If sin 3 A = 1 , then how many distinct values can sin A assume ?

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the equation \( \sin 3A = 1 \) and determine how many distinct values \( \sin A \) can assume, we can follow these steps:

### Step 1: Understand the equation
We start with the equation:
\[ \sin 3A = 1 \]
The sine function equals 1 at specific angles. The general solution for \( \sin x = 1 \) is:
\[ x = \frac{\pi}{2} + 2n\pi, \quad n \in \mathbb{Z} \]
Thus, we can write:
\[ 3A = \frac{\pi}{2} + 2n\pi \]

### Step 2: Solve for \( A \)
To find \( A \), we divide both sides by 3:
\[ A = \frac{\pi}{6} + \frac{2n\pi}{3}, \quad n \in \mathbb{Z} \]

### Step 3: Find the values of \( \sin A \)
Now, we need to find \( \sin A \):
\[ \sin A = \sin\left(\frac{\pi}{6} + \frac{2n\pi}{3}\right) \]
Using the sine addition formula:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
where \( A = \frac{\pi}{6} \) and \( B = \frac{2n\pi}{3} \).

Calculating \( \sin \frac{\pi}{6} = \frac{1}{2} \) and \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \), we have:
\[ \sin A = \frac{1}{2} \cos\left(\frac{2n\pi}{3}\right) + \frac{\sqrt{3}}{2} \sin\left(\frac{2n\pi}{3}\right) \]

### Step 4: Evaluate \( \cos\left(\frac{2n\pi}{3}\right) \) and \( \sin\left(\frac{2n\pi}{3}\right) \)
The values of \( \cos\left(\frac{2n\pi}{3}\right) \) and \( \sin\left(\frac{2n\pi}{3}\right) \) depend on \( n \):
- For \( n = 0 \): \( \cos(0) = 1 \), \( \sin(0) = 0 \) → \( \sin A = \frac{1}{2} \)
- For \( n = 1 \): \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \), \( \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \) → \( \sin A = \frac{1}{2} \cdot \left(-\frac{1}{2}\right) + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = -\frac{1}{4} + \frac{3}{4} = \frac{1}{2} \)
- For \( n = -1 \): \( \cos\left(-\frac{2\pi}{3}\right) = -\frac{1}{2} \), \( \sin\left(-\frac{2\pi}{3}\right) = -\frac{\sqrt{3}}{2} \) → \( \sin A = -\frac{1}{4} - \frac{3}{4} = -1 \)

### Step 5: Collect distinct values
From the calculations, we find:
1. \( \sin A = \frac{1}{2} \)
2. \( \sin A = -1 \)

### Conclusion
Thus, the distinct values that \( \sin A \) can assume are \( \frac{1}{2} \) and \( -1 \). Therefore, the number of distinct values \( \sin A \) can assume is:
\[ \boxed{2} \]