Problem 63 - Thomas Calculus

Question

Show that \(|\cos x-1| \leq|x|\) for all \(x\) -values. (Hint: Consider \(f(t)=\cos t\) on \([0, x].)\)

Step-by-Step Solution

Answer

The inequality \(|\cos x - 1| \leq |x|\) holds for all \(x\).

Step 1: Understand the Problem Statement

We need to prove that the absolute difference between \(\cos x\) and 1 is less than or equal to the absolute value of \(x\). This should be true for all values of \(x\).

Step 2: Consider the Given Hint and Define the Function

The hint suggests considering the function \(f(t) = \cos t\) on the interval \([0, x]\). This will help in analyzing the properties of the cosine function and its derivative over this interval.

Step 3: Evaluate the Derivative

The derivative of \(f(t) = \cos t\) is \(f'(t) = -\sin t\). This tells us how \(f(t)\) changes on the interval. Note that the absolute value of \(-\sin t\) (which is \(\sin t\) itself in absolute value) is always less than or equal to 1.

Step 4: Apply the Mean Value Theorem

The Mean Value Theorem for the interval \([0, x]\) tells us that there exists some \(c \in (0, x)\) such that \(f'(c) = \frac{f(x) - f(0)}{x - 0}\). Substitute the values: \(-\sin c = \frac{\cos x - 1}{x}\).

Step 5: Analyze the Inequality

Rearranging gives \(|\cos x - 1| = |x \sin c|\). Since \(|\sin c| \leq 1\), we have \(|x \sin c| \leq |x|\). Thus, we conclude that \(|\cos x - 1| \leq |x|\).

Step 6: Conclude the Proof

Since every step relied on correctly applying known theorems and properties, the inequality \(|\cos x - 1| \leq |x|\) is valid for all \(x\).

Key Concepts

cosine function

The cosine function, commonly written as \( \cos \theta \), is a fundamental trigonometric function that represents the x-coordinate of a point on the unit circle as the angle \( \theta \) is drawn from the positive x-axis. It gets its name from being the complement of the sine angle. - The function exhibits a wave-like pattern that oscillates between 1 and -1.- It has a period of \( 2\pi \), meaning it repeats every \( 2\pi \) radians.- The cosine function is even, which means \( \cos(-x) = \cos(x) \).When considering the function \( f(t) = \cos t \), as in the exercise, we examine how it behaves over a specific interval, in this case \([0, x]\). The goal is to show that any deviation from its value at 0 (i.e., \( \cos 0 = 1 \)) is less than or equal to the absolute value of the interval length \( x \). To achieve this, we evaluate its changing rate through its derivative.

Mean Value Theorem

The Mean Value Theorem (MVT) is a crucial concept in calculus that connects the derivative of a function to differences in function values over an interval. The theorem states that for a continuous function \( f \) over \([a, b]\), and differentiable on \((a, b)\), there is at least one point \( c \) within \( (a, b) \) where:\[ f'(c) = \frac{f(b) - f(a)}{b - a}\]- This relation suggests that the slope of the tangent at some point \( c \) is equal to the average rate of change over the interval.- In the given problem, the MVT is used for the function \( f(t) = \cos t \) on \([0, x]\).Applying the theorem, we deduce that there exists some \( c \in (0, x) \) such that the derivative \( f'(c) = -\sin c \) equals the difference quotient \( \frac{\cos x - 1}{x} \). This equivalence forms the basis for proving the inequality.

derivative evaluation

Derivative evaluation involves calculating how a function's output changes with a change in input. For a trigonometric function like the cosine function, this step is crucial in understanding its behavior.Given \( f(t) = \cos t \), the derivative is calculated as:\[ f'(t) = -\sin t \]- This derivative \(-\sin t\) tells us the rate at which the cosine function is changing at any point \( t \).- Since the sine function has values constrained between -1 and 1, it implies that the magnitude of \(-\sin t\) isn't greater than 1. In the context of the inequality \(|\cos x - 1| \leq |x|\), calculating the derivative of the cosine function helps us use the MVT to establish this inequality rigorously. It tells us about the bounds on \(|x \sin c|\), confirming that it is contained within \(|x|\), hence supporting the proof by constraining \( |\cos x - 1| \). By consistently evaluating this derivative, we can apply mathematical principles to ensure that our analysis holds for all \( x \).