Find sin 2π ( (27) :)) COS 3 Show your work.
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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cannot use a calculator like a TI-89 or a TI-nSprire CA way, please
![### Problem Statement
Find \( \sin^{-1} \left( \cos \left( \frac{2\pi}{3} \right) \right) \). Show your work.
### Solution
To find \( \sin^{-1} \left( \cos \left( \frac{2\pi}{3} \right) \right) \), follow these steps:
1. **Evaluate \(\cos \left( \frac{2\pi}{3} \right)\):**
2. **Analyze the unit circle** to determine the cosine value at \( \frac{2\pi}{3} \).
- \( \frac{2\pi}{3} \) is located in the second quadrant of the unit circle.
- The reference angle for \( \frac{2\pi}{3} \) is \( \pi - \frac{2\pi}{3} = \frac{\pi}{3} \).
3. **Determine the cosine value** for this angle:
- In the second quadrant, cosine is negative.
- \( \cos \left( \frac{2\pi}{3} \right) = -\cos \left( \frac{\pi}{3} \right) \).
- Since \( \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} \),
\[
\cos \left( \frac{2\pi}{3} \right) = -\frac{1}{2}.
\]
4. **Solve \( \sin^{-1} \left( -\frac{1}{2} \right) \):**
- The inverse sine problem \( \sin^{-1} \left( x \right) \) means finding an angle whose sine is \( x \).
- For \( x = -\frac{1}{2} \),
- The angle \(-\frac{\pi}{6}\) (or in some conventions, \(\frac{11\pi}{6}\)) has a sine of \(-\frac{1}{2}\).
Therefore,
\[
\sin^{-1} \left( \cos \left( \frac{2\pi}{3} \right) \right) = \sin^{-1} \left( -\frac{1}{2} \right) = -\frac{\pi}{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4d6261b1-d641-4c15-8b49-7d4d38863372%2F3f8e27fa-9df0-4f11-bf6f-50a432b6e466%2F3hzppiq_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Find \( \sin^{-1} \left( \cos \left( \frac{2\pi}{3} \right) \right) \). Show your work.
### Solution
To find \( \sin^{-1} \left( \cos \left( \frac{2\pi}{3} \right) \right) \), follow these steps:
1. **Evaluate \(\cos \left( \frac{2\pi}{3} \right)\):**
2. **Analyze the unit circle** to determine the cosine value at \( \frac{2\pi}{3} \).
- \( \frac{2\pi}{3} \) is located in the second quadrant of the unit circle.
- The reference angle for \( \frac{2\pi}{3} \) is \( \pi - \frac{2\pi}{3} = \frac{\pi}{3} \).
3. **Determine the cosine value** for this angle:
- In the second quadrant, cosine is negative.
- \( \cos \left( \frac{2\pi}{3} \right) = -\cos \left( \frac{\pi}{3} \right) \).
- Since \( \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} \),
\[
\cos \left( \frac{2\pi}{3} \right) = -\frac{1}{2}.
\]
4. **Solve \( \sin^{-1} \left( -\frac{1}{2} \right) \):**
- The inverse sine problem \( \sin^{-1} \left( x \right) \) means finding an angle whose sine is \( x \).
- For \( x = -\frac{1}{2} \),
- The angle \(-\frac{\pi}{6}\) (or in some conventions, \(\frac{11\pi}{6}\)) has a sine of \(-\frac{1}{2}\).
Therefore,
\[
\sin^{-1} \left( \cos \left( \frac{2\pi}{3} \right) \right) = \sin^{-1} \left( -\frac{1}{2} \right) = -\frac{\pi}{
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