Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 2 - Acute Angles and Right Triangles - Section 2.2 Trigonometric Functions of Non-Acute Angles - 2.2 Exercises: 21

Answer

$sin$(405)$^{\circ}$ = $\frac{\sqrt2}{2}$ $cos$(405)$^{\circ}$ = $\frac{\sqrt2}{2}$ $tan$(405)$^{\circ}$ = 1 $cot$(405)$^{\circ}$ = 1 $sec$(405)$^{\circ}$ = $\sqrt2$ $sin$(405)$^{\circ}$ = $\sqrt2$

Work Step by Step

405$^{\circ}$ We can solve for the functions by using the coterminal angle. We can find the coterminal angle by adding or subtracting 360$^{\circ}$ as many times as needed. 405$^{\circ}$ - 360$^{\circ}$ = 45$^{\circ}$ $sin$(45)$^{\circ}$ = $\frac{1}{\sqrt2}$ = $\frac{\sqrt2}{2}$ $cos$(45)$^{\circ}$ = $\frac{1}{\sqrt2}$ = $\frac{\sqrt2}{2}$ $tan$(45)$^{\circ}$ = $\frac{1}{1}$ = 1 $cot$(45)$^{\circ}$ = $\frac{1}{1}$ = 1 $sec$(45)$^{\circ}$ = $\frac{\sqrt2}{1}$ = $\sqrt2$ $sin$(45)$^{\circ}$ = $\frac{\sqrt2}{1}$ = $\sqrt2$
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