Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.1 - Verifying Trigonometric Identities - Exercise Set - Page 661: 101


The required solution is $\frac{\sqrt{3}}{2},\frac{1}{2},\frac{1}{2},\frac{\sqrt{3}}{2},0,1$

Work Step by Step

By putting the values according to the trigonometric functions, we get: $\cos {{30}^{\circ }}=\frac{\sqrt{3}}{2}$ $\sin {{30}^{\circ }}=\frac{1}{2}$ $\cos {{60}^{\circ }}=\frac{1}{2}$ $\sin {{60}^{^{\circ }}}=\frac{\sqrt{3}}{2}$ $\cos {{90}^{\circ }}=0$ $\sin {{90}^{\circ }}=1$
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