Answer
a.
$\displaystyle \sin\theta=-\frac{\sqrt{2}}{2}\quad \cos\theta=-\frac{\sqrt{2}}{2}\quad \tan\theta=1$
b.
$\displaystyle \sin\theta=-\frac{\sqrt{3}}{2}\quad \cos\theta=\frac{1}{2}\quad \tan\theta=-\sqrt{3}$
Work Step by Step
We will use the definition of trigonometric functions as given in section 1.3.
a.
We need the distance from the origin, $r$.
$r=\sqrt{x^{2}+y^{2}}$
$r=\sqrt{(-3)^{2}+(-3)^{2}}$
$r=\sqrt{9+9}$
$r=\sqrt{18}$
$r=3\sqrt{2}$
$\displaystyle \sin\theta=\frac{y}{r}=\frac{-3}{3\sqrt{2}}=-\frac{1}{\sqrt{2}}=-\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=-\frac{\sqrt{2}}{2}$
$\displaystyle \cos\theta=\frac{x}{r}=\frac{-3}{3\sqrt{2}}=-\frac{1}{\sqrt{2}}=-\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=-\frac{\sqrt{2}}{2}$
$\displaystyle \tan\theta=\frac{y}{r}=\frac{-3}{-3}=1$
b.
We need the distance from the origin, $r$.
$r=\sqrt{x^{2}+y^{2}}$
$r=\sqrt{1^{2}+(-\sqrt{3})^{2}}$
$r=\sqrt{1+3}$
$r=\sqrt{4}$
$r=2$
$\displaystyle \sin\theta=\frac{y}{r}=\frac{-\sqrt{3}}{2}=-\frac{\sqrt{3}}{2}$
$\displaystyle \cos\theta=\frac{x}{r}=\frac{1}{2}$
$\displaystyle \tan\theta=\frac{y}{x}=\frac{-\sqrt{3}}{1}=-\sqrt{3}$
