Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 1 - Trigonometric Functions - Section 1.4 Using the Definitions of the Trigonometric Functions - 1.4 Exercises: 76

Answer

$\sin\theta=\frac{y}{r}=\frac{-\sqrt {59}}{8}$ $\cos\theta=\frac{x}{r}=\frac{\sqrt 5}{8}$ $\tan\theta=\frac{y}{x}=\frac{-\sqrt {59} }{\sqrt 5}=\frac{-\sqrt {295}}{5}$ $\cot\theta=\frac{x}{y}=\frac{\sqrt 5}{-\sqrt {59}}=\frac{-\sqrt {295}}{59}$ $\sec\theta=\frac{r}{x}=\frac{8}{\sqrt 5}=\frac{8\sqrt 5}{5}$ $\csc\theta=\frac{r}{y}=\frac{8}{-\sqrt {59}}=\frac{-8\sqrt {59}}{59}$

Work Step by Step

1. IV quadrant, x is positive and y is negative $x=\sqrt 5$ and $r=8$ 2. Calculate y using the distance formula $8=\sqrt {(y)^{2}+(\sqrt 5)^{2}}$ $64=5+y^{2}$ $y=-\sqrt {59}$ 3. Insert the values to find the trig functions $\sin\theta=\frac{y}{r}=\frac{-\sqrt {59}}{8}$ $\cos\theta=\frac{x}{r}=\frac{\sqrt 5}{8}$ $\tan\theta=\frac{y}{x}=\frac{-\sqrt {59} }{\sqrt 5}=\frac{-\sqrt {295}}{5}$ $\cot\theta=\frac{x}{y}=\frac{\sqrt 5}{-\sqrt {59}}=\frac{-\sqrt {295}}{59}$ $\sec\theta=\frac{r}{x}=\frac{8}{\sqrt 5}=\frac{8\sqrt 5}{5}$ $\csc\theta=\frac{r}{y}=\frac{8}{-\sqrt {59}}=\frac{-8\sqrt {59}}{59}$
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