Chapter 5.1, Problem 34E

To determine

To find:the terminal point made by different angle given below.

Answer to Problem 34E

  $\left(\text{a}\right)$$\text{P}\left(\frac{\text{3}}{\text{4}}\text{,-}\frac{\sqrt{\text{7}}}{\text{4}}\right)$ , $\left(\text{b}\right)$$\text{P}\left(\frac{\text{3}}{\text{4}}\text{,}\frac{\sqrt{\text{7}}}{\text{4}}\right)$ , $\left(\text{c}\right)$$\text{P}\left(-\frac{\text{3}}{\text{4}}\text{,}\frac{\sqrt{\text{7}}}{\text{4}}\right)$ and $\left(\text{d}\right)$$\text{P}\left(-\frac{\text{3}}{\text{4}}\text{,-}\frac{\sqrt{\text{7}}}{\text{4}}\right)$ .

Explanation of Solution

Given:

Reference number is

  $\begin{array}{l}\left(\text{a}\right)\text{-t }\left(\text{b}\right)\text{4π+t }\\ \left(\text{c}\right)\text{π-t }\left(\text{d}\right)\text{t-π}\end{array}$ .

Concept used:

Terminal point:

The angle made by the end point with nearest horizontal x-axis or the reference plane.

For a unit circle:

  $\text{r=1}$ .

  $\text{x=cosθ}$ .

  $\text{y=sinθ}$ .

In 1^st quadrant all trigonometry ratios are positive.

In 2^nd quadrant only $\text{sinθ and cosecθ}$ is positive.

In 3^rd quadrant only $\text{tanθ and cotθ}$ is positive.

In 4^th quadrant only $\text{cosθ and secθ}$ is positive.

Calculation:

Adding or subtracting multiples of $\text{π}$ will not change the reference number.

The reference number is given by $\left(\frac{3}{4},\frac{\sqrt{7}}{4}\right)$ .

Since the terminal point t is in the quadrant 1^ST.

  $\left(\text{a}\right)$ .Terminal point determined by $\text{-t}$ is in the 4^th quadrant. More specifically, the point will be: $\text{P}\left(\frac{\text{3}}{\text{4}}\text{,-}\frac{\sqrt{\text{7}}}{\text{4}}\right)$ .

Since the point lies in 4^th quadrant where $\text{cosθ}$ is positive and $\text{sinθ}$ is negative therefore x-coordinate is positive and y- coordinate is negative.

For a unit circle:

  $\text{r=1}$ .

  $\text{x=cosθ}$ .

  $\text{y=sinθ}$ .

Since the point lies in 4^th quadrant where $\text{cosθ}$ is positive and $\text{sinθ}$ is negative therefore x-coordinate is positive and y- coordinate is negative.

  $\left(\text{b}\right)$ . Terminal point determined by $\text{4π+t}$ is in the 1^stquadrant. More specifically, the point will be:

  $\text{P}\left(\frac{\text{3}}{\text{4}}\text{,}\frac{\sqrt{\text{7}}}{\text{4}}\right)$ .

since the point lies in 1^st quadrant. where $\text{cosθ}$ is positive and $\text{sinθ}$ is also positive therefore x-coordinate is positive and y- coordinate is positive.

  $\left(\text{c}\right)$ . Terminal point determined by $\text{π-t}$ is in the quadrant no 2^nd more specifically, the point will be:

  $\text{P}\left(-\frac{\text{3}}{\text{4}}\text{,}\frac{\sqrt{\text{7}}}{\text{4}}\right)$ .

Since the point lies in 2^nd quadrant where $\text{cosθ}$ is negative and $\text{sinθ}$ is positive therefore x-coordinate is negative and y- coordinate is positive.

  $\left(\text{d}\right)$ . Terminal point determined by $\text{t-π}$ is in the 3^rdquadrant More specifically, the point will be:

  $\text{P}\left(-\frac{\text{3}}{\text{4}}\text{,-}\frac{\sqrt{\text{7}}}{\text{4}}\right)$ .

Since the point lies in 3^rdquadrant where $\text{cosθ}$ is negative and $\text{sinθ}$ is also negative therefore x-coordinate is negative and y- coordinate is also negative.

Hence, $\left(\text{a}\right)$$\text{P}\left(\frac{\text{3}}{\text{4}}\text{,-}\frac{\sqrt{\text{7}}}{\text{4}}\right)$ , $\left(\text{b}\right)$$\text{P}\left(\frac{\text{3}}{\text{4}}\text{,}\frac{\sqrt{\text{7}}}{\text{4}}\right)$ , $\left(\text{c}\right)$$\text{P}\left(-\frac{\text{3}}{\text{4}}\text{,}\frac{\sqrt{\text{7}}}{\text{4}}\right)$ and $\left(\text{d}\right)$$\text{P}\left(-\frac{\text{3}}{\text{4}}\text{,-}\frac{\sqrt{\text{7}}}{\text{4}}\right)$ .