Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\sin{t} = -\dfrac{1}{3}$ $\cos{t} = \dfrac{2\sqrt{2}}{3}$ $\tan{t} = -\dfrac{\sqrt{2}}{4}$ $\csc{t} =-3$ $\sec{t} =\dfrac{3\sqrt{2}}{4}$ $\cot{t} =-2 \sqrt{2}$
With $P= \left(\dfrac{2\sqrt{2}}{3},-\dfrac{1}{3} \right) = (x,y)$, then $\hspace{15pt}x = \dfrac{2\sqrt{2}}{3} \text{ and } \hspace{10pt} y =-\dfrac{1}{3}$ Thus, $\sin{t} = y$ $\sin{t} = -\dfrac{1}{3}$ $\cos{t} = x$ $\cos{t} = \dfrac{2\sqrt{2}}{3}$ $\tan{t} = \dfrac{y}{x}$ $\tan{t} = \dfrac{-\dfrac{1}{3}}{ \dfrac{2\sqrt{2}}{3} } = -\dfrac{\sqrt{2}}{4}$ $\csc{t} = \dfrac{1}{y}$ $\csc{t} = \dfrac{1}{-\dfrac{1}{3}} = -3$ $\sec{t} = \dfrac{1}{x}$ $\sec{t} = \dfrac{1}{\dfrac{2\sqrt{2}}{3}}= \dfrac{3\sqrt{2}}{4}$ $\cot{t} = \dfrac{x}{y}$ $\cot{t} = \dfrac{\dfrac{2\sqrt{2}}{3} }{-\dfrac{1}{3}} = -2 \sqrt{2}$