## Sunday, October 29, 2006

### Out Of Place Trig. Topic: Geometry/Trigonometry. Level: AMC/AIME.

Problem: (2006-2007 Warm-Up 4) Two sides of a triangle are $4$ and $|\cos{\theta}|$ and the angle between them is $\theta$. If $0 < \theta < \pi$, what is the maximum area of the triangle?

Solution: Well, given two sides and the angle between them, there is only one formula for the area of a triangle that comes to mind. $[ABC] = \frac{1}{2}ab \sin{(\angle C)}$. So plugging this in, we get the area to be

$\frac{1}{2} (4) |\cos{\theta}| \sin{\theta} = 2 \sin{\theta} |\cos{\theta}|$.

By the double-angle identity, this is equal to

$|\sin{(2\theta)}|$

for $0 < \theta < \pi$. But the maximum of this is clearly $1$ and it is obtained when $\theta = \frac{\pi}{4}, \frac{3 \pi}{4}$. QED.

--------------------

Comment: This problem shouldn't have been too hard; finding that specific formula for the area of the triangle could have been derived by drawing an altitude. And the double-angle identity should always be known. Lastly, maximizing the sine function is a given.

--------------------

Practice Problem: (2006 Skyview) If $\sin{x} = \frac{3}{5}$ and $\cos{x} = \frac{4}{5}$, what is $\sin{(3x)}$?

#### 1 comment:

1. Practice Problem:

Use De Moivre's: cis(3x) = (cis x)^3

Multiply out and equate imaginary parts:

sin 3x = 3 cos^2 x sin x - sin^3 x

Thus sin 3x = 117/125 from the given values.