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Precalculus JumpStart

Anthony van Groningen

Section 9.3 The Reference Angle

Associated to every non-quadrantal angle is an acute angle called its reference angle. We will see that the reference angle and the quadrant completely determine its trigonometric function values.

Definition 9.17. The Reference Angle.

Suppose \(\theta\) is a non-quadrantal angle in standard position. The reference angle of \(\theta\text{,}\) denoted by \(\reference{\theta}\text{,}\) is the smallest positive angle formed by the terminal ray of \(\theta\) and the \(x\)-axis.
Study the Definition 9.17 above and each figure below to see how the reference angle is determine in each quadrant for \(0 \lt \theta \lt 2\pi\text{.}\)
Figure 9.18. QII: \(\reference{\theta}=\pi-\theta\text{.}\)
Figure 9.19. QI: \(\reference{\theta}=\theta\text{.}\)
Figure 9.20. QIII: \(\reference{\theta}=\theta-\pi\text{.}\)
Figure 9.21. QIV: \(\reference{\theta}=2\pi-\theta\text{.}\)

Example 9.22. Using the Reference Angle.

Suppose \(\theta = 11\pi/3\text{.}\)
The reference angle is formed by the terminal ray and the positive \(x\)-axis. This smallest such positive angle is \(\reference{\theta} = \pi/3\text{.}\) Drawing and \(\theta\) and \(\reference{\theta}\) both in standard position highlights why its a useful to consider the reference angle.
If \((x,y)\) is a point on the terminal side of \(\reference{\theta}\text{,}\) then after adjusting for the quadrant, \((x,-y)\) is a point on the terminal side of \(\theta\text{.}\) Thus, all the trigonometric function values of \(\theta\) and \(\reference{\theta}\) are the same or opposite and the sign is determined entirely by the quadrant. To illustrate, we know sine is negative in Quadrant IV. So,
\begin{equation*} \sin(11\pi/3) = - \sin(\reference{\theta}) = - \sin(\pi/3) = - \frac{\sqrt{3}}{2}, \end{equation*}
where we obtained the last value by referring to the special triangle Figure 8.24. Since cosine is positive in Quadrant IV,
\begin{equation*} \cos(11\pi/3) = + \cos(\reference{\theta}) = \cos(\pi/3) = \frac{1}{2}. \end{equation*}
Since tangent is negative in Quadrant IV,
\begin{equation*} \tan(11\pi/3) = - \tan(\reference{\theta}) = -\tan(\pi/3) = -\sqrt{3}. \end{equation*}
Thus, the use of the reference angle reduces all computations of trigonometric function values down to that of a positive acute angle.

Example 9.24. Using the Reference Angle.

Use the reference angle together with the special triangles Figure 8.24 and Figure 8.22 to find the sine, cosine, and tangent of each angle.
  1. \(\displaystyle \theta = \frac{7\pi}{6}\)
  2. \(\displaystyle \theta = -\frac{5\pi}{4}\)
  3. \(\displaystyle \theta = 780^\circ\)
Solution.
  1. \(\theta\) is in Quadrant III with reference angle \(\reference{\theta} = \pi/6\text{.}\)
    \begin{gather*} \sin(7\pi/6) = - \sin(\pi/6) = -1/2\\ \cos(7\pi/6) = - \cos(\pi/6) = -\sqrt{3}/2\\ \tan(7\pi/6) = + \tan(\pi/6) = 1/{\sqrt{3}} \end{gather*}
  2. \(\theta\) is in Quadrant II with reference angle \(\reference{\theta} = \pi/4\text{.}\)
    \begin{gather*} \sin\left(-\frac{5\pi}{4}\right) = \sin(\pi/4) = \sqrt{2}/2\\ \cos\left(-\frac{5\pi}{4}\right) = -\cos(\pi/4) = -\sqrt{2}/2\\ \tan\left(-\frac{5\pi}{4}\right) = - \tan(\pi/4) = -1 \end{gather*}
  3. \(\theta = 780^\circ = (60+360+360)^\circ\) is in Quadrant I with reference angle \(\reference{\theta} = 60^\circ\text{.}\)
    \begin{gather*} \sin(780^\circ) = \sin(60^\circ) = \sqrt{3}/2\\ \cos(780^\circ) = \cos(60^\circ) = 1/2\\ \tan(780^\circ) = \tan(60^\circ) = \sqrt{3} \end{gather*}
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