Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.6 Trigonometric Functions - Exercises - Page 140: 1


$$ y=\frac{\sqrt{2}}{2} x+\frac{\sqrt{2}}{2}\left(1-\frac{\pi}{4}\right).$$

Work Step by Step

Since $ y=\sin x $, then $ y'=\cos x $ and hence the slope at $ x=\frac{\pi}{4}$ is $ m=\cos \frac{\pi}{4}=\frac{\sqrt{2}}{2}.$ Now, the tangent line equation is given by $$ y=\frac{\sqrt{2}}{2} x+c.$$ Since the curve and line coincide at $ x=\frac{\pi}{4}$, then $$\sin \frac{\pi}{4}= \frac{\sqrt{2}}{2}\frac{\pi}{4}+c \Longrightarrow c=\frac{\sqrt{2}}{2}\left(1-\frac{\pi}{4}\right) .$$ Hence the equation of the tangent line is given by $$ y=\frac{\sqrt{2}}{2} x+\frac{\sqrt{2}}{2}\left(1-\frac{\pi}{4}\right).$$
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