## Precalculus (6th Edition) Blitzer

The six trigonometric functions of $\theta$ for point $\left( -1,-3 \right)$ are, $\sin \theta =-\frac{3\sqrt{10}}{\sqrt{10}},\cos \theta =-\frac{\sqrt{10}}{10},\tan \theta =3,\csc \theta =-\frac{\sqrt{10}}{3},\sec \theta =-\sqrt{10}$ and $\cot \theta =\frac{1}{3}$.
Consider the point $\left( -1,-3 \right)$. Here, $x=-1$ and $y=-3$. The six trigonometric functions of $\theta$ are defined in terms of a ratio by using the right-angle triangle. In the figure, $y$ is the perpendicular, $x$ is the base and $r$ is the hypotenuse. According to the Pythagoras theorem, the hypotenuse is, $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ Substitute $-1$ for $x$ and $-3$ for $y$. \begin{align} & r=\sqrt{{{\left( -1 \right)}^{2}}+{{\left( -3 \right)}^{2}}} \\ & =\sqrt{1+9} \\ & =\sqrt{10} \end{align} Recall the trigonometric expression of $\sin \theta$. $\sin \theta =\frac{y}{r}$ Substitute $-3$ for $y$ and $\sqrt{10}$ for $r$. \begin{align} & \sin \theta =-\frac{3}{\sqrt{10}} \\ & =-\frac{3}{\sqrt{10}}\cdot \frac{\sqrt{10}}{\sqrt{10}} \\ & =-\frac{3\sqrt{10}}{10} \end{align} Recall the trigonometric expression of $\cos \theta$. $\cos \theta =\frac{x}{r}$ Substitute $-1$ for $x$ and $\sqrt{10}$ for $r$. \begin{align} & \cos \theta =-\frac{1}{\sqrt{10}} \\ & =-\frac{1}{\sqrt{10}}\cdot \frac{\sqrt{10}}{\sqrt{10}} \\ & =-\frac{\sqrt{10}}{10} \end{align} Recall the trigonometric expression of $\tan \theta$. $\tan \theta =\frac{y}{x}$ Substitute $-1$ for $x$ and $-3$ for $y$. \begin{align} & \tan \theta =\frac{-3}{-1} \\ & =3 \end{align} Recall the trigonometric expression of $\csc \theta$. $\csc \theta =\frac{r}{y}$ Substitute $-3$ for $y$ and $\sqrt{10}$ for $r$. $\csc \theta =-\frac{\sqrt{10}}{3}$ Recall the trigonometric expression of $\sec \theta$. $\sec \theta =\frac{r}{x}$ Substitute $-1$ for $x$ and $\sqrt{10}$ for $r$. \begin{align} & \sec \theta =-\frac{\sqrt{10}}{1} \\ & =-\sqrt{10} \end{align} Recall the trigonometric expression of $\cot \theta$. $\cot \theta =\frac{x}{y}$ Substitute $-1$ for $x$ and $-3$ for $y$. \begin{align} & \cot \theta =\frac{-1}{-3} \\ & =\frac{1}{3} \end{align} Thus, the six trigonometric functions of $\theta$ for point $\left( -1,-3 \right)$ are, $\sin \theta =-\frac{3\sqrt{10}}{\sqrt{10}},\cos \theta =-\frac{\sqrt{10}}{10},\tan \theta =3,\csc \theta =-\frac{\sqrt{10}}{3},\sec \theta =-\sqrt{10}$ and $\cot \theta =\frac{1}{3}$.