Question

Find (a) the reference number for each value of $t$ and (b) the terminal point determined by $t$.

$t=\frac{31\pi }{6}$

Short answer

a. The reference number $\overline{t}=\frac{\pi }{6}$.

b. The terminal point for$t=\frac{31\pi }{6}$ is $P\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)$.

Step-by-step solution

Part a. Step 1. State the definition of the reference number.

The reference number$\overline{t}$ associated with $t$, a real number is the shortest distance along the unit circle between the terminal point determined by $t$ and the $x$-axis.

Part a. Step 2. Determine the quadrant where the terminal point lies.

If $0<t<\frac{\pi }{2}$, then $t$ lies in quadrant I and the reference number $\overline{t}=t$.

If $\frac{\pi }{2}<t<\pi$, then $t$ lies in quadrant II and the reference number $\overline{t}=\pi -t$.

If $\pi <t<\frac{3\pi }{2}$, then $t$ lies in quadrant III and the reference number $\overline{t}=t-\pi$.

If $\frac{3\pi }{2}<t<2\pi$, then $t$ lies in quadrant IV and the reference number $\overline{t}=2\pi -t$.

Here, $t=\frac{31\pi }{6}$, which lies in quadrant II.

Part a. Step 3. Find the reference number.

Since $t$ lies in quadrant II, therefore, the reference number will be $\overline{t}=t-5\pi$. Here substitute $\frac{31\pi }{6}$ for $t$ and simplify for $\overline{t}$.

$$\begin{gathered} \overline{t}=t-5\pi \\ =\frac{31\pi }{6}-5\pi \\ =\frac{\pi }{6} \end{gathered}$$

Therefore, the reference number $\overline{t}=\frac{\pi }{6}$.

Part b. Step 1. State the definition of the terminal point on the unit circle.

Starting at the point $\left(1,0\right)$, if $t\ge 0$, then $t$ is the distance along the unit circle in clockwise direction and if $t<0$, then$\left|t\right|$ is the distance along the unit circle in clockwise direction. Here, $t$ is a real number and this distance generates a point$P\left(x,y\right)$ on the unit circle. The point$P\left(x,y\right)$ obtained in this way is called the terminal point determined by the real number $t$.

Part b. Step 2. State the terminal points for some special values of t for reference.

The table below gives the terminal points for some special values of $t$.

$t$	Terminal point determined by $t$
0	$\left(1,0\right)$
$\frac{\pi }{6}$	$\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$
$\frac{\pi }{4}$	$\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$
$\frac{\pi }{3}$	$\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$
$\frac{\pi }{2}$	$\left(0,1\right)$

Part b. Step 3. Simplify and state the conclusion.

Let $P\left(x,y\right)$be the terminal point for $t=\frac{31\pi }{6}$.

Since the reference number for $t=\frac{31\pi }{6}$is $\overline{t}=\frac{\pi }{6}$and the corresponding terminal point for $\overline{t}=\frac{\pi }{6}$is $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$in the first quadrant, note that $t=\frac{31\pi }{6}$ lies in the quadrant II, therefore, its terminal point will have negative $x$-coordinate and positive y-coordinate. Therefore, the terminal point for $t=\frac{31\pi }{6}$is $P\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)$.