Chapter 6, Problem 1CLT

a.

To determine

The sketching of the angle in standard position.

Answer to Problem 1CLT

  

Explanation of Solution

Given information:

Consider the angle $\theta =-{120}^0$ .

  $(a)$ Sketch the angle in standard position.

Calculation:

From the given information, consider an angle is $\theta =-{120}^0$ .

The sketch of the angle $\theta =-{120}^0$ in standard deviation is as follows:

  

The initial side is the positive $x-$ axis, and terminal side is $120$ degrees clockwise from the initial side.

b.

To determine

The conterminal angle in the interval.

Answer to Problem 1CLT

  ${240}^0$ .

Explanation of Solution

Given information:

Consider the angle $\theta =-{120}^0$ .

  $(b)$ Determine a conterminal angle in the interval $\left[0^0{\text{ ,360}}^0\right]$ .

Calculation:

From the given information, consider an angle is $\theta =-{120}^0$ .

The conterminal angle in the interval $\left[0^0{\text{ ,360}}^0\right]$ are as follows:

  

The conterminal angle in the interval $\left[0^0{\text{ ,360}}^0\right]$ is ${240}^0$ or just add $360$ degrees to $-120$ degrees:

  $\begin{array}{l}-{120}^0+{360}^0\\ ={240}^0\end{array}$

Thus, the conterminal angle in the interval $\left[0^0{\text{ ,360}}^0\right]$ is ${240}^0$ .

c.

To determine

The angle in radian measure.

Answer to Problem 1CLT

  $-\frac{2\pi }{3}$.

Explanation of Solution

Given information:

Consider the angle $\theta =-{120}^0$ .

  $(c)$ Rewrite the angle in radian measure as a multiple of $\pi$ .

Calculation:

To convert an angle in degrees to radians, will multiply by $\frac{\pi }{180}$ .

  $-{120}^0=-120\times \frac{\pi }{180}=\frac{120\pi }{180}=-\frac{2\pi }{3}rad$

Thus, the angle in radian measure as a multiple of $\pi$ is $-\frac{2\pi }{3}$.

d.

To determine

The reference angle.

Answer to Problem 1CLT

  ${60}^0$.

Explanation of Solution

Given information:

Consider the angle $\theta =-{120}^0$ .

  $(d)$ The reference angle ${\theta }^{\text{'}}$ .

Calculation:

To find the conterminal angle in $\left[0^0,{360}^0\right)$ , that is part $(b)$ is answer. Since the conterminal angle is in the third quadrant, subtract ${180}^0$ to obtain the reference angle.

  $-{120}^0+{360}^0={240}^0\Rightarrow {\theta }^{\text{'}}={240}^0-{180}^0={60}^0$

Thus, the reference angle is ${60}^0$.

e.

To determine

The exact values of six trigonometric functions.

Answer to Problem 1CLT

  $\begin{array}{l}\sin (-{120}^0)=-\sin {60}^0=-\frac{\sqrt{3}}{2}\\ \cos (-{120}^0)=-\cos {60}^0=-\frac{1}{2}\\ \tan (-{120}^0)=\tan {60}^0=\sqrt{3}\end{array}$

  $\begin{array}{l}\csc (-{120}^0)=\frac{1}{\sin (-{120}^0)}=-\frac{2}{\sqrt{3}}=-\frac{2\sqrt{3}}{3}\\ \sec (-{120}^0)=\frac{1}{\cos (-{120}^0)}=-\frac{2}{1}=-2\\ \cot (-{120}^0)=\frac{1}{\tan (-{120}^0)}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\end{array}$.

Explanation of Solution

Given information:

Consider the angle $\theta =-{120}^0$ .

  $(e)$ Find the exact values of the six trigonometric functions of $\theta$ .

Calculation:

For the basic three trigonometric functions value $(\sin ,\cos ,\tan )$ , will use the reference angle is ${60}^0$ and ASTC to determine whether each trigonometric values should be positive or negative.

  $\begin{array}{l}\sin (-{120}^0)=-\sin {60}^0=-\frac{\sqrt{3}}{2}\\ \cos (-{120}^0)=-\cos {60}^0=-\frac{1}{2}\\ \tan (-{120}^0)=\tan {60}^0=\sqrt{3}\end{array}$

For the other three trigonometric function values, will use the reciprocal of the basic three above trigonometric function values:

  $\begin{array}{l}\csc (-{120}^0)=\frac{1}{\sin (-{120}^0)}=-\frac{2}{\sqrt{3}}=-\frac{2\sqrt{3}}{3}\\ \sec (-{120}^0)=\frac{1}{\cos (-{120}^0)}=-\frac{2}{1}=-2\\ \cot (-{120}^0)=\frac{1}{\tan (-{120}^0)}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\end{array}$