Different ways of solving a contest problem (Part 1)

Precalculus 1 Minute

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.

Here’s the first solution that I received: draw the appropriate triangles for the angle \theta:

3 \sin \theta = \cos \theta

\tan \theta = \displaystyle \frac{1}{3}

Therefore, the angle \theta must lie in either the first or third quadrant, as shown. (Of course, \theta could be coterminal with either displayed angle, but that wouldn’t affect the values of \sin \theta or \cos \theta.)

AHSME problem

In Quadrant I, \sin \theta = \displaystyle \frac{1}{\sqrt{10}} and \cos \theta = \displaystyle \frac{3}{\sqrt{10}}. Therefore,

\sin \theta \cos \theta = \displaystyle \frac{1}{\sqrt{10}} \times \frac{3}{\sqrt{10}} = \displaystyle \frac{3}{10}.

In Quadrant III, \sin \theta = \displaystyle -\frac{1}{\sqrt{10}} and \cos \theta = -\displaystyle \frac{3}{\sqrt{10}}. Therefore,

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

Either way, we can be certain that \sin \theta \cos \theta = \displaystyle \frac{3}{10}.

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Published by John Quintanilla

I'm a Professor of Mathematics and a University Distinguished Teaching Professor at the University of North Texas. For eight years, I was co-director of Teach North Texas, UNT's program for preparing secondary teachers of mathematics and science.

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3 thoughts on “Different ways of solving a contest problem (Part 1)”

  2. Pingback: Different Ways of Solving a Contest Problem: Index | Mean Green Math

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