If `sin (3A-B) = 1 and cos (2A - B) = (sqrt(3))/(2)` , then the value of sin A and cos B are .

Options

A. `(sqrt(3))/(2),0`
B. `(1)/(2) , 0 `
C. `(sqrt(3))/(2) , 1`
D. `(1)/(2) , 1`

Answer

To solve the problem, we need to find the values of \( \sin A \) and \( \cos B \) given the equations:

1. \( \sin(3A - B) = 1 \)
2. \( \cos(2A - B) = \frac{\sqrt{3}}{2} \)

### Step 1: Analyze the first equation
From the first equation, we know that:
\[ \sin(3A - B) = 1 \]
The sine function equals 1 at \( 90^\circ + 360^\circ n \) where \( n \) is any integer. For simplicity, we will consider the principal value:
\[ 3A - B = 90^\circ \]
This gives us our first equation:
\[ 3A - B = 90^\circ \quad \text{(Equation 1)} \]

### Step 2: Analyze the second equation
From the second equation, we have:
\[ \cos(2A - B) = \frac{\sqrt{3}}{2} \]
The cosine function equals \( \frac{\sqrt{3}}{2} \) at \( 30^\circ + 360^\circ n \) or \( 330^\circ + 360^\circ n \). Again, we will consider the principal values:
\[ 2A - B = 30^\circ \quad \text{(Equation 2)} \]
or
\[ 2A - B = 330^\circ \quad \text{(Equation 3)} \]

### Step 3: Solve Equations 1 and 2
Let's first solve Equations 1 and 2:
1. \( 3A - B = 90^\circ \)
2. \( 2A - B = 30^\circ \)

We can subtract Equation 2 from Equation 1:
\[ (3A - B) - (2A - B) = 90^\circ - 30^\circ \]
This simplifies to:
\[ 3A - 2A = 60^\circ \]
\[ A = 60^\circ \]

### Step 4: Substitute \( A \) back to find \( B \)
Now we substitute \( A = 60^\circ \) back into Equation 2:
\[ 2(60^\circ) - B = 30^\circ \]
\[ 120^\circ - B = 30^\circ \]
\[ B = 120^\circ - 30^\circ = 90^\circ \]

### Step 5: Find \( \sin A \) and \( \cos B \)
Now we can find \( \sin A \) and \( \cos B \):
\[ \sin A = \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
\[ \cos B = \cos(90^\circ) = 0 \]

### Final Answer
Thus, the values are:
\[ \sin A = \frac{\sqrt{3}}{2}, \quad \cos B = 0 \]