If sin θ * cos θ = 1/2 and 0° < θ < 90°, find the exact value of: sin θ − cos θ Use standard identities to determine θ and then simplify the result.

Introduction / Context:
This question tests how to use product-to-angle identities and the acute-angle condition to determine an exact trigonometric result. The expression sin θ * cos θ is closely related to sin 2θ through the identity: sin 2θ = 2 sin θ cos θ. Whenever you see sin θ cos θ, converting it into sin 2θ usually simplifies the problem dramatically. Here, sin θ cos θ = 1/2 implies sin 2θ = 1, which fixes 2θ to a standard angle. Because θ is acute (0° < θ < 90°), 2θ lies between 0° and 180°, and sin 2θ = 1 occurs at exactly 90° in that range. So θ must be 45°. Once θ is known, sin θ and cos θ are equal, and their difference becomes 0. This is a typical aptitude trap: people try to compute sin and cos separately without realizing the angle is forced to 45° by the identity and domain restriction.

Given Data / Assumptions:

• sin θ * cos θ = 1/2 • 0° < θ < 90° (θ is acute) • Identity: sin 2θ = 2 sin θ cos θ • Required: sin θ − cos θ

Concept / Approach:
Use sin 2θ = 2 sin θ cos θ to convert the product into a single sine. Solve sin 2θ = 1 in the interval 0° < 2θ < 180°. Then compute sin θ − cos θ using the fact that at θ = 45°, sin θ equals cos θ exactly.

Step-by-Step Solution:
1) Start with the identity: sin 2θ = 2 sin θ cos θ 2) Given sin θ cos θ = 1/2, multiply by 2: 2 sin θ cos θ = 1 3) Therefore: sin 2θ = 1 4) For 0° < θ < 90°, we have 0° < 2θ < 180°. 5) In this range, sin 2θ = 1 occurs at: 2θ = 90° 6) Hence: θ = 45° 7) Now compute the difference: sin 45° − cos 45° = (1/√2) − (1/√2) = 0

Verification / Alternative check:
Check the given condition at θ = 45°: sin θ cos θ = (1/√2)*(1/√2) = 1/2, which matches exactly. Since both sin and cos are equal at 45°, their difference must be 0. This confirms the result without needing any further computation.

Why Other Options Are Wrong:
• √2, 1, 2: would require sin θ and cos θ to be very different, but the condition forces θ = 45°. • -1: impossible here because for acute θ, both sin θ and cos θ are positive and close in magnitude, and at 45° they are equal.

Common Pitfalls:
• Forgetting sin 2θ = 2 sin θ cos θ. • Solving sin 2θ = 1 but selecting an angle outside the allowed range. • Trying to compute sin and cos independently instead of identifying θ first.

Final Answer:
0