## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 2 - Acute Angles and Right Triangles - Section 2.2 Trigonometric Functions of Non-Acute Angles - 2.2 Exercises - Page 62: 25

#### Answer

$sin$(570)$^{\circ}$ = $\frac{-1}{2}$ $cos$(570)$^{\circ}$ = $\frac{-\sqrt3}{2}$ $tan$(570)$^{\circ}$ = $\frac{1}{\sqrt3}$ = $\frac{\sqrt3}{3}$ $cot$(570)$^{\circ}$ = $\frac{\sqrt3}{1}$ = $\sqrt3$ $csc$(570)$^{\circ}$ = $\frac{2}{-1}$ = -2 $sec$(570)$^{\circ}$ = $\frac{2}{-\sqrt3}$ = $\frac{2\sqrt3}{3}$

#### Work Step by Step

570$^{\circ}$ 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. 570$^{\circ}$ - 360$^{\circ}$ = 210$^{\circ}$ Next we find the reference angle: 210$^{\circ}$ - 180$^{\circ}$ = 30$^{\circ}$ -$sin$(30)$^{\circ}$ = $\frac{-1}{2}$ -$cos$(30)$^{\circ}$ = $\frac{-\sqrt3}{2}$ $tan$(30)$^{\circ}$ = $\frac{1}{\sqrt3}$ = $\frac{\sqrt3}{3}$ $cot$(30)$^{\circ}$ = $\frac{\sqrt3}{1}$ = $\sqrt3$ -$csc$(30)$^{\circ}$ = $\frac{2}{-1}$ = -2 -$sec$(30)$^{\circ}$ = $\frac{2}{-\sqrt3}$ = $\frac{2\sqrt3}{3}$

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