If `0 lt beta lt alpha lt pi//4, cos(alpha+beta)=3//5` and ` cos(alpha-beta)=4//5`, then evaluate `sin 2alpha`.

Answer

To solve the problem step by step, we will use the given values of \( \cos(\alpha + \beta) \) and \( \cos(\alpha - \beta) \) to find \( \sin(2\alpha) \).

### Step 1: Use the given values of cosine
We are given:
\[ \cos(\alpha + \beta) = \frac{3}{5} \]
\[ \cos(\alpha - \beta) = \frac{4}{5} \]

### Step 2: Find \( \sin(\alpha + \beta) \)
Using the Pythagorean identity:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
we can find \( \sin(\alpha + \beta) \):
\[ \sin(\alpha + \beta) = \sqrt{1 - \cos^2(\alpha + \beta)} = \sqrt{1 - \left(\frac{3}{5}\right)^2} \]
Calculating this gives:
\[ \sin(\alpha + \beta) = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

### Step 3: Find \( \sin(\alpha - \beta) \)
Similarly, we find \( \sin(\alpha - \beta) \):
\[ \sin(\alpha - \beta) = \sqrt{1 - \cos^2(\alpha - \beta)} = \sqrt{1 - \left(\frac{4}{5}\right)^2} \]
Calculating this gives:
\[ \sin(\alpha - \beta) = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

### Step 4: Use the sine addition formula
Now, we can use the sine addition formula to find \( \sin(2\alpha) \):
\[ \sin(2\alpha) = \sin((\alpha + \beta) + (\alpha - \beta)) = \sin(\alpha + \beta)\cos(\alpha - \beta) + \cos(\alpha + \beta)\sin(\alpha - \beta) \]

### Step 5: Substitute the values
Substituting the values we found:
\[ \sin(2\alpha) = \left(\frac{4}{5}\right) \left(\frac{4}{5}\right) + \left(\frac{3}{5}\right) \left(\frac{3}{5}\right) \]
Calculating this gives:
\[ \sin(2\alpha) = \frac{16}{25} + \frac{9}{25} = \frac{25}{25} = 1 \]

### Final Answer
Thus, the value of \( \sin(2\alpha) \) is:
\[ \sin(2\alpha) = 1 \]

---