What is sin`(alpha+beta) - 2 sin alpha cos beta + sin(alpha- beta)` =

Options

A. 0
B. 2sin`alpha`
C. 2sin`beta`
D. `sin alpha+sin beta`

Answer

To solve the expression \( \sin(\alpha + \beta) - 2 \sin \alpha \cos \beta + \sin(\alpha - \beta) \), we can use trigonometric identities. Let's break it down step by step.

### Step 1: Apply the sine addition and subtraction formulas
We know the following identities:
- \( \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \)
- \( \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \)

Using these identities, we can rewrite the expression:

\[ \sin(\alpha + \beta) + \sin(\alpha - \beta) = (\sin \alpha \cos \beta + \cos \alpha \sin \beta) + (\sin \alpha \cos \beta - \cos \alpha \sin \beta) \]

### Step 2: Combine the terms
Now, combine the terms from the rewritten expression:

\[ = \sin \alpha \cos \beta + \cos \alpha \sin \beta + \sin \alpha \cos \beta - \cos \alpha \sin \beta \]

### Step 3: Simplify the expression
Notice that the \( \cos \alpha \sin \beta \) terms cancel out:

\[ = 2 \sin \alpha \cos \beta \]

Now, we substitute this back into the original expression:

\[ \sin(\alpha + \beta) - 2 \sin \alpha \cos \beta + \sin(\alpha - \beta) = 2 \sin \alpha \cos \beta - 2 \sin \alpha \cos \beta \]

### Step 4: Final simplification
This simplifies to:

\[ = 0 \]

### Conclusion
Thus, the value of the expression \( \sin(\alpha + \beta) - 2 \sin \alpha \cos \beta + \sin(\alpha - \beta) \) is \( 0 \).