Chapter 13.6, Problem 9PPS

To determine

To find the value of $\cos \theta$ and $\sin \theta$ .

Answer to Problem 9PPS

  $\cos \theta =\frac{6}{10}$ , $\sin \theta =-\frac{8}{10}$

Explanation of Solution

Given:

Point: $P\left(\frac{6}{10},-\frac{8}{10}\right)$ on a unit circle.

Concept used:

If the point in the angle’s terminal side is $P\left(x,y\right)$ then the trigonometric functions $\cos \theta$ and $\sin \theta$ can be calculated as:

  $\begin{array}{l}\cos \theta =\frac{x}{r}\\ \sin \theta =\frac{y}{r}\end{array}$

Where $r=\sqrt{x^2+y^2}$ .

Calculation:

Here, $P\left(\frac{6}{10},-\frac{8}{10}\right)$

Since the given circle is a unit circle,

  $r=1$

Calculating $\cos \theta =\frac{x}{r}$

  $\begin{array}{l}=\frac{\frac{6}{10}}{1}\\ =\frac{6}{10}\end{array}$

Now, $\sin \theta =\frac{y}{r}$

  $\begin{array}{l}=\frac{-\frac{8}{10}}{1}\\ =-\frac{8}{10}\end{array}$

Conclusion:

Therefore, $\cos \theta =\frac{6}{10}$ and $\sin \theta =-\frac{8}{10}$ .