Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - Chapter Review - Chapter Test - Page 507: 23

Answer

$\frac{\sqrt 6}{2}$

Work Step by Step

Using trigonometric identities, we have: $sin75^\circ +sin15^\circ=2sin(\frac{75^\circ +15^\circ}{2})cos(\frac{75^\circ -15^\circ}{2})=2sin(\frac{90^\circ}{2})cos(\frac{60^\circ}{2})=2sin(45^\circ)cos(30^\circ)=2(\frac{\sqrt 2}{2})(\frac{\sqrt 3}{2})=\frac{\sqrt 6}{2}$
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