Question:
In a triangle \(\triangle ABC\), if \(a, b,\) and \(c\) are in arithmetic progression, then \(\cos A + 2 \cos B + \cos C =\)
In a triangle \(\triangle ABC\), if \(a, b,\) and \(c\) are in arithmetic progression, then \(\cos A + 2 \cos B + \cos C =\)
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In any triangle where sides are in A.P., specific trigonometric relations like \(\tan(A/2) \tan(C/2) = 1/3\) hold true.
Updated On: May 13, 2026
- \(1\)
- \(2\)
- \(3/2\)
- \(\sqrt{3} + 1\)
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The Correct Option is B
Solution and Explanation
Concept:
For sides in A.P., \(2b = a + c\). We use the Sine Rule (\(a = 2R \sin A\), etc.) and Projection Formulae to relate the angles.
Step 1: Using the A.P. condition.
Given \(a, b, c\) in A.P.: \[ a + c = 2b \]
Step 2: Applying the Sine Rule.
\[ 2R \sin A + 2R \sin C = 2(2R \sin B) \] \[ \sin A + \sin C = 2 \sin B \] Using sum-to-product: \[ 2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right) = 4 \sin \left(\frac{B}{2}\right) \cos \left(\frac{B}{2}\right) \] Since \(\frac{A+C}{2} = 90^\circ - \frac{B}{2}\), then \(\sin \left(\frac{A+C}{2}\right) = \cos \left(\frac{B}{2}\right)\): \[ \cos \left(\frac{A-C}{2}\right) = 2 \sin \left(\frac{B}{2}\right) \]
Step 3: Evaluating the expression.
\(\cos A + \cos C + 2 \cos B\) \[ = 2 \cos \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right) + 2 \cos B \] \[ = 2 \sin \left(\frac{B}{2}\right) [2 \sin \left(\frac{B}{2}\right)] + 2(1 - 2 \sin^2 \left(\frac{B}{2}\right)) \] \[ = 4 \sin^2 \left(\frac{B}{2}\right) + 2 - 4 \sin^2 \left(\frac{B}{2}\right) = 2 \]
Step 1: Using the A.P. condition.
Given \(a, b, c\) in A.P.: \[ a + c = 2b \]
Step 2: Applying the Sine Rule.
\[ 2R \sin A + 2R \sin C = 2(2R \sin B) \] \[ \sin A + \sin C = 2 \sin B \] Using sum-to-product: \[ 2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right) = 4 \sin \left(\frac{B}{2}\right) \cos \left(\frac{B}{2}\right) \] Since \(\frac{A+C}{2} = 90^\circ - \frac{B}{2}\), then \(\sin \left(\frac{A+C}{2}\right) = \cos \left(\frac{B}{2}\right)\): \[ \cos \left(\frac{A-C}{2}\right) = 2 \sin \left(\frac{B}{2}\right) \]
Step 3: Evaluating the expression.
\(\cos A + \cos C + 2 \cos B\) \[ = 2 \cos \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right) + 2 \cos B \] \[ = 2 \sin \left(\frac{B}{2}\right) [2 \sin \left(\frac{B}{2}\right)] + 2(1 - 2 \sin^2 \left(\frac{B}{2}\right)) \] \[ = 4 \sin^2 \left(\frac{B}{2}\right) + 2 - 4 \sin^2 \left(\frac{B}{2}\right) = 2 \]
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