If sinA`=3/5,90^(@)leAlt+180^(@),cosB=1/3`, and `270^(@)leBle360^(@)" , "sin(A+B)=`

Options

A. -0.832
B. -0.554
C. -0.333
D. 0.954

Explanation

To solve the problem, we need to find the value of \( \sin(A + B) \) given \( \sin A = \frac{3}{5} \) (with \( 90^\circ \leq A < 180^\circ \)) and \( \cos B = \frac{1}{3} \) (with \( 270^\circ \leq B < 360^\circ \)).

### Step-by-Step Solution:

1. **Identify the Quadrants**:
- Since \( A \) is in the second quadrant, \( \sin A \) is positive and \( \cos A \) is negative.
- Since \( B \) is in the fourth quadrant, \( \cos B \) is positive and \( \sin B \) is negative.

2. **Find \( \cos A \)**:
- We know \( \sin A = \frac{3}{5} \).
- Using the Pythagorean identity:
\[ \cos^2 A = 1 - \sin^2 A \]
\[ \cos^2 A = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \]
- Therefore, \( \cos A = -\sqrt{\frac{16}{25}} = -\frac{4}{5} \) (negative in the second quadrant).

3. **Find \( \sin B \)**:
- We know \( \cos B = \frac{1}{3} \).
- Using the Pythagorean identity:
\[ \sin^2 B = 1 - \cos^2 B \]
\[ \sin^2 B = 1 - \left(\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9} \]
- Therefore, \( \sin B = -\sqrt{\frac{8}{9}} = -\frac{2\sqrt{2}}{3} \) (negative in the fourth quadrant).

4. **Apply the Sine Addition Formula**:
- The formula for \( \sin(A + B) \) is:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
- Substituting the values we found:
\[ \sin(A + B) = \left(\frac{3}{5}\right)\left(\frac{1}{3}\right) + \left(-\frac{4}{5}\right)\left(-\frac{2\sqrt{2}}{3}\right) \]
\[ = \frac{3}{15} + \frac{8\sqrt{2}}{15} \]
\[ = \frac{3 + 8\sqrt{2}}{15} \]

5. **Final Result**:
- Thus, the value of \( \sin(A + B) \) is:
\[ \sin(A + B) = \frac{3 + 8\sqrt{2}}{15} \]