Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 3 - Section 3.3 - Definition III: Circular Functions - 3.3 Problem Set - Page 143: 8


$\cos{225^o} = x = -\dfrac{\sqrt2}{2}$

Work Step by Step

RECALL: In a unit circle $\cos {t}=x \\\sin{t}=y \\\tan{t} = \dfrac{y}{x}, x\ne0 \\\cot{t} = \dfrac{x}{y}, y \ne 0 \\\sec{t} = \dfrac{1}{x}, x\ne0 \\\csc{t} = \dfrac{1}{y}, y \ne0$ Using the definition above and the unit circle in Figure 5 on page 137 of this book, then: Point on the unit circle: $\left(-\dfrac{\sqrt2}{2}, -\dfrac{\sqrt2}{2}\right)$ Thus, $\cos{225^o} = x = -\dfrac{\sqrt2}{2}$
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