If `alpha` and `beta` satisfy `sinalpha cos beta=-1/2`, then the greatest value of `2cos alpha sin beta` is ……….

Answer

To solve the problem, we need to find the greatest value of \(2 \cos \alpha \sin \beta\) given that \(\sin \alpha \cos \beta = -\frac{1}{2}\).

### Step 1: Use the given equation
We start with the equation:
\[ \sin \alpha \cos \beta = -\frac{1}{2} \]

### Step 2: Express \(2 \cos \alpha \sin \beta\)
We want to find the maximum value of \(2 \cos \alpha \sin \beta\). We can use the identity:
\[ 2 \cos \alpha \sin \beta = \sin(\alpha + \beta) - \sin(\alpha - \beta) \]

### Step 3: Set up the equations
Let’s denote:
\[ x = \sin(\alpha + \beta) \]
\[ y = \sin(\alpha - \beta) \]
Thus, we have:
\[ 2 \cos \alpha \sin \beta = x - y \]

### Step 4: Find maximum values of \(x\) and \(y\)
To maximize \(x - y\), we need to find the maximum values of \(x\) and \(y\).

From the given condition \(\sin \alpha \cos \beta = -\frac{1}{2}\), we can deduce that:
1. \(\sin \alpha\) can take values from \(-1\) to \(1\).
2. \(\cos \beta\) can also take values from \(-1\) to \(1\).

### Step 5: Analyze the maximum
To find the maximum of \(2 \cos \alpha \sin \beta\), we can consider the extreme values of \(\sin \alpha\) and \(\cos \beta\) that satisfy the equation \(\sin \alpha \cos \beta = -\frac{1}{2}\).

### Step 6: Substitute values
Assuming \(\sin \alpha = -1\) and \(\cos \beta = 1\), we have:
\[ -1 \cdot 1 = -1 \quad \text{(not valid)} \]
Assuming \(\sin \alpha = -\frac{1}{2}\) and \(\cos \beta = 1\), we have:
\[ -\frac{1}{2} \cdot 1 = -\frac{1}{2} \quad \text{(valid)} \]

### Step 7: Find \(2 \cos \alpha \sin \beta\)
From the equation:
\[ 2 \cos \alpha \sin \beta = 1 - (-1) = 2 \]

### Final Result
Thus, the greatest value of \(2 \cos \alpha \sin \beta\) is:
\[ \boxed{1} \]