If `A,Bin(0,pi//2),sinA=(4)/(5)andcos(A+B)=-(12)/(13)`, then `sinB=?`

Options

A. `(61)/(65)`
B. `(8)/(65)`
C. `(63)/(65)`
D. `(5)/(13)`

Answer

To solve the problem step by step, we will use the given information and trigonometric identities.

### Step 1: Identify Given Values
We are given:
- \( \sin A = \frac{4}{5} \)
- \( \cos(A + B) = -\frac{12}{13} \)

### Step 2: Find \( \cos A \)
Using the Pythagorean identity:
\[ \sin^2 A + \cos^2 A = 1 \]
Substituting the value of \( \sin A \):
\[ \left(\frac{4}{5}\right)^2 + \cos^2 A = 1 \]
\[ \frac{16}{25} + \cos^2 A = 1 \]
\[ \cos^2 A = 1 - \frac{16}{25} = \frac{9}{25} \]
Taking the square root:
\[ \cos A = \frac{3}{5} \quad (\text{since } A \in (0, \frac{\pi}{2})) \]

### Step 3: Find \( \sin(A + B) \)
Using the identity:
\[ \sin^2(A + B) + \cos^2(A + B) = 1 \]
Substituting the value of \( \cos(A + B) \):
\[ \sin^2(A + B) + \left(-\frac{12}{13}\right)^2 = 1 \]
\[ \sin^2(A + B) + \frac{144}{169} = 1 \]
\[ \sin^2(A + B) = 1 - \frac{144}{169} = \frac{25}{169} \]
Taking the square root:
\[ \sin(A + B) = \frac{5}{13} \quad (\text{since } A + B \in (0, \frac{\pi}{2})) \]

### Step 4: Use the Sine Difference Formula
To find \( \sin B \), we can use the sine difference formula:
\[ \sin B = \sin(A + B - A) = \sin(A + B) \cos A - \cos(A + B) \sin A \]
Substituting the known values:
\[ \sin B = \left(\frac{5}{13}\right) \left(\frac{3}{5}\right) - \left(-\frac{12}{13}\right) \left(\frac{4}{5}\right) \]

### Step 5: Calculate \( \sin B \)
Calculating each term:
\[ \sin B = \frac{5 \cdot 3}{13 \cdot 5} + \frac{12 \cdot 4}{13 \cdot 5} \]
\[ = \frac{15}{65} + \frac{48}{65} \]
\[ = \frac{15 + 48}{65} = \frac{63}{65} \]

### Final Answer
Thus, the value of \( \sin B \) is:
\[ \sin B = \frac{63}{65} \]