Problem 45 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the Mean Value Theorem to show that $$ |\sin x-\sin y| \leq|x-y| $$

Step-by-Step Solution

Answer

\(|\sin x - \sin y| \leq |x - y|\) by applying the Mean Value Theorem.

Step 1: Recall the Mean Value Theorem

The Mean Value Theorem states that if a function \( f \) is continuous on a closed interval \( [a, b] \) and differentiable on the open interval \( (a, b) \), then there exists a point \( c \) in \( (a, b) \) such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \] In our case, the function is \( f(x) = \sin x \).

Step 2: Apply the Mean Value Theorem

Consider the function \( f(x) = \sin x \) on the closed interval \([y, x]\). According to the Mean Value Theorem, there exists some \( c \in (y, x) \) such that:\[ \sin x - \sin y = \cos(c) \cdot (x - y) \]

Step 3: Use Properties of Cosine Function

We know that \( |\cos(c)| \leq 1 \) for any real number \( c \). Therefore, we can modify the equation from Step 2 to \[ |\sin x - \sin y| = |\cos(c)| \cdot |x - y| \leq |x - y| \]

Step 4: Conclude the Proof Using the Inequality

Since \(|\cos(c)| \leq 1\), the inequality simplifies to:\[ |\sin x - \sin y| \leq |x - y| \]This confirms the statement using the Mean Value Theorem.

Key Concepts

Inequalities

Inequalities can be a bit tricky to understand at first, but they’re essential in math. An inequality compares two values, showing if one is less than, greater than, or simply not equal to the other. In your exercise, you look at the inequality \[|\sin x - \sin y| \leq |x-y|\]. It means that the absolute difference between the sine of two angles is less than or equal to the difference between those two angles. Why is this important? It helps us understand the behavior of functions, like how fast or slow they change.

• You may run into terms like 'strict inequality' (like < or >) and 'non-strict inequality' (like ≤ or ≥). These show whether equality is part of the solution.
• Inequalities have rules, much like equations. For example, you can add or multiply both sides by the same positive number without changing the inequality.
• Understanding inequalities allows you to grasp bigger concepts in calculus and analysis.

Trigonometric Functions

Trigonometric functions, like sine and cosine, arise from the relationships of the angles and sides of a right triangle. The sine function, \( \sin \theta \), gives the ratio of the length of the opposite side to the hypotenuse of a right triangle with angle \( \theta \). These functions are periodic and oscillate, always staying within a particular range based on their nature.

• The sine function oscillates between -1 and 1. This bounded nature contributes to many properties of sine, including the inequalities discussed in your problem.
• Besides sine, cosine and tangent are the other basic trigonometric functions. Understanding these can help you model wave-like scenarios, such as sound waves or light waves.
• Trigonometric identities, like \( \sin^2 x + \cos^2 x = 1\), are crucial for solving many mathematical problems efficiently.

Continuity

Continuity is one of those concepts that might sound complicated but is simple once you break it down. A function is continuous if you can draw it without lifting your pencil. In mathematical terms, this means that for every input close to a specific point, the output is also close to the function’s value at that point.

• Continuity is important because it guarantees that functions behave in a predictable way between known values.
• In the context of the Mean Value Theorem, the function \( \sin x \) is continuous on any interval you choose because it does not 'break' at any point.
• Understanding how continuity works can help you understand limits, another central concept in calculus.

Differentiability

Differentiability takes the idea of continuity a step further. If a function is differentiable, it is also continuous. Differentiability means that a function has a derivative at every point in its domain, or in simpler terms, it has a defined slope everywhere.

• A differentiable function doesn’t just smoothly connect all its parts; each little segment has a slope—meaning you can compute changes instantaneously.
• For the Mean Value Theorem to apply, like in your problem, the function \( \sin x \) needs to be differentiable over the interval \( (y, x) \). This allows you to find points where the derivative matches the average rate of change over the interval.
• The derivative of \( \sin x \) is \( \cos x \), and knowing this derivative helps you understand the rate at which \( \sin x \) changes around any given point.

Understanding differentiability ultimately allows you to tackle more complex calculus problems, paving the way for analyzing the behavior of more diverse functions.