## Trigonometry (11th Edition) Clone

a. $\displaystyle \sin\theta=-\frac{\sqrt{2}}{2}\quad \cos\theta=-\frac{\sqrt{2}}{2}\quad \tan\theta=1$ b. $\displaystyle \sin\theta=-\frac{\sqrt{3}}{2}\quad \cos\theta=\frac{1}{2}\quad \tan\theta=-\sqrt{3}$
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, $r$. $r=\sqrt{x^{2}+y^{2}}$ $r=\sqrt{1^{2}+(-\sqrt{3})^{2}}$ $r=\sqrt{1+3}$ $r=\sqrt{4}$ $r=2$ $\displaystyle \sin\theta=\frac{y}{r}=\frac{-\sqrt{3}}{2}=-\frac{\sqrt{3}}{2}$ $\displaystyle \cos\theta=\frac{x}{r}=\frac{1}{2}$ $\displaystyle \tan\theta=\frac{y}{x}=\frac{-\sqrt{3}}{1}=-\sqrt{3}$