Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 1: Functions - Section 1.3 - Trigonometric Functions - Exercises 1.3 - Page 28: 43

Answer

$\dfrac{\sqrt{6}+\sqrt{2}}{4}$

Work Step by Step

Addition formula $$\sin{(A+B)}= \sin{A} \cos{B} +\cos{A} \sin{B}$$ $\therefore \sin{\left(\dfrac{\pi}{4}+\dfrac{\pi}{3}\right)} = \sin{\left(\dfrac{\pi}{4} \right)} \cos{\left(\dfrac{\pi}{3} \right)} + \cos{\left(\dfrac{\pi}{4} \right)} \sin{\left(\dfrac{\pi}{3} \right)}$ $\sin{\left(\dfrac{\pi}{4}+\dfrac{\pi}{3}\right)} =\dfrac{\sqrt{2}}{2} \times \dfrac{1}{2}+ \dfrac{\sqrt{2}}{2} \times \dfrac{\sqrt{3}}{2}$ $\sin{\left(\dfrac{\pi}{4}+\dfrac{\pi}{3}\right)} = \dfrac{\sqrt{6}+\sqrt{2}}{4}$
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