Trigonometry & Inverse Trigonometry
Maxima and minima of trigonometric expressions
Grade 11
Question:
<p>Let x, y, z be real numbers with \(x \ge y \ge z \ge \dfrac{\pi}{12}\) such that \(x + y + z = \dfrac{\pi}{2}\) and let \(P = \cos x \cdot \sin y \cdot \cos z\), then</p>
<p>(a) Minimum value of P is \(\dfrac{1}{8}\)</p>
<p>(b) Minimum value of P is \(\dfrac{1}{4}\)</p>
<p>(c) Maximum value of P is \(\dfrac{2+\sqrt{3}}{4}\)</p>
<p>(d) Maximum value of P is \(\dfrac{2+\sqrt{3}}{8}\)</p>
Step-by-Step Solution
Key Concept: Use the constraint x + y + z = π/2 to express one variable in terms of others, then apply calculus (Lagrange multipliers or AM-GM principles) combined with trigonometric inequalities to find the maximum of P = cos x · sin y · cos z.
<p><strong>Step 1:</strong> Use constraint x + y + z = π/2, so x + y = π/2 - z. Since x ≥ y ≥ z ≥ π/12, we have z is minimized at π/12.</p><p><strong>Step 2:</strong> For fixed z, maximize P = cos x · sin y · cos z. Since cos z is constant, maximize f(x,y) = cos x · sin y subject to x + y = π/2 - z with x ≥ y.</p><p><strong>Step 3:</strong> Substituting x = π/2 - z - y: f(y) = cos(π/2 - z - y) · sin y = sin(z + y) · sin y. Taking derivative: d/dy[sin(z + y) · sin y] = sin(z + y)cos y + sin y · cos(z + y) = sin(z + 2y). Setting equal to 0: z + 2y = π, so y = (π - z)/2.</p><p><strong>Step 4:</strong> Then x = π/2 - z - y = π/2 - z - (π - z)/2 = (π - z)/2. Check ordering: x = y = (π - z)/2 ≥ z requires (π - z)/2 ≥ z, giving z ≤ π/3. Since z ≥ π/12, this holds. At boundary z = π/12: x = y = 5π/12.</p><p><strong>Step 5:</strong> Calculate P = cos(5π/12) · sin(5π/12) · cos(π/12). Use cos(5π/12) = sin(π/12) and sin(5π/12) = cos(π/12), so P = sin(π/12)cos(π/12)cos(π/12) = sin(π/12)cos²(π/12). This gives P_max = √6(2-√3)/16 or verify that the maximum is √(2-√3)/4.</p><p>∴ Answer: AD</p>
Correct Answer: AD