Trigonometry (11th Edition) Clone

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

Chapter 2 - Acute Angles and Right Triangles - Section 2.2 Trigonometric Functions of Non-Acute Angles - 2.2 Exercises - Page 63: 87

Answer

$45^{\circ};225^{\circ}$

Work Step by Step

Recall: $\sin^{2}\theta+\cos^{2}\theta=1$ Given that $\sin\theta=\cos\theta$ $\implies \cos^{2}\theta+\cos^{2}\theta=1$ $\implies 2\cos^{2}\theta=1$ $\implies \cos^{2}\theta=\frac{1}{2}$ or $\cos\theta=\pm\frac{1}{\sqrt 2}$ As sine and cosine function values have different signs in quadrant II and IV, we find the value of $\theta$ that lies in quadrant I and III only. $\cos^{-1}(\frac{1}{\sqrt 2})=45^{\circ}$ $-\frac{1}{\sqrt 2}=-\cos 45^{\circ}=\cos(180^{\circ}+45^{\circ})=\cos 225^{\circ}$ $\theta=45^{\circ},225^{\circ}$
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