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: 45

Answer

The values of $x$ where the tangent line is horizontal are $x=\dfrac{\pi}{4},\dfrac{3\pi}{4},\dfrac{5\pi}{4},\dfrac{7\pi}{4}$

Work Step by Step

The value of the derivative is the slope of the tangent line. Since the tangent line is horizontal, so the slope of the tangent will be zero. Since slope and derivative are the same. Derivative of $y = \sin {x} \cos {x}$ will also be zero. That is, $\dfrac{dy}{dx}=\dfrac{d}{dx}\left(\sin{x}\cos{x}\right)=0$ Now use the product rule to solve for x. $\dfrac{d}{dx}\left(\sin{x}\right)\cos{x}+\dfrac{d}{dx}\left(\cos{x}\right)\sin{x}=0$ $\implies \cos{x}\times\cos{x}+(-\sin{x})\sin{x}=0$ $\implies \cos^{2}{x}-\sin^{2}{x}=0$ Now factorize and solve. $(\cos{x}+\sin{x})(\cos{x}-\sin{x})=0$ $\implies (\cos{x}+\sin{x})=0$ or $(\cos{x}-\sin{x})=0$ $\implies \cos{x}=-\sin{x}$ or $\cos{x}=\sin{x}$ Now divide by $\cos{x}$ in both the equation. $\implies \dfrac{-\sin{x}}{\cos{x}}=1$ or $\dfrac{\sin{x}}{\cos{x}}=1$ $\implies \tan{x}=-1$ or $\tan{x}=1$ Since, $x$ is between $0$ and $2\pi$. If $\tan{x}=1$ Then, $x=\dfrac{\pi}{4},\dfrac{5\pi}{4}$ If $\tan{x}=-1$ Then, $x=\dfrac{3\pi}{4},\dfrac{7\pi}{4}$
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