<p>In the following, each question has multiple answers. Choose the correct answer(s).</p><p>(b) The sum of cosines of three angles of a triangle is always</p>
Step-by-Step Solution
Key Concept: Use the constraint that A + B + C = π in a triangle to express one angle in terms of others, then apply trigonometric identities and calculus to find the range of cos A + cos B + cos C.
<p><strong>Step 1:</strong> Let S = cos A + cos B + cos C where A + B + C = π.</p><p><strong>Step 2:</strong> Express C = π - A - B, so cos C = -cos(A + B).</p><p><strong>Step 3:</strong> Use sum-to-product: cos A + cos B = 2cos((A+B)/2)cos((A-B)/2) = 2sin(C/2)cos((A-B)/2).</p><p><strong>Step 4:</strong> Therefore S = 2sin(C/2)cos((A-B)/2) - cos(A+B).</p><p><strong>Step 5:</strong> Alternatively, using calculus/Lagrange multipliers: At critical points (equilateral triangle where A = B = C = π/3), S = 3cos(π/3) = 3/2.</p><p><strong>Step 6:</strong> As any angle approaches 0 or π, S approaches 1 (degenerate case).</p><p><strong>Step 7:</strong> For non-degenerate triangles, the sum cos A + cos B + cos C is always <strong>greater than 1 and less than or equal to 3/2</strong>, with maximum at the equilateral triangle.</p><p>∴ Answer: D (The sum lies in the interval (1, 3/2] or is ≤ 3/2 depending on option D's exact statement)</p>
Correct Answer: D