## Precalculus (6th Edition) Blitzer

$\sin \theta =\dfrac{y}{r}=-\dfrac{2\sqrt {2}}{3} \\ \cos \theta =\dfrac{x}{r}=\dfrac{1}{3} \\ \tan \theta =\dfrac{y}{x}=-2 \sqrt 2$ and $\csc \theta =\dfrac{r}{y}=-\dfrac{ 3 \sqrt 2}{ 4} \\ \sec \theta =\dfrac{r}{x}=3 \\ \cot \theta =\dfrac{x}{y}=-\dfrac{\sqrt 2}{4}$
Here, $x=1; r=3$ and $y=\sqrt {(3)^2-(1)^2}=\sqrt {9-1}=2 \sqrt 2$ Note that we must be in quadrant IV because we have positive cosine and negative tangent. Thus, we use the negative value of $y$: $y=-2\sqrt 2$ The trigonometric ratios are as follows: $\sin \theta =\dfrac{y}{r}=-\dfrac{2\sqrt {2}}{3} \\ \cos \theta =\dfrac{x}{r}=\dfrac{1}{3} \\ \tan \theta =\dfrac{y}{x}=-2 \sqrt 2$ and $\csc \theta =\dfrac{r}{y}=-\dfrac{ 3 \sqrt 2}{ 4} \\ \sec \theta =\dfrac{r}{x}=3 \\ \cot \theta =\dfrac{x}{y}=-\dfrac{\sqrt 2}{4}$