If `cos(alpha + beta)=0`, then sin`(alpha-beta)` can be reduced to

Options

A. `cosbeta`
B. `cos2beta`
C. `sinalpha`
D. `sin2alpha`

Explanation

To solve the problem step by step, we start with the given condition and manipulate it using trigonometric identities.

### Step-by-Step Solution:

1. **Given Condition**: We start with the equation:
\[ \cos(\alpha + \beta) = 0 \]

2. **Understanding the Cosine Function**: We know that the cosine function equals zero at specific angles. The primary angle where this occurs is:
\[ \cos(90^\circ) = 0 \]
Therefore, we can equate:
\[ \alpha + \beta = 90^\circ \]

3. **Isolating Alpha**: From the equation above, we can express \(\alpha\) in terms of \(\beta\):
\[ \alpha = 90^\circ - \beta \]

4. **Finding \(\sin(\alpha - \beta)\)**: We need to find \(\sin(\alpha - \beta)\). Substituting \(\alpha\) into this expression gives:
\[ \sin(\alpha - \beta) = \sin((90^\circ - \beta) - \beta) \]
Simplifying the expression inside the sine function:
\[ = \sin(90^\circ - 2\beta) \]

5. **Using the Sine Identity**: We can use the identity for sine:
\[ \sin(90^\circ - \theta) = \cos(\theta) \]
Applying this identity here:
\[ \sin(90^\circ - 2\beta) = \cos(2\beta) \]

6. **Final Result**: Therefore, we conclude that:
\[ \sin(\alpha - \beta) = \cos(2\beta) \]

### Final Answer:
\[ \sin(\alpha - \beta) = \cos(2\beta) \]