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Chapter 19: Problem 1816

If \(A=\cos ^{4} \theta+\sin ^{2} \theta, \forall \theta \in R\) then \(A\) lies in the interval (a) \([1,2]\) (b) \([(3 / 4), 1]\) (c) \([(13 / 16), 1]\) (d) \([(3 / 4),(13 / 16)]\)

Short Answer

Expert verified
The interval in which A lies is \((c) \; [(13/16), 1]\).

Step by step solution

01

To find the interval of A, let's re-write the expression in terms of only one trigonometric function. We'll use the identity: \[\sin^2{\theta} = 1 - \cos^2{\theta}\] Now, substitute this identity in the original expression: \[A = \cos^4{\theta} + (1 - \cos^2{\theta})\] #Step 2: Compile the expression for easy evaluation

The next step is to compile the expression, so it can be easily analyzed. \[A = \cos^4{\theta} - \cos^2{\theta} + 1\] #Step 3: Find the critical points of the expression
02

To determine the range of A, we need to find the critical points where the derivative is equal to zero or does not exist. Let's use chain rule to find the derivative of A with respect to θ. \[A'(\theta) = \frac{d}{d\theta}(\cos^4{\theta} - \cos^2{\theta} + 1)\] #Step 4: Apply chain rule and simplify the expression

Applying the chain rule: \[A'(\theta) = 4\cos^3{\theta}\cdot (-\sin{\theta}) - 2\cos{\theta}\cdot (-\sin{\theta})\] Now, let's simplify the expression by factoring out the common term -\(\sin{\theta}\): \[A'(\theta) = \sin{\theta}\cdot[2\cos{\theta} - 4\cos^3{\theta}]\] #Step 5: Find values of θ for the critical points
03

Now, let's find the values of θ for which A'(θ) = 0. This is possible for two cases: 1. sin(θ) = 0 2. 2cos(θ) - 4cos^3(θ) = 0 For case 1: sin(θ) = 0 This occurs when θ is a multiple of π: θ = nπ, where n is an integer. For case 2: 2cos(θ) - 4cos^3(θ) = 0 Factor out 2cos(θ) from the equation: 2cos(θ)(1 - 2cos^2(θ)) = 0 This equation is satisfied when cos(θ) = 0 or 1 - 2cos^2(θ) = 0. The latter case implies that \(cos^{2} \theta= \frac{1}{2}\), θ can be \(\pm\frac{\pi}{4}, \pm\frac{5\pi}{4}... \) #Step 6: Evaluate the expression for the critical points and identify the interval

Plug in these critical points of θ (nπ, ±π/4, and ±5π/4) into the expression for A to find the maximum and minimum possible values: 1. For θ = nπ: \[A = \cos^4{nπ} - \cos^2{nπ} + 1 = 1\] 2. For θ = ±π/4: \[A = \cos^4{(\pm \pi/4)} - \cos^2{(\pm \pi/4)} + 1 = \frac{13}{16}\] 3. For θ = ±5π/4: \[A = \cos^4{(\pm 5\pi/4)} - \cos^2{(\pm 5\pi/4)} + 1 = \frac{13}{16}\] From these values, we can conclude that A can attain values between the minimum value 13/16 and the maximum value 1. Therefore, the interval for A is: \[\boxed{(c) \; [(13/16), 1]}\

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Most popular questions from this chapter

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