If `alpha+beta=90^(@),alpha=2beta`, then find the value of `cos^(2)alpha+sin^(2)beta`.

Options

A. `(1)/(2)`
B. `(3)/(4)`
C. `(3)/(2)`
D. `(4)/(3)`

Answer

To solve the problem, we need to find the value of \( \cos^2 \alpha + \sin^2 \beta \) given that \( \alpha + \beta = 90^\circ \) and \( \alpha = 2\beta \).

### Step-by-Step Solution:

1. **Set up the equations**:
We have two equations:
\[ \alpha + \beta = 90^\circ \quad \text{(1)} \]
\[ \alpha = 2\beta \quad \text{(2)} \]

2. **Substitute equation (2) into equation (1)**:
Replace \( \alpha \) in equation (1) with \( 2\beta \):
\[ 2\beta + \beta = 90^\circ \]
This simplifies to:
\[ 3\beta = 90^\circ \]

3. **Solve for \( \beta \)**:
Divide both sides by 3:
\[ \beta = 30^\circ \]

4. **Find \( \alpha \)**:
Use equation (2) to find \( \alpha \):
\[ \alpha = 2\beta = 2 \times 30^\circ = 60^\circ \]

5. **Calculate \( \cos^2 \alpha \) and \( \sin^2 \beta \)**:
- First, find \( \cos^2 \alpha \):
\[ \cos^2 \alpha = \cos^2 60^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
- Next, find \( \sin^2 \beta \):
\[ \sin^2 \beta = \sin^2 30^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

6. **Add \( \cos^2 \alpha \) and \( \sin^2 \beta \)**:
\[ \cos^2 \alpha + \sin^2 \beta = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \]

### Final Answer:
The value of \( \cos^2 \alpha + \sin^2 \beta \) is \( \frac{1}{2} \).

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