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Use the Mean Value Theorem to prove the inequality

\(\left| {\sin a - \sin b} \right| \le \left| {a - b} \right|\)for all \(a\) and \(b\).

Short Answer

Expert verified

It is proved that \(\left| {\sin a - \sin b} \right| \le \left| {a - b} \right|\) for all \(a\) and \(b\).

Step by step solution

01

The mean value theorem

Let \(f\) be a function that is continuous on the closed interval \(\left( {a,b} \right)\) and differentiable on the open interval \(\left( {a,b} \right)\). Then there is a number \(c \in \left( {a,b} \right)\) such as that \(f'\left( c \right) = \frac{{f\left( b \right) - f\left( b \right)}}{{b - a}}\).

02

Proof

Suppose the function \(f\left( x \right) = \sin x\) on \(\left( {a,b} \right)\) . So, it is clear that \(f\left( x \right)\) is continuous on \(\left( {a,b} \right)\) and differentiable on \(\left( {a,b} \right)\). Therefore, this function satisfies the conditions for Mean Value Theorem.

By Mean Value Theorem, there exists a number \(c \in \left( {a,b} \right)\) such that

\(\begin{aligned}{c}f'\left( c \right) &= \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\\\cos \left( c \right) &= \frac{{\sin b - \sin a}}{{b - a}}\end{aligned}\)

It is known that \( - 1 < \cos c < 1\) or \(\left| {\cos c} \right| \le 1\). That implies:

\(\begin{array}{c}\left| {\frac{{\sin b - \sin a}}{{b - a}}} \right| \le 1\\\frac{{\left| {\sin b - \sin a} \right|}}{{\left| {b - a} \right|}} \le 1\\\left| {\sin b - \sin a} \right| \le \left| {b - a} \right|\end{array}\)

Solve further to get;

\(\left| {\sin a - \sin b} \right| \le \left| {a - b} \right|\).

Hence, it is proved that \(\left| {\sin a - \sin b} \right| \le \left| {a - b} \right|\).

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