Question

Find the terminal point$P\left(x,y\right)$ on the unit circle determined by the given value of $t$.

$t=\frac{11\pi }{6}$

Short answer

The terminal point on the unit circle is determined by $t=\frac{11\pi }{6}$ is $P\left(x,y\right)=\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)$.

Step-by-step solution

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 a clockwise direction, and if $t<0$, then $\left|t\right|$ is the distance along the unit circle in a 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$t$.

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	$\left(1,0\right)$
$\frac{\pi }{6}$	$\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$
$\frac{\pi }{4}$	$\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$
$\frac{\pi }{3}$	$\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$
$\frac{\pi }{2}$	$\left(0,1\right)$

Step 3. Simplify and state the conclusion.

Let $P$ be the terminal point determined by $\frac{11\pi }{6}$, and let $Q$ be the terminal point determined by $\frac{\pi }{6}$. Since $P$ is in the IV quadrant (because $\frac{11\pi }{6}=2\pi -\frac{\pi }{6}$), therefore, the point $P$ has the same coordinates as $Q$ except for the sign, i.e., the $x$-coordinate will be positive and it's $y$-coordinate will be negative. Since the terminal point for $t=\frac{\pi }{6}$is $Q\left(x,y\right)=\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$, thus, the terminal point for $t=\frac{11\pi }{6}$ will be$P\left(x,y\right)=\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)$ .