If `sin x = 1/3and cos y = 1/3`, then what is the value of `sin (x + y)` ?

Options

A. `2//3`
B. `4//9`
C. `5//9`
D. 1

Answer

To find the value of \( \sin(x + y) \) given that \( \sin x = \frac{1}{3} \) and \( \cos y = \frac{1}{3} \), we can use the sine addition formula:

\[ \sin(x + y) = \sin x \cos y + \cos x \sin y \]

### Step 1: Find \( \sin x \) and \( \cos y \)

We are given:
- \( \sin x = \frac{1}{3} \)
- \( \cos y = \frac{1}{3} \)

### Step 2: Find \( \cos x \)

Using the Pythagorean identity:
\[ \sin^2 x + \cos^2 x = 1 \]
Substituting \( \sin x \):
\[ \left(\frac{1}{3}\right)^2 + \cos^2 x = 1 \]
\[ \frac{1}{9} + \cos^2 x = 1 \]
\[ \cos^2 x = 1 - \frac{1}{9} = \frac{8}{9} \]
Taking the square root:
\[ \cos x = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3} \]

### Step 3: Find \( \sin y \)

Using the Pythagorean identity again:
\[ \sin^2 y + \cos^2 y = 1 \]
Substituting \( \cos y \):
\[ \sin^2 y + \left(\frac{1}{3}\right)^2 = 1 \]
\[ \sin^2 y + \frac{1}{9} = 1 \]
\[ \sin^2 y = 1 - \frac{1}{9} = \frac{8}{9} \]
Taking the square root:
\[ \sin y = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3} \]

### Step 4: Substitute values into the sine addition formula

Now we can substitute \( \sin x \), \( \cos y \), \( \cos x \), and \( \sin y \) into the sine addition formula:
\[ \sin(x + y) = \sin x \cos y + \cos x \sin y \]
\[ = \left(\frac{1}{3}\right) \left(\frac{1}{3}\right) + \left(\frac{2\sqrt{2}}{3}\right) \left(\frac{2\sqrt{2}}{3}\right) \]
\[ = \frac{1}{9} + \frac{4 \cdot 2}{9} = \frac{1}{9} + \frac{8}{9} = \frac{9}{9} = 1 \]

### Final Answer:
\[ \sin(x + y) = 1 \]