University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.3 - Trigonometric Functions - Exercises - Page 27: 6

Answer

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Work Step by Step

sin(-3π/2) = 1 sin(-π/3) = -$\frac{\sqrt 3}{2}$ sin(-π/6) = $\frac{-1}{2}$ sin(π/4) = $\frac{\sqrt 2}{2}$ sin(5π/6) = 1/2 cos(-3π/2) = 0 cos(-π/3) = $\frac{1}{2}$ cos(-π/6) = $\frac{\sqrt 3}{2}$ cos(π/4) = $\frac{\sqrt 2}{2}$ cos(5π/6) = -$\frac{\sqrt 3}{2}$ tan(-3π/2) = UND tan(-π/3) = -$\sqrt 3$ tan(-π/6) = $\frac{\sqrt 3}{3}$ tan(π/4) = 1 tan(5π/6) = -$\frac{\sqrt 3}{3}$ cot(-3π/2) = 0 cot(-π/3) = -$\frac{\sqrt 3}{3}$ cot(-π/6) = -$\sqrt 3$ cot(π/4) = 1 cot(5π/6) = -$\sqrt 3$ sec(-3π/2) = UND sec(-π/3) = 2 sec(-π/6) = 2$\frac{\sqrt 3}{3}$ sec(π/4) = $\sqrt 2$ sec(5π/6) = -2$\frac{\sqrt 3}{3}$ csc(-3π/2) = 1 csc(-π/3) = -2$\frac{\sqrt 3}{3}$ csc(-π/6) = -2 csc(π/4) = $\sqrt 2$ csc(5π/6) = 2
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