# Trigo - Showing

• Jan 12th 2013, 04:42 PM
fActor
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
• Jan 12th 2013, 06:41 PM
chiro
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.
• Jan 13th 2013, 04:13 PM
Soroban
Re: Trigo - Showing
Hello, fActor!

Quote:

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

Quote:

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:

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

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

Quote:

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:

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