If cos `(2pi)/3 =sin alpha`,which could be `alpha` ?

Options

A. `pi/6`
B. `-pi/6`
C. `(2pi)/3`
D. `-(2pi)/3`

Answer

To solve the problem, we need to find the value of \( \alpha \) such that \( \cos \left( \frac{2\pi}{3} \right) = \sin \alpha \).

### Step 1: Calculate \( \cos \left( \frac{2\pi}{3} \right) \)
The angle \( \frac{2\pi}{3} \) is in the second quadrant. We can express it as:
\[ \cos \left( \frac{2\pi}{3} \right) = \cos \left( \pi - \frac{\pi}{3} \right) \]
Using the cosine identity \( \cos(\pi - \theta) = -\cos(\theta) \), we have:
\[ \cos \left( \frac{2\pi}{3} \right) = -\cos \left( \frac{\pi}{3} \right) \]
Since \( \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} \), it follows that:
\[ \cos \left( \frac{2\pi}{3} \right) = -\frac{1}{2} \]

### Step 2: Set the equation
Now we have:
\[ \sin \alpha = -\frac{1}{2} \]

### Step 3: Find \( \alpha \)
The sine function is negative in the third and fourth quadrants. The angles where \( \sin \alpha = -\frac{1}{2} \) are:
\[ \alpha = -\frac{\pi}{6} \quad \text{(4th quadrant)} \]
and
\[ \alpha = \frac{7\pi}{6} \quad \text{(3rd quadrant)} \]

### Step 4: Check the options
From the options provided:
- Option A: \( \frac{\pi}{6} \) (not correct)
- Option B: \( -\frac{\pi}{6} \) (correct)
- Option C: \( \frac{2\pi}{3} \) (not correct)
- Option D: \( -\frac{2\pi}{3} \) (not correct)

Thus, the correct answer is:
\[ \alpha = -\frac{\pi}{6} \]

### Final Answer:
The value of \( \alpha \) that satisfies the equation \( \cos \left( \frac{2\pi}{3} \right) = \sin \alpha \) is:
\[ \alpha = -\frac{\pi}{6} \]