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

If , what is ?

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 :

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

In Quadrant I, and . Therefore,

.

In Quadrant III, and . Therefore,

.

Either way, we can be certain that .

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## howardat58

/ September 20, 2015Here’s another:

3sin(t)=cos(t)

9sinsq(t)=cossq(t)=1-sinsq(t)

10sinsq(t)=1

sinsq(t)=1/10

and with 3sin(t)=cos(t)

sin(t)cos(t)=3sinsq(t)=3/10

## John Quintanilla

/ September 22, 2015Indeed, this was one of the other solutions I received: https://meangreenmath.com/2015/09/22/different-ways-of-solving-a-contest-problem-part-3/