Question

Find the terminal point on the unit circle determined by $\frac{2\mathrm{\pi }}{3}$ radians $?$

Answer 1

Find the terminal point on the unit circle determined by $\frac{2\mathrm{\pi }}{3}$ radians.

Any point $\mathrm{P}(\mathrm{x},\mathrm{y})$ on the unit circle of radius $\mathrm{r}$ and center at origin is given by $(\mathrm{r}\cos \mathrm{\theta },\mathrm{r}\sin \mathrm{\theta })$, where $\mathrm{\theta }$ is a parameter.

Here, $\mathrm{\theta }=\frac{2\mathrm{\pi }}{3}$ and $\mathrm{r}=1$

Since $\frac{2\mathrm{\pi }}{3}$ is on the second quadrant, the sine is positive and the cosine is negative.

substitute $\mathrm{\theta }=\frac{2\mathrm{\pi }}{3}$ and $\mathrm{r}=1$in $(\mathrm{r}\cos \mathrm{\theta },\mathrm{r}\sin \mathrm{\theta })$:

$\begin{array}{rcl}\mathrm{P}(\mathrm{x},\mathrm{y})& =& \left[1\times \cos \frac{2\mathrm{\pi }}{3},1\times \sin \frac{2\mathrm{\pi }}{3}\right]\\ & =& \left[\cos \frac{2\mathrm{\pi }}{3},\sin \frac{2\mathrm{\pi }}{3}\right]\end{array}$

Subtract $\frac{\mathrm{\pi }}{2}$ to get the equivalent angle on the first quadrant:

$\begin{array}{lcc}\Rightarrow \mathrm{P}(\mathrm{x},\mathrm{y})=\left[\cos \left(\frac{2\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{2}\right),\sin \left(\frac{2\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{2}\right)\right]& & \\ \Rightarrow \mathrm{P}(\mathrm{x},\mathrm{y})=\left(\cos \frac{\mathrm{\pi }}{6},\sin \frac{\mathrm{\pi }}{6}\right)& & \\ \Rightarrow \mathrm{P}(\mathrm{x},\mathrm{y})=\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)\left[\because \cos \frac{\mathrm{\pi }}{6}=-\frac{\sqrt{3}}{2},\sin \frac{\mathrm{\pi }}{6}=\frac{1}{2}\right]& & \end{array}$

Hence, the terminal point on the unit circle determined by $\frac{2\mathrm{\pi }}{3}$ radians is$\begin{array}{l}\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)\end{array}$.