2
$\begingroup$

In my Trig class we have begun working on graphing the trig functions and working with radians and I'm trying to wrap my head around them.

At the moment I'm having trouble understanding radian measures and how to find where certain points lie on the unit circle and how to know what quadrant they are in.

For example, we are to find the reference angle of $\frac{5\pi}{6}$. My book says it terminates in QII.

This may be a dumb question, but how does one figure this out? What am I missing? How do you know that $\frac{\pi}{2} < \frac{5\pi}{6} < \pi$?

Thanks in advance...

  • 0
    Well you know $1/2 < 5/6 < 1.$ Since $\pi > 0,$ you get $\pi/2 < (5/6) \pi < \pi.$2011-03-06
  • 0
    The reference angle is always from the $x$-axis. If $\theta = \frac{2\pi}{3}$ for example, the canonical reference angle is $\pi - \frac{2\pi}{3} = \frac{\pi}{3}$ (answering the question, where is my angle from the $x$-axis?), and you use that to figure out values as if they lay in Quadrant I, and then afterward apply the mnemonic "All Students Take Calculus" to figure out the sign of the end result: All trig functions are positive in Quadrant I, sine is the only one positive in Quadrant II; tangent is the only one positive in Quadrant III; and cosine is the only one positive in Quadrant IV.2011-03-06
  • 0
    At least I think that's what you mean by your question.2011-03-06
  • 0
    I probably should have said "from the portion $x$-axis that lies adjacent to the quadrant the angle 'sits' in." The intuitive way of looking at it. E.g. the reference angle of $\frac{5\pi}{3}$ is $2\pi - \frac{5\pi}{3} = \frac{\pi}{3}$.2011-03-06

3 Answers 3