Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Section 5.6 Half-Angle Identities - 5.6 Exercises - Page 239: 77

Answer

$csc~18^{\circ} = \sqrt{5}+1$

Work Step by Step

$csc~x = \frac{1}{sin~x}$ We can find the value of $csc~18^{\circ}$: $csc~18^{\circ} = \frac{1}{sin~18^{\circ}}$ $csc~18^{\circ} = \frac{1}{\frac{\sqrt{5}-1}{4}}$ $csc~18^{\circ} = \frac{4}{\sqrt{5}-1}$ $csc~18^{\circ} = \frac{4}{\sqrt{5}-1}~\frac{\sqrt{5}+1}{\sqrt{5}+1}$ $csc~18^{\circ} = \frac{4~(\sqrt{5}+1)}{4}$ $csc~18^{\circ} = \sqrt{5}+1$
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