Problem 83 - Calculus of a Single Variable

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, b according to the Mean Value Theorem.

Step 1: Understand the Problem

First, the inequality \(|\cos a-\cos b|\leq |a-b|\) is given. The mission for this exercise is to prove that it is true for all a, b.

Step 2: Applying Trigonometric Identity

Use the trigonometric identity \(\cos a - \cos b = -2\sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)\)

Step 3: Taking Absolute Value

Taking the absolute value of both sides: \(|\cos a - \cos b| = |-2\sin(\frac{a+b}{2})\sin(\frac{a-b}{2})|\). Because the absolute value of any number is always positive, Therefore, we have \(|\cos a- \cos b| = 2|\sin(\frac{a+b}{2})\sin(\frac{a-b}{2})|\)

Step 4: Using the Fact that Absolute Value of Sine is Less than or Equals to One

We know that \(|\sin x| \leq 1\). Therefore, we can say that \(|\cos a- \cos b| \leq 2|\sin(\frac{a-b}{2})|\)

Step 5: Using Mean Value Theorem

According to Mean Value Theorem, there exists a number 'c' between a and b such that \(\sin(c)=\sin(\frac{a-b}{2})\), \(\frac{a-b}{2}=\cos(c)\), hence \(|\sin(\frac{a-b}{2})|=|\frac{a-b}{2}|\). Therefore, \(2|\sin(\frac{a-b}{2})|\leq 2|a-b|\)

Step 6: Final Inequality

So we have \(|\cos a- \cos b| \leq 2|\sin(\frac{a-b}{2})| \leq 2|a-b|\), thus \(|\cos a - \cos b|\leq |a-b|\). The inequality holds true for all a, b