Problem 28 - Single Variable Calculus

Question

Use the Mean Value Theorem to prove that \(|\sin a-\sin b| \leq|a-b|\) for all real numbers \(a\) and \(b\).

Step-by-Step Solution

Answer

To prove the inequality \(|\sin a - \sin b| \leq |a - b|\) for all real numbers \(a\) and \(b\), we apply the Mean Value Theorem to the sine function \(f(x) = \sin x\). We find its derivative \(f'(x) = \cos x\) and then establish that there exists an intermediate point \(c\) such that \(\cos c = \frac{\sin b - \sin a}{b - a}\). Taking the absolute value on both sides, we have \(|\cos c| = \left|\frac{\sin b - \sin a}{b - a}\right|\). Since \(|\cos c| \leq 1\), multiplying both sides by \(|b - a|\) yields \(|\sin b-\sin a| \leq |a-b|\), proving the inequality.

Step 1: Apply the Mean Value Theorem to the sine function

We are given \(|\sin a - \sin b| \leq |a - b|\). To use the Mean Value Theorem, let's consider the sine function \(f(x) = \sin x\). Since \(\sin x\) is continuous and differentiable on the entire real line, we can apply the Mean Value Theorem to the interval \([a, b]\), which gives us: There exists a point \(c\) in the interval \((a, b)\) such that \[f'(c) = \frac{f(b) - f(a)}{b - a}\]

Step 2: Find the derivative of the sine function

To apply the Mean Value Theorem, we need to find the derivative of the sine function. The derivative of \(f(x) = \sin x\) with respect to \(x\) is: \[f'(x) = \cos x\] Now, let's replace \(f'(x)\) with \(\cos x\) in our Mean Value Theorem equation: \[\cos c = \frac{\sin b - \sin a}{b - a}\]

Step 3: Take the absolute value of both sides

We want to prove an inequality, so let's take the absolute value of both sides of the equation: \[ |\cos c| = \left|\frac{\sin b - \sin a}{b - a}\right| \]

Step 4: Use the inequality property of the absolute value of the cosine function

Since the range of the cosine function is \([-1, 1]\), the absolute value of the cosine function is always in the range \([0, 1]\). Therefore, we have: \[|\cos c| \leq 1\]

Step 5: Prove the inequality

Now, let's multiply both sides of the previous equation by the absolute value of \((b - a)\): \[ |\cos c||b - a| \leq |b - a| \] From Step 3, we know that \( |\cos c| = \left|\frac{\sin b - \sin a}{b - a}\right|\). Therefore, we can write: \[ \left|\frac{\sin b - \sin a}{b - a}\right||b - a| \leq |b - a| \] Finally, multiplying both sides by the absolute value of \((b - a)\), we get: \[ |\sin b - \sin a| \leq |b - a| \] This is the inequality we wanted to prove. The proof is complete.

Key Concepts

Sine Function

The sine function, denoted by \(\sin x\), is a fundamental concept in trigonometry, and it's essential for understanding the behavior of waves and oscillating systems. This periodic function describes the relationship between the angle of a right-angled triangle and the ratio of the length of the side opposite the angle to the hypotenuse. Graphically, the sine function creates a wave-like pattern, repeating every \(2\pi\) radians, with values that oscillate between -1 and 1. \
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\ Understanding the sine function is crucial when applying mathematical theorems, such as the Mean Value Theorem. It's also worth mentioning that the sine function is differentiable, which means we can find its rate of change at any point along its domain, leading us to the concept of the derivative of the sine function.

Derivative of Sine Function

When we talk about the derivative of the sine function, we're looking at how the function \(\sin x\) changes at any given point. The derivative of \(\sin x\) with respect to \(x\) is \(\cos x\), which is another trigonometric function that relates to the sine function and the unit circle. \
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\ In the context of the Mean Value Theorem, we use the derivative \(\cos x\) to determine the average rate of change of the sine function over an interval. This average rate of change is what ties the behavior of the sine and cosine functions to the inequality involving the absolute value of the differences of their values, which is at the heart of the exercise we are discussing. Understanding how to differentiate functions, and in particular, the sine function, is key to solving a wide range of problems in calculus.

Absolute Value Inequalities

Absolute value inequalities are expressions that describe a range of values for which the absolute value of a difference or sum falls within a certain interval. The absolute value, denoted by \( |x| \), represents the distance of a number \(x\) from zero on the number line, regardless of the direction. \
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\ In the context of our exercise, absolute value inequalities are used to provide bounds on the possible values of trigonometric functions. For instance, knowing that \( |\cos x| \leq 1 \) for all real numbers \(x\) allows us to constrain the behavior of the sine function in the inequality we aim to prove. When combined with the Mean Value Theorem, which guarantees the existence of a certain value within an interval that satisfies a condition based on the derivative, we establish a link between the derivatives of trigonometric functions and the absolute values of their differences. This understanding enables us to solve complex inequalities and understand the relationships between trigonometric values.