Question

If \( x + y = \frac{\pi}{2} \), then the maximum value of \( \sin x \cdot \sin y \) is

• A. \( \frac{1}{2} \)
• B. \( -\frac{1}{2} \)
• C. \( -\frac{1}{\sqrt{2}} \)
• D. \( \frac{1}{\sqrt{2}} \)

Hint:

To maximize \( \sin x \cdot \cos x \), use the fact that the maximum value of \( \sin 2x \) is 1.

Answer 1

The Correct Option is A

Step 1: Use the given condition.
We are given that \( x + y = \frac{\pi}{2} \). This implies \( y = \frac{\pi}{2} - x \). We are asked to find the maximum value of \( \sin x \cdot \sin y \).
Step 2: Use the trigonometric identity.
Using the identity \( \sin \left( \frac{\pi}{2} - x \right) = \cos x \), we get: \[ \sin x \cdot \sin y = \sin x \cdot \cos x. \] This is a standard trigonometric expression, and its maximum value occurs when \( x = \frac{\pi}{4} \), at which point: \[ \sin \left( \frac{\pi}{4} \right) \cdot \cos \left( \frac{\pi}{4} \right) = \frac{1}{2}. \]
Step 3: Conclusion.
Thus, the maximum value of \( \sin x \cdot \sin y \) is \( \frac{1}{2} \), corresponding to option (A).