Explanation Given Information: The provided trigonometric function is f ( x ) = sin 2 x + sin x over the interval ( 0 , 2 π ) . Formula used: Sine derivative Rule: d d x [ sin u ] = cos u d u d x General power derivative rule: d d x x n = n x n − 1 Write trigonometric reciprocal identity. tan θ = sin θ cos θ Condition for critical points: d y d x = f ′ ( x ) = 0 If function f ( x ) is increasing, then f ′ ( x ) > 0 and if function is decreasing, then f ′ ( x ) < 0 . From first derivative Test, If f ′ ( x ) changes positive to negative, then function f ( x ) has relative maxima and, if f ′ ( x ) changes negative to positive then function f ( x ) has relative minima. Calculation: Consider the provided trigonometric function: f ( x ) = sin 2 x + sin x Differentiate the above function with respect to x using Sine, and Power derivative rules. f ′ ( x ) = d d x [ sin 2 x + sin x ] = d d x [ sin 2 x ] + d d x [ sin x ] = 2 sin x d d x [ sin x ] + cos x = 2 sin x cos x + cos x Apply f ′ ( x ) = 0 in order to find critical points. 2 sin x cos x + cos x = 0 cos x ( 2 sin x + 1 ) = 0 Further solve. cos x = 0 And, 2 sin x + 1 = 0 2 sin x = − 1 sin x = − 1 2 Since, the value of the cos θ is 0 at θ = π 2 , and θ = 3 π 2 , and the value of the sin θ is − 1 2 at θ = 7 π 6 , and θ = 11 π 6 . So, cos x = 0 cos x = cos π 2 x = π 2 , cos x = 0 cos x = cos 3 π 2 x = 3 π 2 , sin x = − 1 2 sin x = sin 7 π 6 x = 7 π 6 And, sin x = − 1 2 sin x = sin 11 π 6 x = 11 π 6 Thus, critical points are x = π 2 , x = 7 π 6 , x = 3 π 2 and x = 11 π 6 . Now, rewrite the first derivative of the function. f ′ ( x ) = 2 sin x cos x + cos x At x = π 3 . f ′ ( π 3 ) = 2 sin π 3 cos π 3 + cos π 3 = 2 ( 3 2 ) ( 1 2 ) + 1 2 = 1.366 = + ve > 0 At x = 2 π 3 . f ′ ( 2 π 3 ) = 2 sin 2 π 3 cos 2 π 3 + cos 2 π 3 = 2 ( 3 2 ) ( − 1 2 ) + ( − 1 2 ) = − 1.366 = − ve < 0 At x = 5 π 4 . f ′ ( 5 π 4 ) = 2 sin 5 π 4 cos 5 π 4 + cos 5 π 4 = 2 ( − 2 2 ) ( − 2 2 ) + ( − 2 2 ) = 0.29289 = + ve > 0 At x = 5 π 3 . f ′ ( 5 π 3 ) = 2 sin 5 π 3 cos 5 π 3 + cos 5 π 3 = 2 ( − 3 2 ) ( 1 2 ) + ( 1 2 ) = − 0.29289 = − ve < 0 At x = 23 π 12 . f ′ ( 23 π 12 ) = 2 sin 23 π 12 cos 23 π 12 + cos 23 π 12 = 2 ( − 0.2588 ) ( 0.9659 ) + 0.9659 = 0

Calculus: An Applied Approach (MindTap Course List)

10th Edition

ISBN: 9781305860919

Author: Ron Larson

Publisher: Cengage Learning

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Chapter 8, Problem 95RE

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To calculate: The relative extrema of the trigonometric function f(x)=sin2x+sinx over the interval (0,2π).

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Chapter 8, Problem 94RE

Chapter 8, Problem 96RE

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Calculus: An Applied Approach (MindTap Course List)

Ch. 8.1 - For each angle, find a coterminal angle such that...Ch. 8.1 - Prob. 2CP Ch. 8.1 - Prob. 3CP Ch. 8.1 - Prob. 4CP Ch. 8.1 - Prob. 1SWU Ch. 8.1 - Prob. 2SWU Ch. 8.1 - Prob. 3SWU Ch. 8.1 - Prob. 4SWU Ch. 8.1 - Prob. 5SWU Ch. 8.1 - Prob. 6SWU

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The relative extrema of the trigonometric function f ( x ) = sin 2 x + sin x over the interval ( 0 , 2 π ) .

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To calculate: The relative extrema of the trigonometric function f(x)=sin2x+sinx over the interval (0,2π).

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Chapter 8 Solutions