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Chapter 3: Problem 84

Prove that \(|\sin a-\sin b| \leq|a-b|\) for all \(a\) and \(b\).

Short Answer

Expert verified
Applying Mean Value Theorem to the function \(\sin x\) on the interval [a,b], there must exist at least one point c in (a,b) such that \(\cos c = \frac{\sin b - \sin a}{b - a}\). Since \( |\cos c| \leq 1\) for all permitted c, we have \( |\frac{\sin b - \sin a}{b - a}| \leq 1\). Multiplying both sides of the inequality by the non-negative quantity \( |b - a| \), we get \( |\sin b - \sin a| \leq |b - a|\).

Step by step solution

01

Preliminary thoughts

First, observe that both \(|a-b|\) and \(|\sin a - \sin b|\) are non-negative quantities. We also need to note that both \(a-b\) and \(\sin a - \sin b\) are continuous functions, which is a pre-requisite for applying the Mean Value Theorem.
02

Applying Mean Value Theorem

The Mean Value Theorem (MVT) states that if f is a function continuous on a closed interval [a, b] and differentiable on an open interval (a, b), then there exists at least one point c in (a, b) at which the derivative of the function equals the average change in y-coordinates per unit change in x-coordinates between the points (a, f(a)) and (b, f(b)). In mathematical terms, \(f'(c) = \frac{f(b)-f(a)}{b-a}\). Applying MVT to the function \(f(x) = \sin x\) on the interval [a,b], there must exist at least one point c in (a,b) such that \(\cos c = \frac{\sin b - \sin a}{b - a}\). Since \( |\cos c| \leq 1\) for all permitted c, we have \( |\frac{\sin b - \sin a}{b - a}| \leq 1\).
03

Final rearrangement

Finally, multiplying both sides of the inequality by the non-negative quantity \( |b - a| \), we get \( |\sin b - \sin a| \leq |b - a|\). The right side of the inequality is what we want, thus concluding our proof.

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