# Chapter 2 - Acute Angles and Right Triangles - Section 2.2 Trigonometric Functions of Non-Acute Angles - 2.2 Exercises - Page 62: 32

$\sin$(-1020) = $\frac{\sqrt3}{2}$ $\cos$(-1020) = $\frac{1}{2}$ $\tan$(-1020) = $\sqrt3$ $\cot$(-1020) = $\frac{2\sqrt3}{3}$ $\csc$(-1020) = 2 $\sec$(-1020) = $\frac{\sqrt3}{3}$

#### Work Step by Step

-1020$^{\circ}$ We must first fine the coterminal angle: -1020$^{\circ}$ + 360$^{\circ}$ = -660$^{\circ}$ -660$^{\circ}$ + 360$^{\circ}$ = -300$^{\circ}$ -300$^{\circ}$ + 360$^{\circ}$ = 60$^{\circ}$ -150 is in Quadrant I. Therefore all the trigonometric functions are positive. The reference angle is: $\theta$$^{1}$ = 60$^{\circ}$ $\sin$(60) = $\frac{\sqrt3}{2}$ $\cos$(60) = $\frac{1}{2}$ $\tan$(60) = $\sqrt3$ $\cot$(60) = $\frac{2\sqrt3}{3}$ $\csc$(60) = 2 $\sec$(60) = $\frac{\sqrt3}{3}$

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