9Tt sin 8. Find the exact value of: sin

Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.
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How do I find the exact value ?
**Problem**

2) Find the exact value of: 

\[
\sin^{-1} \left[ \sin \left( \frac{9\pi}{8} \right) \right].
\]

**Solution**

The function \(\sin^{-1}(x)\), also known as arcsin, returns the angle \(\theta\) in the range \([- \frac{\pi}{2}, \frac{\pi}{2}]\) such that \(\sin(\theta) = x\).

To solve \(\sin^{-1} \left[ \sin \left( \frac{9\pi}{8} \right) \right]\), we must first determine the angle whose sine is equal to \(\sin \left( \frac{9\pi}{8} \right)\) and that lies within the principal range of arcsin.

First, observe that:

1. \(\frac{9\pi}{8}\) is greater than \(\pi\) and less than \(\frac{3\pi}{2}\). This means \(\frac{9\pi}{8}\) lives in the third quadrant, where sine is negative.
2. The reference angle corresponding to \(\frac{9\pi}{8}\) is \(\frac{9\pi}{8} - \pi = \frac{\pi}{8}\).
3. Since sine has a period of \(2\pi\), the sine value of \(\frac{9\pi}{8}\) is negative and equals \(-\sin \left( \frac{\pi}{8} \right)\).

We need an equivalent angle within \([- \frac{\pi}{2}, \frac{\pi}{2}]\) that has the same sine value. As \(\sin \left( \frac{9\pi}{8} \right) = -\sin \left( \frac{\pi}{8} \right)\),

- The angle with sine \(-\sin \left( \frac{\pi}{8} \right)\) in the range \([- \frac{\pi}{2}, \frac{\pi}{2}]\) is \(-\frac{\pi}{8}\).

Therefore, 

\[
\sin^{-1} \left[ \sin \left( \frac{9\pi}{8} \right) \right] = -\frac{\pi}{8}.
\
Transcribed Image Text:**Problem** 2) Find the exact value of: \[ \sin^{-1} \left[ \sin \left( \frac{9\pi}{8} \right) \right]. \] **Solution** The function \(\sin^{-1}(x)\), also known as arcsin, returns the angle \(\theta\) in the range \([- \frac{\pi}{2}, \frac{\pi}{2}]\) such that \(\sin(\theta) = x\). To solve \(\sin^{-1} \left[ \sin \left( \frac{9\pi}{8} \right) \right]\), we must first determine the angle whose sine is equal to \(\sin \left( \frac{9\pi}{8} \right)\) and that lies within the principal range of arcsin. First, observe that: 1. \(\frac{9\pi}{8}\) is greater than \(\pi\) and less than \(\frac{3\pi}{2}\). This means \(\frac{9\pi}{8}\) lives in the third quadrant, where sine is negative. 2. The reference angle corresponding to \(\frac{9\pi}{8}\) is \(\frac{9\pi}{8} - \pi = \frac{\pi}{8}\). 3. Since sine has a period of \(2\pi\), the sine value of \(\frac{9\pi}{8}\) is negative and equals \(-\sin \left( \frac{\pi}{8} \right)\). We need an equivalent angle within \([- \frac{\pi}{2}, \frac{\pi}{2}]\) that has the same sine value. As \(\sin \left( \frac{9\pi}{8} \right) = -\sin \left( \frac{\pi}{8} \right)\), - The angle with sine \(-\sin \left( \frac{\pi}{8} \right)\) in the range \([- \frac{\pi}{2}, \frac{\pi}{2}]\) is \(-\frac{\pi}{8}\). Therefore, \[ \sin^{-1} \left[ \sin \left( \frac{9\pi}{8} \right) \right] = -\frac{\pi}{8}. \
Expert Solution
Step 1

Given:

Trigonometry homework question answer, step 1, image 1

The inverse sine function is defined in the interval -π/2,π/2.

From the above expression,

Trigonometry homework question answer, step 1, image 2

The angle π+π/8 lies in the 3rd quadrant, where the value of the sine function is negative.

Trigonometry homework question answer, step 1, image 3

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