Trigonometry & Inverse Trigonometry
Trigonometric Inequalities
Grade 11
Question:
<p>In the following incomplete sentences, fill in the blanks so that the resulting sentences may become true.</p><p>(a) For all real \(x\), the value of \(\sin^2 x \cdot \cos^2 x \leq k\), \(k\) being the least possible. Then \(k =\) ______.</p>
Step-by-Step Solution
Key Concept: Use the AM-GM inequality or calculus to find the maximum value of sin²x·cos²x. The maximum occurs when sin²x = cos²x = 1/2, giving the product as 1/4.
<p><strong>Step 1:</strong> Let f(x) = sin²x · cos²x. We need to find the maximum value since k is the least upper bound.</p><p><strong>Step 2:</strong> Using the constraint sin²x + cos²x = 1, let sin²x = t where 0 ≤ t ≤ 1. Then cos²x = 1 - t.</p><p><strong>Step 3:</strong> So f = t(1-t) = t - t². To find maximum, differentiate: df/dt = 1 - 2t = 0, giving t = 1/2.</p><p><strong>Step 4:</strong> When t = 1/2, sin²x = 1/2 and cos²x = 1/2, so sin²x · cos²x = (1/2)(1/2) = 1/4.</p><p><strong>Step 5:</strong> Verify: At t = 1/2, d²f/dt² = -2 < 0, confirming this is a maximum. Thus sin²x · cos²x ≤ 1/4 for all real x.</p><p>∴ Answer: <strong>k = 1/4</strong></p>
Correct Answer: 1/4