Chapter 4.3, Problem 23E

Solution

To choose: The point on the terminal side of the angle $\theta =\frac{7\pi }{6}$ .

The point on the terminal side of the angle $\theta =\frac{7\pi }{6}$ is $\left(-\sqrt{3},-1\right)$ .

Given information: The given angle is $\theta =\frac{7\pi }{6}$ .

Formula used: Consider a right triangle with $\theta =\frac{7\pi }{6}$ as the angle at the standard position, $x$ as the base of the triangle, $y$ as the perpendicular of the triangle, and $r$ as the distance from the origin to the terminal side of the angle making it the hypotenuse of the triangle.

  $\tan \theta =\frac{y}{x}$

Calculation:

Substitute $\frac{7\pi }{6}$ for $\theta$ in the tangent formula and calculate the ratio $\frac{y}{x}$ .

  $\begin{array}{c}\tan \frac{7\pi }{6}=\frac{y}{x}\\ \tan \left(\pi +\frac{\pi }{6}\right)=\frac{y}{x}\\ \tan \left(\frac{\pi }{6}\right)=\frac{y}{x}\\ \frac{1}{\sqrt{3}}=\frac{y}{x}\end{array}$

Calculate the ratio of the two coordinates for point a. The point a is $\left(-\sqrt{3},-1\right)$ .

  $\begin{array}{c}\frac{y}{x}=\frac{-1}{-\sqrt{3}}\\ \frac{y}{x}=\frac{1}{\sqrt{3}}\end{array}$

Calculate the ratio of the two coordinates for point b. The point b is $\left(-1,\sqrt{3}\right)$ .

  $\begin{array}{c}\frac{y}{x}=\frac{\sqrt{3}}{-1}\\ \frac{y}{x}=-\sqrt{3}\end{array}$

Calculate the ratio of the two coordinates for point c. The point c is $\left(-\sqrt{3},1\right)$ .

  $\begin{array}{c}\frac{y}{x}=\frac{1}{-\sqrt{3}}\\ \frac{y}{x}=-\frac{1}{\sqrt{3}}\end{array}$

It can be observed that value of $\frac{y}{x}$ is equal to $\frac{1}{\sqrt{3}}$ for the point a. So, the point a is on the terminal side of $\theta$ .

Therefore, the point on the terminal side of the angle $\theta =\frac{7\pi }{6}$ is $\left(-\sqrt{3},-1\right)$ .