Consider the system in ordered pairs `(x,y)` of real numbers `sin x+sin y=sin(x+y)`,`|x|+|y|=1`. The number of ordered pairs `(x,y)` satisfying the system is

Answer

To solve the system of equations given by \( \sin x + \sin y = \sin(x+y) \) and \( |x| + |y| = 1 \), we will proceed step by step.

### Step 1: Simplify the first equation
We start with the equation:
\[ \sin x + \sin y = \sin(x+y) \]
Using the sine addition formula, we can express \( \sin(x+y) \) as:
\[ \sin(x+y) = \sin x \cos y + \cos x \sin y \]
Thus, we rewrite the equation as:
\[ \sin x + \sin y = \sin x \cos y + \cos x \sin y \]

### Step 2: Rearranging the equation
Rearranging gives:
\[ \sin x - \sin x \cos y + \sin y - \cos x \sin y = 0 \]
Factoring out common terms, we have:
\[ \sin x (1 - \cos y) + \sin y (1 - \cos x) = 0 \]

### Step 3: Analyzing the factors
This equation can be satisfied if either:
1. \( \sin x = 0 \) or
2. \( \sin y = 0 \)

### Step 4: Solving for \( \sin x = 0 \)
If \( \sin x = 0 \), then:
\[ x = n\pi \quad (n \in \mathbb{Z}) \]

### Step 5: Solving for \( \sin y = 0 \)
If \( \sin y = 0 \), then:
\[ y = m\pi \quad (m \in \mathbb{Z}) \]

### Step 6: Using the second equation
Now we apply the second equation \( |x| + |y| = 1 \).

#### Case 1: \( x = 0 \)
If \( x = 0 \), then:
\[ |y| = 1 \implies y = 1 \text{ or } y = -1 \]
This gives us the points:
- \( (0, 1) \)
- \( (0, -1) \)

#### Case 2: \( y = 0 \)
If \( y = 0 \), then:
\[ |x| = 1 \implies x = 1 \text{ or } x = -1 \]
This gives us the points:
- \( (1, 0) \)
- \( (-1, 0) \)

### Step 7: Collecting all solutions
Combining all the points we have found, we get:
1. \( (0, 1) \)
2. \( (0, -1) \)
3. \( (1, 0) \)
4. \( (-1, 0) \)

### Conclusion
Thus, the total number of ordered pairs \( (x, y) \) satisfying the system is:
\[ \boxed{4} \]