Chapter 9.3, Problem 1GP

Solution

To find: the values of six trigonometric function for the given angle.

  $\begin{array}{c}\sin \theta =-\frac{\sqrt{2}}{2},\cos \theta =\frac{\sqrt{2}}{2},\tan \theta =-1\\ \csc \theta =-\sqrt{2},\sec \theta =\sqrt{2},\cot \theta =-1\end{array}$

Given:

  

Concept used:

If $P\left(x,y\right)$ is the point on the terminal side of an angle $\theta$ in the standard position, then

  $\sin \theta =\frac{y}{r},\cos \theta =\frac{x}{r},\tan \theta =\frac{y}{x},\csc \theta =\frac{r}{y},\sec \theta =\frac{r}{x},\tan \theta =\frac{x}{y}$ , where $r=\sqrt{x^2+y^2}$ .

Calculation:

Here, the point on the terminal side of an angle $\theta$ is $\left(3,-3\right)$ . That is,

  $\left(x,y\right)=\left(3,-3\right)$ .

Now,

  $\begin{array}{c}r=\sqrt{{\left(3\right)}^2+{\left(-3\right)}^2}\\ =\sqrt{18}\\ =3\sqrt{2}\end{array}$

Therefore,

  $\begin{array}{c}\sin \theta =-\frac{3}{3\sqrt{2}}=-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}\\ \cos \theta =\frac{3}{3\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\\ \tan \theta =\frac{-3}{3}=-1\\ \csc \theta =-\frac{3\sqrt{2}}{3}=-\sqrt{2}\\ \sec \theta =\frac{3\sqrt{2}}{3}=\sqrt{2}\\ \cot \theta =\frac{3}{-3}=-1\end{array}$