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: 27

Answer

-$sin$(1305)$^{\circ}$ = $\frac{-\sqrt2}{2}$ -$cos$(1305)$^{\circ}$ = $\frac{-\sqrt2}{2}$ $tan$(1305)$^{\circ}$ = 1 $cot$(1305)$^{\circ}$ = 1 -$csc$(1305)$^{\circ}$ = -$\sqrt2$ -$sec$(1305)$^{\circ}$ = -$\sqrt2$

Work Step by Step

1305$^{\circ}$ 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. 1305$^{\circ}$ - 360$^{\circ}$ = 945$^{\circ}$ 945$^{\circ}$ - 360$^{\circ}$ = 585$^{\circ}$ 585$^{\circ}$ - 360$^{\circ}$ = 225$^{\circ}$ Next we must find the reference angle: 225$^{\circ}$ - 180$^{\circ}$ = 45$^{\circ}$ -$sin$(45)$^{\circ}$ = $\frac{-\sqrt2}{2}$ -$cos$(45)$^{\circ}$ = $\frac{-\sqrt2}{2}$ $tan$(45)$^{\circ}$ = $\frac{-\sqrt2}{-\sqrt2}$ = 1 $cot$(45)$^{\circ}$ = $\frac{-\sqrt2}{-\sqrt2}$ = 1 -$csc$(45)$^{\circ}$ = $\frac{-\sqrt2}{1}$ = -$\sqrt2$ -$sec$(45)$^{\circ}$ = $\frac{-\sqrt2}{1}$ = -$\sqrt2$
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