Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.4 Double-Angle and Half-Angle Identities - 7.4 Exercises: 69

Answer

$\tan \dfrac {147}{2}=\tan 73.5$

Work Step by Step

$\sqrt {\dfrac {1-\cos 147}{1+\cos 147}}==\sqrt {\dfrac {1-\cos \left( 2\times \dfrac {147}{2}\right) }{1+\cos \left( 2\times \dfrac {147}{2}\right) }}=\sqrt {\dfrac {1-\left( \cos ^{2}\dfrac {147}{2}-\sin ^{2}\dfrac {147}{2}\right) }{1+\left( \cos ^{2}\dfrac {147}{2}-\sin ^{2}\dfrac {147}{2}\right) }}=\sqrt {\dfrac {2\sin ^{2}\dfrac {147}{2}}{2\cos ^{2}\dfrac {147}{2}}}=\tan \dfrac {147}{2}=\tan 73.5$
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