Thomas' Calculus 13th Edition

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

Chapter 3: Derivatives - Section 3.5 - Derivatives of Trigonometric Functions - Exercises 3.5 - Page 141: 35

Answer

See graph and explanations.
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Work Step by Step

See graph. Step 1. With $y=sin(x)$, we have $y'=cos(x)$ which give the slope of tangent lines. Step 2. At $x=-\pi$, $y=sin(\pi)=0$ and $m_1=y'=cos(-\pi)=-1$, tangent line equation: $y=-x-\pi$ Step 3. At $x=0$, $y=sin(0)=0$ and $m_2=y'=cos(0)=1$, tangent line equation: $y=x$ Step 2. At $x=3\pi/2$, $y=sin(3\pi/2)=-1$ and $m_3=y'=cos(3\pi/2)=0$, tangent line equation: $y=-1$
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