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Chapter 4: Problem 110

Determine whether the statement is true or false. Justify your answer. $$\sin \frac{5 \pi}{6}=\frac{1}{2} \quad \rightarrow \quad \arcsin \frac{1}{2}=\frac{5 \pi}{6}$$

Short Answer

Expert verified
The statement is false, because while \( \sin \frac{5 \pi}{6}=\frac{1}{2} \), the inverse sine of \( \frac{1}{2} \) is \( \frac{\pi}{6} \), not \( \frac{5 \pi}{6} \).

Step by step solution

01

Understand the Problem Statement

The problem involves a trigonometric function (\(\sin\)) and its inverse (\(\arcsin\)). The statement given is \(\sin \frac{5 \pi}{6}=\frac{1}{2} \rightarrow \arcsin \frac{1}{2}=\frac{5 \pi}{6}\). If \( \sin \frac{5 \pi}{6}=\frac{1}{2} \), it doesn't necessarily mean that \( \arcsin \frac{1}{2}=\frac{5 \pi}{6} \), because the range of inverse sine is \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
02

Calculate \(\sin \frac{5 \pi}{6}\)

The sine of \( \frac{5 \pi}{6} \) equals \( \frac{1}{2} \). This can be found from the unit circle or a table of trigonometric values. So the first half of the statement is true.
03

Calculate \(\arcsin \frac{1}{2}\)

The arcsine or inverse sine of \( \frac{1}{2} \) is \( \frac{\pi}{6} \), not \( \frac{5 \pi}{6} \), because the range of the inverse sine function is \( - \frac{\pi}{2} \leq x \leq \frac{\pi}{2} \). So the second half of the statement is false.

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