Is there a simple rationale for learning “reference angle” I can give a 9th grader?

I'm helping a 9th grader review for his Algebra 2 Regents exam (New York State). They need to know how to find the "reference angle." (I did read Why teach reference angles?.)

I haven't found a simple reason that he will be able to relate to at his current stage, for why we would need or want the reference angle.

If there is one, please explain.

If there isn't one, that's okay, he understands that the educational system does involve some senseless hoops to go through.

Historical context could be interesting.

• Presumably this is in the context of learning to use the unit circle to describe sines and cosines of angles with measure larger than 90°, no? – mweiss Jun 12 '18 at 18:25
• Obviously not an answer to your question, but before calculators were available (when I learned trigonometry, in 1974), this never really needed justification. In fact, the tables typically only went up to 45 degrees, so you also had to know how to utilize co-functions. – Dave L Renfro Jun 12 '18 at 19:01
• Yes, especially if solving sides/angles of an obtuse triangle. The tables only went up to 45 degrees, so if you wanted to find the value (technically, an approximate value) for the cosine of 68 degrees, then you know this equals the sine of 22 degrees, and you look that up in the tables at the back of the book. And if you wanted the cosine of 117 degrees, then you know that's negative the cosine of 180 - 117 = 63 degrees, which in turn is negative the sine of 27 degrees (or just take the y-axis reference angle to begin with, which switches you from cosine to sine, but the negative sign stays). – Dave L Renfro Jun 12 '18 at 22:16
• Reference triangles provide the conceptual glue between two distinct notions: trigonometric ratios (SOH CAH TOA and all that) on the one hand, and trigonometric functions on the other. Trigonometric ratios don't really make sense for angles larger than 90° because you can't have an obtuse angle in a right triangle, so the way we make sense of $\sin \theta$ with $\theta \ge 90^\circ$ is to interpret it as the $y$-coordinate of a point on the unit circle. But this is also (up to a sign) the trigonometric ratio of $\sin \phi$ where $\phi$ is the corresponding reference angle. – mweiss Jun 13 '18 at 3:28
• @aparente001 Looking at one question in isolation may be missing the point. The reason you learn reference angles is because it makes it possible for you to do something else (e.g. find the sine of an angle not in the range $[0, 90°]$). The prerequisite skill (find the reference angle for $\theta = 135^\circ$) and the application of it (find the value of $\cos 135^\circ$) may be assessed in two separate questions but that doesn't mean your explanation for the first question cannot appeal to the second question for motivation. – mweiss Jun 13 '18 at 3:48