 \;=\;10$