Question

In a triangle $\triangle ABC$, if $a$, $b$, and $c$ are in arithmetic progression, then $\cos A + 2\cos B + \cos C =$

• A. 1
• B. 2
• C. $\dfrac{3}{2}$
• D. $\sqrt{3}+1$

Hint:

For A.P. side problems, quickly verify with an equilateral triangle ($A=B=C=60^\circ$): $\cos 60^\circ + 2\cos 60^\circ + \cos 60^\circ = 0.5 + 1 + 0.5 = 2$. \checkmark

Answer 1

The Correct Option is B

Step 1: Concept
If $a, b, c$ are in A.P., then $2b = a + c$. By the Sine Rule, $a = 2R\sin A$, so equivalently $2\sin B = \sin A + \sin C$.

Step 2: Meaning
Using sum-to-product identities: $4\sin\frac{B}{2}\cos\frac{B}{2} = 2\sin\frac{A+C}{2}\cos\frac{A-C}{2}$. Since $\frac{A+C}{2} = 90^\circ - \frac{B}{2}$, this gives $2\sin\frac{B}{2} = \cos\frac{A-C}{2}$.

Step 3: Analysis
Expand the target expression: \[\cos A + \cos C + 2\cos B = 2\cos\!\tfrac{A+C}{2}\cos\!\tfrac{A-C}{2} + 2\!\left(1-2\sin^2\!\tfrac{B}{2}\right).\] \[= 2\sin\!\tfrac{B}{2}\cdot 2\sin\!\tfrac{B}{2} + 2 - 4\sin^2\!\tfrac{B}{2} = 4\sin^2\!\tfrac{B}{2} + 2 - 4\sin^2\!\tfrac{B}{2} = 2.\]

Step 4: Conclusion
Therefore $\cos A + 2\cos B + \cos C = 2$. Final Answer: (B)