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: 77

Answer

$\sin\theta=\frac{y}{r}=\frac{\sqrt {15}}{4}$ $\cos\theta=\frac{x}{r}=\frac{-1}{4}$ $\tan\theta=\frac{y}{x}=\frac{\sqrt {15} }{-1}=-\sqrt {15}$ $\cot\theta=\frac{x}{y}=\frac{-1}{\sqrt {15}}=\frac{-\sqrt {15}}{15}$ $\sec\theta=\frac{r}{x}=\frac{4}{-1}=-4$ $\csc\theta=\frac{r}{y}=\frac{4}{\sqrt {15}}=\frac{4\sqrt {15}}{15}$

Work Step by Step

1. II quadrant : x negative and y is positive, because sec is negative and sin is positive. x=-1 and r=4 2. Calculate y using distance formula $4=\sqrt {(y)^{2}+(-1)^{2}}$ $16=1+y^{2}$ $y=\sqrt {15}$ 3. Insert the values to find trig functions $\sin\theta=\frac{y}{r}=\frac{\sqrt {15}}{4}$ $\cos\theta=\frac{x}{r}=\frac{-1}{4}$ $\tan\theta=\frac{y}{x}=\frac{\sqrt {15} }{-1}=-\sqrt {15}$ $\cot\theta=\frac{x}{y}=\frac{-1}{\sqrt {15}}=\frac{-\sqrt {15}}{15}$ $\sec\theta=\frac{r}{x}=\frac{4}{-1}=-4$ $\csc\theta=\frac{r}{y}=\frac{4}{\sqrt {15}}=\frac{4\sqrt {15}}{15}$
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