**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.

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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.

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

Practice Problem:

ReplyDeleteUse 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.