Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 5 - Trigonometric Functions - 5.3 Trigonometric Functions Values and Angle Measures - 5.3 Exercises - Page 533: 73

Answer

$\frac{\sqrt 2}{2}$, $ \frac{\sqrt 2}{2}$, $ 1$, $ 1$, $ \sqrt 2$, and $ \sqrt 2$.

Work Step by Step

Given $\theta=405^\circ=360^\circ+45^\circ$ (quadrant I), we have $sin\theta =sin45^\circ=\frac{\sqrt 2}{2}$, $cos\theta =cos45^\circ=\frac{\sqrt 2}{2}$, $tan\theta =tan45^\circ=1$, $cot\theta = \frac{1}{tan\theta}=1$, $sec\theta = \frac{1}{cos\theta}=\sqrt 2$, and $csc\theta = \frac{1}{sin\theta}=\sqrt 2$.
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