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: Use the identity cos A + cos B = 2cos((A+B)/2)cos((A-B)/2) and the constraint A+B+C=π to express everything in terms of C, then apply sin²(C/2) = (1-cosC)/2 to find the relationship.
<p><strong>Step 1:</strong> Apply sum-to-product formula: cos A + cos B = 2cos((A+B)/2)cos((A-B)/2)</p><p><strong>Step 2:</strong> Since A+B+C=π, we have A+B=π-C, so (A+B)/2 = π/2 - C/2</p><p><strong>Step 3:</strong> Therefore cos((A+B)/2) = cos(π/2 - C/2) = sin(C/2)</p><p><strong>Step 4:</strong> The equation becomes: 2sin(C/2)cos((A-B)/2) = 4sin²(C/2)</p><p><strong>Step 5:</strong> Since sin(C/2) > 0, divide by 2sin(C/2): cos((A-B)/2) = 2sin(C/2)</p><p><strong>Step 6:</strong> Using sin²(C/2) = (1-cosC)/2 and cos(A-B)/2 ≤ 1, we need 2sin(C/2) ≤ 1, so sin(C/2) ≤ 1/2, giving C ≤ π/3</p><p><strong>Step 7:</strong> For equality in cos((A-B)/2) = 2sin(C/2), when A=B: cos(0) = 1 = 2sin(C/2), so sin(C/2) = 1/2, thus C = π/3</p><p><strong>Step 8:</strong> This gives A = B = π/3, making it an equilateral triangle. The triangle is isosceles (A=B) and equilateral (A=B=C=π/3)</p><p>∴ Answer: A,C</p>
Correct Answer: A,C