If `sinalpha+cosbeta=2(o^(@)lebetaltalphale90^(@))` , then `sin((2alpha+beta)/(3))=`

Options

A. `"sin"(alpha)/(2)`
B. `"cos"(alpha)/(3)`
C. `"sin"(alpha)/(3)`
D. `"cos"(2alpha)/(3)`

Answer

To solve the problem step by step, we begin with the given equation:

**Given:**
\[ \sin \alpha + \cos \beta = 2 \]
with the constraints:
\[ 0^\circ \leq \beta < \alpha \leq 90^\circ \]

### Step 1: Analyze the equation
The maximum value of \(\sin \alpha\) is 1 (which occurs when \(\alpha = 90^\circ\)) and the maximum value of \(\cos \beta\) is also 1 (which occurs when \(\beta = 0^\circ\)). Therefore, the only way for \(\sin \alpha + \cos \beta\) to equal 2 is if:
\[ \sin \alpha = 1 \quad \text{and} \quad \cos \beta = 1 \]

### Step 2: Determine values of \(\alpha\) and \(\beta\)
From the above analysis:
- \(\sin \alpha = 1\) implies \(\alpha = 90^\circ\)
- \(\cos \beta = 1\) implies \(\beta = 0^\circ\)

### Step 3: Substitute values into the expression
Now we need to find:
\[ \sin\left(\frac{2\alpha + \beta}{3}\right) \]

Substituting the values we found:
\[ 2\alpha = 2 \times 90^\circ = 180^\circ \]
\[ \beta = 0^\circ \]

Thus,
\[ 2\alpha + \beta = 180^\circ + 0^\circ = 180^\circ \]

### Step 4: Calculate the sine value
Now, substitute this into the sine function:
\[ \sin\left(\frac{180^\circ}{3}\right) = \sin(60^\circ) \]

### Step 5: Find the sine of 60 degrees
We know:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

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