If `alphaandbeta` are such that `tanalpha=2tanbeta`, then what is `sin(alpha+beta)` equal to ?

Options

A. 1
B. `2sin(alpha-beta)`
C. `sin(alpha-beta)`
D. `3sin(alpha-beta)`

Explanation

To solve the problem where \( \tan \alpha = 2 \tan \beta \) and we need to find \( \sin(\alpha + \beta) \), we can follow these steps:

### Step 1: Express \( \tan \alpha \) in terms of \( \tan \beta \)
Given:
\[ \tan \alpha = 2 \tan \beta \]

### Step 2: Use the sine and cosine definitions of tangent
Recall that:
\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \quad \text{and} \quad \tan \beta = \frac{\sin \beta}{\cos \beta} \]
Thus, we can write:
\[ \frac{\sin \alpha}{\cos \alpha} = 2 \cdot \frac{\sin \beta}{\cos \beta} \]

### Step 3: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ \sin \alpha \cos \beta = 2 \sin \beta \cos \alpha \]

### Step 4: Rearrange to find \( \sin(\alpha + \beta) \)
We know from the sine addition formula:
\[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \]
From our previous equation, we can substitute \( \sin \alpha \cos \beta \):
\[ \sin(\alpha + \beta) = 2 \sin \beta \cos \alpha + \cos \alpha \sin \beta \]
This simplifies to:
\[ \sin(\alpha + \beta) = \sin \beta (2 \cos \alpha + \cos \alpha) = \sin \beta (3 \cos \alpha) \]

### Step 5: Express \( \sin(\alpha + \beta) \) in terms of \( \sin \beta \) and \( \cos \alpha \)
Thus, we have:
\[ \sin(\alpha + \beta) = 3 \sin \beta \cos \alpha \]

### Final Result
Therefore, the value of \( \sin(\alpha + \beta) \) is:
\[ \sin(\alpha + \beta) = 3 \sin \beta \cos \alpha \]