/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 Prove that \(|\cos a-\cos b| \le... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
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 by step solution

01

Use the Mean Value Theorem

By the Mean value theorem, we have that there exists some \(c\) between \(a\) and \(b\) such that \(\cos a - \cos b = (a - b)(-\sin c)\).
02

Apply the absolute value to both sides

Applying the absolute value to both sides we get: \(|\cos a-\cos b| = |a-b||-\sin c|\)
03

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.
04

Use the property of inequality

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

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