Different ways of solving a contest problem (Part 2)

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 triangles. Here’s a second solution that I received: begin by squaring both sides and using a Pythagorean trig identity.

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

We use the Pythagorean identity again to find \sin \theta:

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

\displaystyle \frac{9}{10} = 1 - \sin^2 \theta

\sin^2 \theta = \displaystyle \frac{1}{10}

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

Therefore, we know that

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

so the answer is either \displaystyle \frac{3}{10} or \displaystyle -\frac{3}{10}. However, this was a multiple-choice contest problem and \displaystyle -\frac{3}{10} was not listed as a possible answer, and so the answer must be \displaystyle \frac{3}{10}.

green lineFor a contest problem, the above logic makes perfect sense. However, the last step definitely plays to the fact that this was a multiple-choice problem, and the concluding step would not have been possible had \displaystyle -\frac{3}{10} been given as an option.


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