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Chapter 6: Problem 6

Use the Mean Value Theorem to prove that \(|\sin x-\sin y| \leq|x-y|\) for all \(x, y\) in \(\mathbb{R}\).

Short Answer

Expert verified
Used Mean Value Theorem (MVT) and properties of trigonometric functions, particularly the cosine function, to prove that \(|\sin x-\sin y| \leq|x-y|\) for all \(x, y\) in \(\mathbb{R}\).

Step by step solution

01

Declare the function and Interval

Select the function \(f(x) = \sin x\), and let the interval be \([x, y]\) for \(x, y\) in \(\mathbb{R}\) and \(x < y\). The function is continuous and differentiable on the chosen interval, which are preconditions for applying the MVT.
02

Apply the Mean Value Theorem

Apply the MVT. According to MVT, there exists at least one point \(c\) in the interval \((x, y)\) such that \(f'(c) = \frac{f(y) - f(x)}{y - x}\). So, \(f'(c) = \frac{\sin y - \sin x}{y - x}\).
03

Compute the Derivative

The derivative of the function \(f(x) = \sin x\) is \(f'(x) = \cos x\). Therefore \(f'(c) = \cos c\). Hence, \(\cos c = \frac{\sin y - \sin x}{y - x}\).
04

Apply the properties of Cosine function

Remember that in general, \(-1 \leq \cos c \leq 1\). We can now use this property to establish the inequality \(-1 \leq \frac{\sin y - \sin x}{y - x} \leq 1\). Multiplying all parts of the inequality by \(|y - x|\), we get that \(-|y-x| \leq \sin y - \sin x \leq |y - x|\). Taking absolute values, we finally obtain that \(|\sin y - \sin x| \leq |y - x|\), which is the required proof.

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