Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 575: 5

Answer

The six trigonometric functions of $\theta $ for point $\left( 3,-3 \right)$ are, $\sin \theta =-\frac{\sqrt{2}}{2},\cos \theta =\frac{\sqrt{2}}{2},\tan \theta =-1,\csc \theta =-\sqrt{2},\sec \theta =\sqrt{2}$ and $\cot \theta =-1$.

Work Step by Step

Consider the point $\left( 3,-3 \right)$. Here, $x=3$ and $y=-3$. The six trigonometric functions of $\theta $ are defined in terms of a ratio. According to the Pythagoras theorem, the hypotenuse is, $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ Substitute $3$ for $x$ and $-3$ for $y$. $\begin{align} & r=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( -3 \right)}^{2}}} \\ & =\sqrt{9+9} \\ & =\sqrt{18} \\ & =3\sqrt{2} \end{align}$ Recall the trigonometric expression of $\sin \theta $. $\sin \theta =\frac{y}{r}$ Substitute $-3$ for $y$ and $3\sqrt{2}$ for $r$. $\begin{align} & \sin \theta =-\frac{3}{3\sqrt{2}} \\ & =-\frac{1}{\sqrt{2}} \\ & =-\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}} \\ & =\frac{\sqrt{2}}{2} \end{align}$ Recall the trigonometric expression of $\cos \theta $. $\cos \theta =\frac{x}{r}$ Substitute $3$ for $x$ and $3\sqrt{2}$ for $r$. $\begin{align} & \cos \theta =\frac{3}{3\sqrt{2}} \\ & =\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}} \\ & =\frac{\sqrt{2}}{2} \end{align}$ Recall the trigonometric expression of $\tan \theta $. $\tan \theta =\frac{y}{x}$ Substitute $3$ for $x$ and $-3$ for $y$. $\begin{align} & \tan \theta =\frac{-3}{3} \\ & =-1 \end{align}$ Recall the trigonometric expression of $\csc \theta $. $\csc \theta =\frac{r}{y}$ Substitute $-3$ for $y$ and $3\sqrt{2}$ for $r$. $\begin{align} & \csc \theta =\frac{3\sqrt{2}}{-3} \\ & =-\sqrt{2} \end{align}$ Recall the trigonometric expression of $\sec \theta $. $\sec \theta =\frac{r}{x}$ Substitute $3$ for $x$ and $3\sqrt{2}$ for $r$. $\begin{align} & \sec \theta =\frac{3\sqrt{2}}{3} \\ & =\sqrt{2} \end{align}$ Recall the trigonometric expression of $\cot \theta $. $\cot \theta =\frac{x}{y}$ Substitute $3$ for $x$ and $-3$ for $y$. $\begin{align} & \cot \theta =\frac{3}{-3} \\ & =-1 \end{align}$ Thus, the six trigonometric functions of $\theta $ for point $\left( 3,-3 \right)$ are, $\sin \theta =-\frac{\sqrt{2}}{2},\cos \theta =\frac{\sqrt{2}}{2},\tan \theta =-1,\csc \theta =-\sqrt{2},\sec \theta =\sqrt{2}$ and $\cot \theta =-1$.
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