Trigonometry (11th Edition) Clone

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

Chapter 6 - Quiz (Sections 6.1-6.3) - Page 282: 2a



Work Step by Step

$$y=\sin^{-1}(-\frac{\sqrt2}{2})$$ *Remember: $\sin^{-1}A=x$ means that $\sin x=A$ Therefore, to find $y$, we need to recall which value of $y$ would have $\sin y=-\frac{\sqrt2}{2}$. $\sin y\lt0$ shows that we can find a value of $y$ in either quadrant III or IV of the unit circle. The choice is arbitrary. I would choose quadrant IV here. In quadrant IV, $-\frac{\pi}{4}$ is the value of the angle with which $\sin(-\frac{\pi}{4})=-\frac{\sqrt2}{2}$ Therefore, $$y=-\frac{\pi}{4}$$
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