Step 1: Understanding the Concept:
In triangle geometry, many identities connect the sides \( (a, b, c) \) and the angles \( (A, B, C) \).
When sides are in Arithmetic Progression (A.P.), it means that the difference between consecutive sides is constant, which satisfies the condition \( 2b = a + c \).
Using the Sine Rule, we can translate this side relationship into an angular relationship.
Additionally, trigonometric sum-to-product identities are used to simplify the expression involving cosines.
The final result will be a constant value independent of specific side lengths.
Step 2: Key Formula or Approach:
Condition for A.P.: \( a + c = 2b \).
Sine Rule: \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \).
Sum-to-product identity: \( \cos A + \cos C = 2 \cos(\frac{A+C}{2}) \cos(\frac{A-C}{2}) \).
Angle sum property: \( A + B + C = 180^\circ \).
Step 3: Detailed Explanation:
Given \( a, b, c \) are in A.P., we have:
\[ a + c = 2b \]
Substitute \( a = 2R \sin A, b = 2R \sin B, c = 2R \sin C \):
\[ 2R \sin A + 2R \sin C = 2(2R \sin B) \]
\[ \sin A + \sin C = 2 \sin B \]
Apply sum-to-product on the left:
\[ 2 \sin(\frac{A+C}{2}) \cos(\frac{A-C}{2}) = 2(2 \sin\frac{B}{2} \cos\frac{B}{2}) \]
Since \( \frac{A+C}{2} = 90^\circ - \frac{B}{2} \), then \( \sin(\frac{A+C}{2}) = \cos\frac{B}{2} \).
\[ 2 \cos\frac{B}{2} \cos(\frac{A-C}{2}) = 4 \sin\frac{B}{2} \cos\frac{B}{2} \]
Divide by \( 2\cos\frac{B}{2} \):
\[ \cos(\frac{A-C}{2}) = 2 \sin\frac{B}{2} \]
Now, simplify the target expression \( X = \cos A + \cos C + 2 \cos B \):
\[ X = 2 \cos(\frac{A+C}{2}) \cos(\frac{A-C}{2}) + 2 \cos B \]
Replace \( \cos(\frac{A+C}{2}) \) with \( \sin\frac{B}{2} \):
\[ X = 2 \sin\frac{B}{2} [2 \sin\frac{B}{2}] + 2(1 - 2 \sin^2\frac{B}{2}) \]
\[ X = 4 \sin^2\frac{B}{2} + 2 - 4 \sin^2\frac{B}{2} \]
The terms involving \( \sin^2\frac{B}{2} \) cancel each other perfectly.
\[ X = 2 \]
Step 4: Final Answer:
The value of the expression \( \cos A + 2 \cos B + \cos C \) is 2.
This is Option (B).