Chapter 6.2, Problem 19AYU

To determine

To find: The exact values of the six trigonometric functions of t.

Answer to Problem 19AYU

Solution:

  $\sin t=-\frac{1}{3}$ , $\cos t= \frac{2\sqrt{2}}{3}$ , $\tan t=-\frac{\sqrt{2}}{4}$, $\csc t=-3$ , $\sec t=\frac{3\sqrt{2}}{4}$ and $\cot t =-2\sqrt{2}$

Explanation of Solution

Given:

P = (x, y) is the point on the unit circle that corresponds to a real number t and $P=\left(\frac{2\sqrt{2}}{3}, -\frac{1}{3}\right)$ .

Calculation:

  $P=\left(\frac{2\sqrt{2}}{3}, -\frac{1}{3}\right)=\left(x, y\right)$ , $x=\frac{2\sqrt{2}}{3}$ and $y=-\frac{1}{3}$ we have,

  $\sin t=y=-\frac{1}{3}$

  $\cos t=x= \frac{2\sqrt{2}}{3}$

  $\tan t=\frac{y}{x}=\frac{-\frac{1}{3}}{\frac{2\sqrt{2}}{3}}=-\frac{\sqrt{2}}{4}$

  $\csc t=\frac{1}{y}=\frac{1}{-\frac{1}{3} }=-3$

  $\sec t=\frac{1}{\frac{2\sqrt{2}}{3}}=\frac{3}{2\sqrt{2}}=\frac{3\sqrt{2}}{4}$

  $\cot t =\frac{x}{y}=\frac{\frac{2\sqrt{2}}{3}}{-\frac{1}{3} }=-2\sqrt{2}$