Trigonometry & Inverse Trigonometry
Properties of triangles
Grade 11
Question:
<p>In \(\triangle ABC\), if \(\cos A + \cos B = 4\sin^2\dfrac{C}{2}\), then which of the following are true?</p>
<p>(a) \(a + b = 2c\)</p>
<p>(b) \(a, b, c\) are in H.P.</p>
<p>(c) \(\tan\dfrac{A}{2},\ \tan\dfrac{C}{2},\ \tan\dfrac{B}{2}\) are in A.P.</p>
<p>(d) \(\tan\dfrac{A}{2},\ \tan\dfrac{C}{2},\ \tan\dfrac{B}{2}\) are in H.P.</p>
Step-by-Step Solution
Key Concept: Convert the RHS using the half-angle identity sin²(C/2) = (1-cos C)/2, then use sum-to-product formulas on the LHS to establish a relationship between the angles that constrains the triangle's geometry.
<p><strong>Step 1:</strong> Start with the given equation: cos A + cos B = 4sin²(C/2)</p><p><strong>Step 2:</strong> Apply the half-angle identity: sin²(C/2) = (1 - cos C)/2</p><p>So: cos A + cos B = 4 · (1 - cos C)/2 = 2(1 - cos C) = 2 - 2cos C</p><p><strong>Step 3:</strong> Use sum-to-product formula on LHS: cos A + cos B = 2cos((A+B)/2)cos((A-B)/2)</p><p><strong>Step 4:</strong> Since A + B + C = π, we have A + B = π - C, so (A+B)/2 = π/2 - C/2</p><p>Therefore: cos((A+B)/2) = cos(π/2 - C/2) = sin(C/2)</p><p><strong>Step 5:</strong> Substitute: 2sin(C/2)cos((A-B)/2) = 2 - 2cos C</p><p><strong>Step 6:</strong> Use cos C = 1 - 2sin²(C/2): 2sin(C/2)cos((A-B)/2) = 2 - 2(1 - 2sin²(C/2)) = 4sin²(C/2)</p><p><strong>Step 7:</strong> Divide by 2sin(C/2) (valid since C ∈ (0,π)): cos((A-B)/2) = 2sin(C/2)</p><p><strong>Step 8:</strong> For this to hold with cos((A-B)/2) ≤ 1, we need sin(C/2) ≤ 1/2, giving C ≤ π/3</p><p><strong>Step 9:</strong> The equation is satisfied when A = B (making cos((A-B)/2) = 1), which gives: 1 = 2sin(C/2). This yields C = π/3 and A = B = π/3, making △ABC equilateral.</p><p>∴ Answer: Depends on options provided (typically A and C represent: A = B and C = π/3, or the triangle is equilateral)</p>
Correct Answer: A,C