Problem 29 - Essential Calculus Early Transcendentals

Question

Use the Mean Value Theorem to prove the inequality $$|\sin a-\sin b| \leqslant|a-b| \quad\( for all \)a\( and \)b$$

Step-by-Step Solution

Answer

The inequality holds because the derivative \( |\cos(c)| \leq 1 \).

Step 1: Recall the Mean Value Theorem

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

Step 2: Define the Function and Its Derivative

Consider the function \( f(x) = \sin x \), which is continuous and differentiable everywhere. Its derivative is given by \[f'(x) = \cos x.\]

Step 3: Apply the Mean Value Theorem

By applying the Mean Value Theorem to \( f(x) = \sin x \) over the interval \([a, b]\), we have:\[\cos(c) = \frac{\sin b - \sin a}{b - a},\]where \( c \) is some point in \( (a, b) \).

Step 4: Evaluate the Derivative's Absolute Value

Recognize that the absolute value of \( \cos(c) \) satisfies \[|\cos(c)| \leq 1\]for any \( c \).

Step 5: Express the Inequality Using the Mean Value Theorem

From \( \cos(c) = \frac{\sin b - \sin a}{b - a} \), we can multiply both sides by \( |b-a| \) to get: \[|\sin b - \sin a| = |\cos(c)||b-a|.\]

Step 6: Simplify the Inequality

Using the property \( |\cos(c)| \leq 1 \), we deduce that:\[|\sin b - \sin a| \leq |b-a|,\]as required.

Key Concepts

Trigonometric Functions

Trigonometric functions are a pivotal part of mathematics that deal with the angles and sides of triangles. The most common trigonometric functions are sine, cosine, and tangent. These functions are periodic, which means they repeat their values in regular intervals.
In the context of this problem, we are using the sine function, denoted as \( \sin x \). It's crucial to understand how the sine function behaves:

• \( \sin x \) is a continuous function, meaning it has no breaks or jumps.
• The values of \( \sin x \) range between -1 and 1.
• The function is smooth and differentiable across its entire domain.

Our task involves proving an inequality involving sine values at two different angles \( a \) and \( b \), and how these compare with the linear distance between \( a \) and \( b \). Understanding the behavior of sine is fundamental to applying the Mean Value Theorem effectively in this instance.

Continuous Function

A continuous function is a function without any interruptions, holes, or jumps in its graph. This means you can draw it without lifting your pencil off the paper. Continuity is an essential property if we want to apply certain theorems in calculus, such as the Mean Value Theorem.
For the sine function \( f(x) = \sin x \):

• It is continuous across all real numbers, meaning it seamlessly connects from one point to another.
• To check continuity, you can verify that the limit of \( f(x) \) as \( x \) approaches any given point is equal to \( f \) evaluated at that point.
• This characteristics allow us to use the Mean Value Theorem on \( \sin x \) over any interval.

Continuous functions are predictable, which is why they enable powerful tools like calculus to function effectively.

Derivative

The derivative provides us with the rate at which a function changes. When you differentiate a function, you are essentially finding its slope at any given point. For trigonometric functions, knowing the derivative is especially useful in understanding their behavior.
For our function \( f(x) = \sin x \):

• The derivative is \( f'(x) = \cos x \).
• This tells us the slope or rate of change of \( \sin x \) at any point \( x \).
• The derivative, \( \cos x \), is bounded by values between -1 and 1, which affects how much \( \sin x \) can change over a small interval.

By applying these properties in the context of the Mean Value Theorem, we can leverage the derivative to establish relationships, such as inequalities, involving \( \sin x \).

Inequality Proof

Proving inequalities often involves showing that one side of an expression does not exceed the other under the given conditions. In our problem, we're proving an inequality involving the sine function using the Mean Value Theorem as our tool.
The Mean Value Theorem helps us by stating that for a function that is continuous over \([a, b]\) and differentiable over \((a, b)\), there exists a point \( c \) such that:

• \( f'(c) = \frac{f(b) - f(a)}{b - a} \)
• Here, applying \( f(x) = \sin x \), we found \( \cos(c) = \frac{\sin b - \sin a}{b - a} \).

Recognizing that \( \cos(c) \) is bounded such that \(|\cos(c)| \leq 1\), allows for simplification:

• By multiplying the inequality by \(|b-a|\), we derive \( |\sin b - \sin a| = |\cos(c)| |b-a| \).
• That leads us to conclude that \(|\sin b - \sin a| \leq |b-a|\).

Through this logical chain, we prove the inequality utilizing trigonometric properties and calculus, showcasing the power and utility of these mathematical techniques.