Problem 61 - Essential Calculus

Question

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

Step-by-Step Solution

Answer

The inequality \(|\cos a-\cos b| \leq|a-b|\) holds true for all \(a\) and \(b\), as proven through the Mean Value Theorem and the properties of the sine function.

Step 1: Use the Mean Value Theorem

By the Mean value theorem, we have that there exists some \(c\) between \(a\) and \(b\) such that \(\cos a - \cos b = (a - b)(-\sin c)\).

Step 2: Apply the absolute value to both sides

Applying the absolute value to both sides we get: \(|\cos a-\cos b| = |a-b||-\sin c|\)

Step 3: Identify the range of sine function

The sine function ranges from -1 to 1. Hence, \(|-\sin c|\) can be anywhere between 0 and 1 inclusive.

Step 4: Use the property of inequality

By the properties of inequality, \(|a-b||-\sin c| \leq |a-b|\). hence we can conclude \(|\cos a - \cos b| \leq |a - b|\)