if `cos alpha = sin beta` , and `alpha = (5pi)/6`, which could be a value of `beta` ?

Options

A. `-pi/6`
B. `pi/6`
C. `-pi/3`
D. `pi/3`

Answer

To solve the problem where \( \cos \alpha = \sin \beta \) and \( \alpha = \frac{5\pi}{6} \), we can follow these steps:

### Step 1: Find \( \cos \alpha \)
Given \( \alpha = \frac{5\pi}{6} \), we need to find \( \cos \left( \frac{5\pi}{6} \right) \).

### Step 2: Use the cosine identity
We can express \( \frac{5\pi}{6} \) as:
\[ \frac{5\pi}{6} = \pi - \frac{\pi}{6} \]
Using the cosine identity \( \cos(\pi - x) = -\cos(x) \), we have:
\[ \cos\left(\frac{5\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) \]

### Step 3: Calculate \( \cos\left(\frac{\pi}{6}\right) \)
The value of \( \cos\left(\frac{\pi}{6}\right) \) is:
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]
Thus,
\[ \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} \]

### Step 4: Set up the equation
Since \( \cos \alpha = \sin \beta \), we can write:
\[ \sin \beta = -\frac{\sqrt{3}}{2} \]

### Step 5: Find possible values of \( \beta \)
The sine function is negative in the third and fourth quadrants. The angles where \( \sin \beta = -\frac{\sqrt{3}}{2} \) are:
\[ \beta = -\frac{\pi}{3} \quad \text{(in the fourth quadrant)} \]
and
\[ \beta = \frac{4\pi}{3} \quad \text{(in the third quadrant)} \]

### Step 6: Identify the correct option
From the options provided, \( -\frac{\pi}{3} \) is one of the possible values of \( \beta \).

### Final Answer
Thus, the value of \( \beta \) could be:
\[ \beta = -\frac{\pi}{3} \]