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Chapter 5: Problem 44

Explain why $$ |\sin \theta| \leq|\tan \theta| $$ for all \(\theta\) such that \(\tan \theta\) is defined.

Short Answer

Expert verified
The given inequality can be rewritten as \(|\sin \theta| \leq \left|\frac{\sin \theta}{\cos \theta}\right|\). By assuming \(\cos \theta \neq 0\) and multiplying both sides by \(|\cos \theta|\), we get \(|\sin \theta| \cdot |\cos \theta| \leq \left|\sin \theta\right|\). Dividing by \(|\sin \theta|\) when \(\sin \theta \neq 0\), we have that \(|\cos \theta| \leq 1\), which is always true. The inequality \(|\sin \theta| \leq |\tan \theta|\) holds for all \(\theta\) where \(\tan \theta\) is defined.

Step by step solution

01

Rewrite using sine and cosine

We can rewrite the inequality using the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). The given inequality becomes: \[ |\sin \theta| \leq \left|\frac{\sin \theta}{\cos \theta}\right| \]
02

Remove sine from both sides

Since we wish to prove the inequality \(|\sin \theta| \leq |\tan \theta|\) for all values of \(\theta\) such that \(\tan \theta\) is defined, we must assume \(\cos \theta \neq 0\). Thus, we can multiply both sides by \(|\cos \theta|\) without changing the inequality: \[ |\sin \theta| \cdot |\cos \theta| \leq \left|\sin \theta\right| \]
03

Simplify

Since both sides of the inequality have \(\sin \theta\), we can divide by \(|\sin \theta|\), as long as \(\sin \theta \neq 0\). The inequality becomes: \[ |\cos \theta| \leq 1 \]
04

Address the cases when sine or cosine are 0

Now let's consider the cases when either or both \(\sin \theta = 0\) or \(\cos \theta = 0\). If \(\sin \theta = 0\), the inequality \(|\sin \theta| \leq |\tan \theta|\) holds, because the left side of the inequality is \(0\) and \(\tan \theta\) is always nonnegative. If \(\cos \theta = 0\), then \(\tan \theta\) is undefined, so there is no need to check the inequality.
05

Conclusion

We have shown that the inequality \(|\sin \theta| \leq |\tan \theta|\) holds for all \(\theta\) where \(\tan \theta\) is defined.

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