Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 1 - Trigonometric Functions - Section 1.4 Using the Definitions of the Trigonometric Functions - 1.4 Exercises - Page 38: 79

Answer

$\sin\theta=\frac{y}{r}=\frac{\sqrt {2}}{6}$ $\cos\theta=\frac{x}{r}=\frac{-\sqrt {34}}{6}$ $\tan\theta=\frac{y}{x}=\frac{\sqrt {2} }{-\sqrt {34}}=\frac{-\sqrt {17}}{17}$ $\cot\theta=\frac{x}{y}=\frac{-\sqrt {34}}{\sqrt {2}}=-\sqrt {17}$ $\sec\theta=\frac{r}{x}=\frac{6}{-\sqrt {34}}=\frac{-3\sqrt {34}}{17}$ $\csc\theta=\frac{r}{y}=\frac{6}{\sqrt {2}}=3\sqrt 2$

Work Step by Step

1. Quadrant II, x is negative and y is positive 2. $y=\sqrt 24$ and $r=6$ 3. Find x using distance formula $6=\sqrt {(x)^{2}+(\sqrt 2)^{2}}$ $36=2+x^{2}$ $x=-\sqrt {34}$ 4. Plug values in to find the trig functions $\sin\theta=\frac{y}{r}=\frac{\sqrt {2}}{6}$ $\cos\theta=\frac{x}{r}=\frac{-\sqrt {34}}{6}$ $\tan\theta=\frac{y}{x}=\frac{\sqrt {2} }{-\sqrt {34}}=\frac{-\sqrt {17}}{17}$ $\cot\theta=\frac{x}{y}=\frac{-\sqrt {34}}{\sqrt {2}}=-\sqrt {17}$ $\sec\theta=\frac{r}{x}=\frac{6}{-\sqrt {34}}=\frac{-3\sqrt {34}}{17}$ $\csc\theta=\frac{r}{y}=\frac{6}{\sqrt {2}}=3\sqrt 2$
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