If `sin^(2)alpha+cos^(2)beta=2`, then find `sin((2alpha+beta)/(3))`.

Options

A. `sin""(alpha)/(3)`
B. `cos""(2alpha)/(3)`
C. `cos""(alpha)/(3)`
D. none of these

Answer

To solve the problem, we need to find the value of \(\sin\left(\frac{2\alpha + \beta}{3}\right)\) given that \(\sin^2 \alpha + \cos^2 \beta = 2\).

### Step-by-Step Solution:

1. **Understanding the Equation**:
We are given:
\[ \sin^2 \alpha + \cos^2 \beta = 2 \]
The maximum value of \(\sin^2 \alpha\) is 1 and the maximum value of \(\cos^2 \beta\) is also 1. Therefore, for the sum to equal 2, both \(\sin^2 \alpha\) and \(\cos^2 \beta\) must be at their maximum values:
\[ \sin^2 \alpha = 1 \quad \text{and} \quad \cos^2 \beta = 1 \]

2. **Finding Angles**:
From \(\sin^2 \alpha = 1\), we deduce:
\[ \sin \alpha = 1 \implies \alpha = 90^\circ \]
From \(\cos^2 \beta = 1\), we deduce:
\[ \cos \beta = 1 \implies \beta = 0^\circ \]

3. **Substituting Values**:
Now, we substitute \(\alpha\) and \(\beta\) into the expression we need to evaluate:
\[ \sin\left(\frac{2\alpha + \beta}{3}\right) = \sin\left(\frac{2 \times 90^\circ + 0^\circ}{3}\right) \]

4. **Calculating the Argument**:
Simplifying the argument:
\[ 2 \times 90^\circ = 180^\circ \implies \frac{180^\circ + 0^\circ}{3} = \frac{180^\circ}{3} = 60^\circ \]

5. **Finding the Sine Value**:
Now we find:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

### Final Answer:
Thus, the value of \(\sin\left(\frac{2\alpha + \beta}{3}\right)\) is:
\[ \frac{\sqrt{3}}{2} \]