If `x sin^(3) alpha + y cos ^(3) alpha = sin alpha cos alpha ` and `x sin alpha = y cos alpha`, then `x^(2) + y^(2) =`

Options

A. 0
B. 1
C. 2
D. none

Answer

To solve the problem step by step, we start with the given equations:

1. **Given Equations**:
\[ x \sin^3 \alpha + y \cos^3 \alpha = \sin \alpha \cos \alpha \quad \text{(1)} \]
\[ x \sin \alpha = y \cos \alpha \quad \text{(2)} \]

2. **From Equation (2)**, we can express \(y\) in terms of \(x\):
\[ y = \frac{x \sin \alpha}{\cos \alpha} \]

3. **Substituting \(y\) in Equation (1)**:
Replace \(y\) in Equation (1):
\[ x \sin^3 \alpha + \left(\frac{x \sin \alpha}{\cos \alpha}\right) \cos^3 \alpha = \sin \alpha \cos \alpha \]
Simplifying the second term:
\[ x \sin^3 \alpha + x \sin \alpha \cos^2 \alpha = \sin \alpha \cos \alpha \]

4. **Factoring out \(x\)**:
\[ x (\sin^3 \alpha + \sin \alpha \cos^2 \alpha) = \sin \alpha \cos \alpha \]

5. **Using the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\)**:
We can rewrite \(\sin^3 \alpha\) as \(\sin \alpha (\sin^2 \alpha)\):
\[ x \sin \alpha (\sin^2 \alpha + \cos^2 \alpha) = \sin \alpha \cos \alpha \]
This simplifies to:
\[ x \sin \alpha = \sin \alpha \cos \alpha \]

6. **Dividing both sides by \(\sin \alpha\)** (assuming \(\sin \alpha \neq 0\)):
\[ x = \cos \alpha \]

7. **Finding \(y\)**:
Substitute \(x\) back into Equation (2):
\[ x \sin \alpha = y \cos \alpha \implies \cos \alpha \sin \alpha = y \cos \alpha \]
Dividing both sides by \(\cos \alpha\) (assuming \(\cos \alpha \neq 0\)):
\[ y = \sin \alpha \]

8. **Finding \(x^2 + y^2\)**:
Now we have:
\[ x = \cos \alpha \quad \text{and} \quad y = \sin \alpha \]
Therefore,
\[ x^2 + y^2 = \cos^2 \alpha + \sin^2 \alpha = 1 \]

Thus, the final answer is:
\[ \boxed{1} \]