Answer
$-\frac{\sqrt 2}{2},-\frac{\sqrt 2}{2},1$
Work Step by Step
Use the figure given in the exercise, we have $sin\theta=\frac{-3}{\sqrt {(-3)^2+(-3)^2}}=-\frac{\sqrt 2}{2}$, $cos\theta=\frac{-3}{\sqrt {(-3)^2+(-3)^2}}=-\frac{\sqrt 2}{2}$, and $tan\theta=\frac{-3}{-3}=1$
