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 62: 37

Answer

$\sin$ 1305$^{\circ}$ = $\frac{-\sqrt2}{2}$

Work Step by Step

$\sin$ 1305$^{\circ}$ First, lets find the coterminal angle. 1305$^{\circ}$ - 3(360$^{\circ}$) = 225$^{\circ}$ Next, we find the reference angle. $\theta$$^{1}$ = 225$^{\circ}$ - 180$^{\circ}$ $\theta$$^{1}$ = 45$^{\circ}$ Since the coterminal angle is in Quadrant III, $\sin$ is negative. Therefore: $\sin$ 45$^{\circ}$ = $\frac{-1}{\sqrt2}$ = $\frac{-\sqrt2}{2}$
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