## Precalculus (6th Edition) Blitzer

The six trigonometric functions of $\theta$ for point $\left( 2,3 \right)$ are, $\sin \theta =\frac{3\sqrt{13}}{13},\cos \theta =\frac{2\sqrt{13}}{13},\tan \theta =\frac{3}{2},\csc \theta =\frac{\sqrt{13}}{3},\sec \theta =\frac{\sqrt{13}}{2}$ and $\cot \theta =\frac{2}{3}$.
According to the Pythagoras theorem in a right angle triangle, the hypotenuse is given as, ${{r}^{2}}={{x}^{2}}+{{y}^{2}}$ Consider the point $\left( 2,3 \right)$. Here, $x=2$ and $y=3$. The six trigonometric functions of $\theta$ are defined in terms of the ratio. According to the Pythagoras theorem, the hypotenuse is, $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ Substitute $2$ for $x$ and $3$ for $y$. \begin{align} & r=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 3 \right)}^{2}}} \\ & =\sqrt{4+9} \\ & =\sqrt{13} \end{align} Recall the trigonometric expression of $\sin \theta$. $\sin \theta =\frac{y}{r}$ Substitute $3$ for $y$ and $\sqrt{13}$ for $r$. \begin{align} & \sin \theta =\frac{3}{\sqrt{13}} \\ & =\frac{3}{\sqrt{13}}.\frac{\sqrt{13}}{\sqrt{13}} \\ & =\frac{3\sqrt{13}}{13} \end{align} Recall the trigonometric expression of $\cos \theta$. $\cos \theta =\frac{x}{r}$ Substitute $2$ for $x$ and $\sqrt{13}$ for $r$. \begin{align} & \cos \theta =\frac{2}{\sqrt{13}} \\ & =\frac{2}{\sqrt{13}}.\frac{\sqrt{13}}{\sqrt{13}} \\ & =\frac{2\sqrt{13}}{13} \end{align} Recall the trigonometric expression of $\tan \theta$. $\tan \theta =\frac{y}{x}$ Substitute $2$ for $x$ and $3$ for $y$. $\tan \theta =\frac{3}{2}$ Recall the trigonometric expression of $\csc \theta$. $\csc \theta =\frac{r}{y}$ Substitute $3$ for $y$ and $\sqrt{13}$ for $r$. $\csc \theta =\frac{\sqrt{13}}{3}$ Recall the trigonometric expression of $\sec \theta$. $\sec \theta =\frac{r}{x}$ Substitute $2$ for $x$ and $\sqrt{13}$ for $r$. $\sec \theta =\frac{\sqrt{13}}{2}$ Recall the trigonometric expression of $\cot \theta$. $\cot \theta =\frac{x}{y}$ Substitute $2$ for $x$ and $3$ for $y$. $\cot \theta =\frac{2}{3}$ Thus, the six trigonometric functions of $\theta$ for point $\left( 2,3 \right)$ are, $\sin \theta =\frac{3\sqrt{13}}{13},\cos \theta =\frac{2\sqrt{13}}{13},\tan \theta =\frac{3}{2},\csc \theta =\frac{\sqrt{13}}{3},\sec \theta =\frac{\sqrt{13}}{2}$ and $\cot \theta =\frac{2}{3}$.