Question

\(a_1,a_2,a_3\) are in arithmetic progression in which common difference is \(2\) and the first term is \(2\). If \[ \cos a_1+\cos a_2+\cos a_3 = \frac{\sin\alpha}{\sin\beta}\cos\gamma, \] then \(\alpha+\beta=\)

• A. \(2\gamma\)
• B. \(\gamma\)
• C. \(\dfrac{\gamma}{2}\)
• D. \(\gamma+2\)

Hint:

For three cosine terms in arithmetic progression, \[ \boxed{\cos(A-d)+\cos A+\cos(A+d) =\cos A\,(1+2\cos d).} \] Then use \[ \boxed{\sin3x=\sin x\,(1+2\cos2x)} \] to simplify further.

Answer 1

The Correct Option is B

Step 1: Write the A.P. terms. Since the first term is \(2\) and the common difference is \(2\), \[ a_1=2,\qquad a_2=4,\qquad a_3=6. \] Hence, \[ \cos a_1+\cos a_2+\cos a_3 = \cos2+\cos4+\cos6. \]

Step 2: Simplify the sum. Using \[ \cos A+\cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}, \] we get \[ \cos2+\cos6 = 2\cos4\cos2. \] Therefore, \[ \cos2+\cos4+\cos6 = \cos4(2\cos2+1). \] Again, \[ 2\cos2+1 = \frac{\sin3}{\sin1}. \] Hence, \[ \cos2+\cos4+\cos6 = \frac{\sin3}{\sin1}\cos4. \] Thus, \[ \alpha=3,\qquad \beta=1,\qquad \gamma=4. \]

Step 3: Find the required value. Therefore, \[ \alpha+\beta = 3+1 = 4 = \gamma. \] Hence, \[ \boxed{\gamma}. \] Thus, \[ \boxed{(B)} \] is the correct answer.