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 1.3 Trigonometric Functions - 1.3 Exercises - Page 27: 8

Answer

$-\frac{\sqrt 2}{2}$

Work Step by Step

$r=\sqrt {x^{2}+y^{2}}=\sqrt {(-3)^{2}+(-3)^{2}}=3\sqrt 2$ $\sin\theta=\frac{y}{r}=\frac{-3}{3\sqrt 2}=-\frac{1}{\sqrt 2}=-\frac{\sqrt 2}{2}$
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