Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 7 - Applications of Trigonometric Functions - Section 7.4 Area of a Triangle - 7.4 Assess Your Understanding - Page 567: 60

Answer

$sin(t) =\frac{\sqrt 2}{3}$, $cos(t) =-\frac{\sqrt 7}{3}$, $tan(t) =-\frac{\sqrt {14}}{7}$, $cot(t) =-\frac{\sqrt {14}}{2}$, $sec(t) =-\frac{3\sqrt {7}}{7}$, $csc(t) =\frac{3\sqrt {2}}{2}$.

Work Step by Step

Given point $P$ on the unit circle, we have $x=-\frac{\sqrt 7}{3}, y=\frac{\sqrt 2}{3}$, we have: $sin(t)=y=\frac{\sqrt 2}{3}$, $cos(t)=x=-\frac{\sqrt 7}{3}$, $tan(t)=y/x=-\frac{\sqrt {14}}{7}$, $cot(t)=x/y=-\frac{\sqrt {14}}{2}$, $sec(t)=1/x=-\frac{3\sqrt {7}}{7}$, $csc(t)=1/y=\frac{3\sqrt {2}}{2}$.
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