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: 2


$$ y=-\frac{\sqrt{3}}{2} x+\frac{1}{2}\left(1+\frac{\sqrt{3}\pi}{3}\right).$$

Work Step by Step

Since $ y=\cos x $, then $ y'=-\sin x $ and hence the slope at $ x=\frac{\pi}{3}$ is $ m=-\sin \frac{\pi}{3}=-\frac{\sqrt{3}}{2}.$ Now, the tangent line equation is given by $$ y=-\frac{\sqrt{3}}{2} x+c.$$ Since the curve and line coincide at $ x=\frac{\pi}{3}$, then $$\cos \frac{\pi}{3}= -\frac{\sqrt{3}}{2}\frac{\pi}{3}+c \Longrightarrow c=\frac{1}{2}\left(1+\frac{\sqrt{3}\pi}{3}\right)$$ Hence the equation of the tangent line is given by $$ y=-\frac{\sqrt{3}}{2} x+\frac{1}{2}\left(1+\frac{\sqrt{3}\pi}{3}\right).$$
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