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: Understanding the Concept:
Arithmetic Progression (A.P.) implies that the middle term is the average of the surrounding terms.
For sides \( a, b, c \), this means \( 2b = a + c \).
We need to evaluate a trigonometric sum of cosines for a triangle.
Properties of triangles, such as \( A + B + C = \pi \), and the Sine Rule (\( a = 2R \sin A \)) are key here.
Key Formula or Approach:
1. A.P. Condition: \( a + c = 2b \).
2. Sine Rule: \( a = 2R \sin A \), \( b = 2R \sin B \), \( c = 2R \sin C \).
3. Sum-to-Product Identity: \( \sin A + \sin C = 2 \sin\left(\frac{A+C}{2}\right) \cos\left(\frac{A-C}{2}\right) \).
4. Triple-Cosine Relation: \( \cos A + \cos C = 2 \cos\left(\frac{A+C}{2}\right) \cos\left(\frac{A-C}{2}\right) \).
Step 2: Detailed Explanation:
Given \( a, b, c \) are in A.P., we have:
\[ a + c = 2b \] Using the Sine Rule, we substitute the sides with their sine equivalents:
\[ 2R \sin A + 2R \sin C = 2(2R \sin B) \] Dividing by \( 2R \):
\[ \sin A + \sin C = 2 \sin B \] Applying the sum-to-product identity:
\[ 2 \sin\left(\frac{A+C}{2}\right) \cos\left(\frac{A-C}{2}\right) = 2 \sin B \] Since \( A + B + C = \pi \), we have \( A + C = \pi - B \), thus \( \frac{A+C}{2} = \frac{\pi}{2} - \frac{B}{2} \).
This gives \( \sin\left(\frac{A+C}{2}\right) = \sin\left(\frac{\pi}{2} - \frac{B}{2}\right) = \cos\left(\frac{B}{2}\right) \).
Also, we use the double angle formula \( \sin B = 2 \sin\frac{B}{2} \cos\frac{B}{2} \).
Substituting these back:
\[ 2 \cos\frac{B}{2} \cos\frac{A-C}{2} = 2 (2 \sin\frac{B}{2} \cos\frac{B}{2}) \] Assuming \( \cos\frac{B}{2} \neq 0 \) (valid as \( 0<B<\pi \)):
\[ \cos\frac{A-C}{2} = 2 \sin\frac{B}{2} \] Now we look at the required expression \( E = \cos A + 2 \cos B + \cos C \):
\[ E = (\cos A + \cos C) + 2 \cos B \] \[ E = 2 \cos\frac{A+C}{2} \cos\frac{A-C}{2} + 2 \cos B \] Substitute \( \cos\frac{A+C}{2} = \sin\frac{B}{2} \) and our derived relation \( \cos\frac{A-C}{2} = 2 \sin\frac{B}{2} \):
\[ E = 2 (\sin\frac{B}{2}) (2 \sin\frac{B}{2}) + 2 \cos B \] \[ E = 4 \sin^2\frac{B}{2} + 2 \cos B \] Using the identity \( \cos B = 1 - 2 \sin^2\frac{B}{2} \):
\[ E = 4 \sin^2\frac{B}{2} + 2 (1 - 2 \sin^2\frac{B}{2}) \] \[ E = 4 \sin^2\frac{B}{2} + 2 - 4 \sin^2\frac{B}{2} = 2 \] Step 3: Final Answer:
The sum \( \cos A + 2 \cos B + \cos C \) is equal to 2.
This corresponds to Option (B).