# Why teach reference angles?

Reference angles are most useful for limited trigonometric tables with values from 0 to $\frac{\pi}{2}$. They can also help illustrate the periodicity of trigonometric functions, but I feel this is done more effectively with graphing. Should we still teach them, and if so how to make them more relevant?

• What are reference angles? – Ben Crowell Nov 23 '17 at 23:18

For students who will only be using right triangle trig, reference angles are definitely important.

For example, I (sometimes) teach a one-credit course on trig for forestry students, and their main application is moving along (or describing) a path connecting a series of points on a map. Note that they do not measure angles counterclockwise, relative to the positive $x$-axis as we do in math, but rather clockwise, relative to North (called an azimuth).

Thus, a movement such as "travel $68$ feet at $213.5^{\circ}$ boils down to their understanding of a reference angle to make a right triangle. Of course, this angle would be in reference to any convenient cardinal direction so they can determine, say, a northing or easting for that leg of the path.

Now, it would be awkward to try to force these students to learn a new convention for angles, as azimuths are the standard in their field. So, I stay in their world, having them sketch their particular path on the plane and using a reference angle to draw a right triangle, etc.

• It's exciting to me to hear about this application. I'd love a little bit more detail. Could you could link to something? – Sue VanHattum Nov 23 '17 at 4:47
• @SueVanHattum I don't have any particular links to external material, but you could search for "surveying, traverse, azimuth". – Nick C Nov 24 '17 at 5:57

I disagree with your statement on the usefulness of reference angles, they have more use in the plane of the unit circle. So to answer the question specifically, yes we should still teach them and one idea on relevance is below in a Physics I problem.

Trigonometric functions in a complete unit circle are not one to one so the concept of a reference angle is needed to orient the inverse functions to the coordinate plane.

Teaching physics primarily and mathematics secondary when using the trigonometric functions to find components in the forward direction this idea is not needed, but when solving for the angle and using the inverse trigonometric functions the understanding of a reference angle is very important.

For example:

If I have 4 forces

$F_1= 12N@30^{\circ}$, $F_2=12N@150^{\circ}$, $F_3=12N@210^{\circ}$ and $F_4=12N@300^{\circ}$

The resultant force can be found by adding the components in each co-ordinate direction:

$\Sigma F_x=12 N \cos{30^{\circ}}+12N \cos{150^{\circ}}+12N \cos{210^{\circ}}+12N\cos{300^{\circ}}$

$\Sigma F_y=12 N \sin{30^{\circ}}+12N \sin{150^{\circ}}+12N \sin{210^{\circ}}+12N\sin{300^{\circ}}$

This solves nicely with the large angles keeping track of the positive and negative values of the components nicely:

$\Sigma F_x \approx -4.4 N$

$\Sigma F_y \approx -4.4 N$

The Magnitude of the Force is found:

$F_R=\left[(F_x)^2+(F_y)^2\right]^{\frac{1}{2}}$

$F_R\approx 6.2 N$

When we want to find the direction of this vector we use: $\theta = \arctan{\frac{F_y}{F_x}}$

$\theta = 45^{\circ}$

This is a reference angle. The student needs to apply the concept of the reference angle and recognize that because both components of the vector were negative that this would be in the Third Quadrant and the angle relative to the positive x axis is $\theta=225^{\circ}$.

So because of the nature of the inverse trigonometric functions when mapping from an answer to an $\arcsin$, $\arccos$ and $\arctan$ students need to understand the concept. It is worth spending time on the concept especially as the concept of angle is increased beyond its initial geometric definitions.

• +1 for the mention of the output of arctan(x). – Nick C Nov 9 '17 at 20:58

Reference angles connect triangles (especially the 2 special triangles, 30-60-90 and 45-45-90) to the unit circle. If I want to think about the sine of 120 degrees, knowing that its reference angle is 60 degrees allows me to figure out the sine from the 30-60-90 triangle.