Different ways of solving a contest problem (Part 3)

The following problem appeared on the American High School Mathematics Examination (now called the AMC 12) in 1988:

If $3 \sin \theta = \cos \theta$ , what is $\sin \theta \cos \theta$ ?

When I presented this problem to a group of students, I was pleasantly surprised by the amount of creativity shown when solving this problem.

Yesterday, I presented a solution using a Pythagorean identity, but I was unable to be certain if the final answer was a positive or negative without drawing a picture. Here’s a third solution that also use a Pythagorean trig identity but avoids this difficulty. Again, I begin by squaring both sides.

$9 \sin^2 \theta = \cos^2 \theta$

$9 (1 - \cos^2 \theta) = \cos^2 \theta$

$9 - 9 \cos^2 \theta = \cos^2 \theta$

$9 = 10 \cos^2 \theta$

$\displaystyle \frac{9}{10} = \cos^2 \theta$

$\displaystyle \pm \frac{3}{\sqrt{10}} = \cos \theta$

Yesterday, I used the Pythagorean identity again to find $\sin \theta$ . Today, I’ll instead plug back into the original equation $3 \sin \theta = \cos \theta$ :

$3 \sin \theta = \cos \theta$

$3 \sin \theta = \displaystyle \frac{3}{\sqrt{10}}$

$\sin \theta = \displaystyle \pm \frac{1}{\sqrt{10}}$

Unlike the example yesterday, the signs of $\sin \theta$ and $\cos \theta$ must agree. That is, if $\cos \theta = \displaystyle \frac{3}{\sqrt{10}}$ , then $\sin \theta = \displaystyle \frac{1}{\sqrt{10}}$ must also be positive. On the other hand, if $\cos \theta = \displaystyle -\frac{3}{\sqrt{10}}$ , then $\sin \theta = \displaystyle -\frac{1}{\sqrt{10}}$ must also be negative.

If they’re both positive, then

$\sin \theta \cos \theta = \displaystyle \left( \frac{1}{\sqrt{10}} \right) \left( \frac{3}{\sqrt{10}} \right) =\displaystyle \frac{3}{10}$ ,

and if they’re both negative, then

$\sin \theta \cos \theta = \displaystyle \left( -\frac{1}{\sqrt{10}} \right) \left( -\frac{3}{\sqrt{10}} \right) = \displaystyle \frac{3}{10}$ .

Either way, the answer must be $\displaystyle \frac{3}{10}$ .

This is definitely superior to the solution provided in yesterday’s post, as there’s absolutely no doubt that the product $\sin \theta \cos \theta$ must be positive.

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Different ways of solving a contest problem (Part 3)

Published by John Quintanilla

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Published by John Quintanilla

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