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. Since $A$ and $B$ are acute:

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

• For $B = \frac{\pi}{4}$, $\sin B = \frac{\sqrt{2}}{2}$ and $\sin^2 B = \frac{1}{2}$.
• Therefore, $\sin A = \frac{1}{2} \implies A = \frac{\pi}{6}.$

3. Substitute $A = \frac{\pi}{6}$ and $B = \frac{\pi}{4}$ into the second condition:

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

4. Verify:

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

- This holds true.

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