Finding Reference Angles in Precalculus?

Question

I'm reviewing for an exam, and having some trouble with reference angles depending on the quadrant they lie in. For example, my book shows the following:

I get the part about subtracting 12pi/6, but what I'm having issues with is deciding whether to add or subtract when the angle is NOT in quadrant IV.

For example:

Based on the book's other example, I would subract 12pi/6 from 31pi/6 until I get a result less than 12pi/6.

31 - 12 = 19

19 - 12 = 7

7pi/6 I believe would lie in Quadrant III, as 2pi is a full rotation, and 1pi is half, and 7pi/6 is a little over 1/2.

My question though, is compared to Quadrant IV, do I know subtract or add 7pi/6 to 12pi/6?

Being Quadrant III I figured I would still subtract, resulting in 12pi/6 - 7pi/6 = 5pi/6 but apparently the correct answer is pi/6???

Can someone please explain this to me?

Answer 1

In quadrant III, subtract $\pi$ from the value.

In quadrant II, subtract the value from $\pi$.

You don't need to subtract anything if it's in quadrant I!

Note: I may have misunderstood the question - I'm not sure if this is right.

Answer 2

After you get the angle at or below $2\pi$ the reference angle is just the angle $0\le x\le\frac{\pi}{2}$ that is the distance between your angle and the x axis.

So in quadrant I it is just x.

Quadrant II it is $\pi -x$

Quadrant III it is $x-\pi$

Quadrant IV it is $2\pi - x$

Some example reference angles in degrees...

30 is 30. 90 is 90. 95 is 85. 150 is 30. 190 is 10. 350 is 10. Just think of the unit circle and figure out how big the angle is between your angle and the x axis. The red angles in the image below.