Question

If \( A \) and \( B \) are acute angles such that \( \sin A = \sin^2 B \) and \( 2\cos^2 A = 3\cos^2 B \), then \( (A, B) \) is:

• A. \( \left(\frac{\pi}{6}, \frac{\pi}{4}\right) \)
• B. \( \left(\frac{\pi}{6}, \frac{\pi}{6}\right) \)
• C. \( \left(\frac{\pi}{4}, \frac{\pi}{6}\right) \)
• D. \( \left(\frac{\pi}{4}, \frac{\pi}{4}\right) \)

Hint:

For trigonometric equations, use known values of sin and cos for standard angles to test conditions

Answer 1

The Correct Option is A

1. Given:

\[ \sin A = \sin^2 B \quad \text{and} \quad 2 \cos^2 A = 3 \cos^2 B. \]

2. Because \( A \) and \( B \) are acute:

\[ \sin A = \sin^2 B \implies A = \arcsin(\sin^2 B). \]

• When \( B = \frac{\pi}{4} \), \( \sin B = \frac{\sqrt{2}}{2} \) and \( \sin^2 B = \frac{1}{2} \).
• Consequently, \( \sin A = \frac{1}{2} \implies A = \frac{\pi}{6}. \)

3. Inserting \( A = \frac{\pi}{6} \) and \( B = \frac{\pi}{4} \) into the second equation:

\[ 2 \cos^2 \frac{\pi}{6} = 3 \cos^2 \frac{\pi}{4}. \]

4. Verification:

• \( \cos^2 \frac{\pi}{6} = \frac{3}{4}, \quad \cos^2 \frac{\pi}{4} = \frac{1}{2}. \)
• \( 2 \cdot \frac{3}{4} = 3 \cdot \frac{1}{2}. \)

- This is true.

Therefore, \( (A, B) = \left( \frac{\pi}{6}, \frac{\pi}{4} \right). \)