If angles A, B and C are in AP, then what is sin A + 2 sin B + sin C equal to ?

Options

A. `4 sin B cos^(2)((A-C)/(2))`
B. `4 sin B cos^(2)((A-C)/(4))`
C. `4 sin (2B) cos^(2)((A-C)/(2))`
D. `4 sin (2B) cos^(2)((A-C)/(4))`

Explanation

To solve the problem, we need to find the value of \( \sin A + 2 \sin B + \sin C \) given that angles \( A, B, \) and \( C \) are in arithmetic progression (AP).

### Step-by-step Solution:

1. **Understanding the AP Condition**:
Since \( A, B, C \) are in AP, we can express this relationship mathematically:
\[ 2B = A + C \]

2. **Using the Triangle Sum Property**:
We know that in any triangle, the sum of the angles is \( 180^\circ \):
\[ A + B + C = 180^\circ \]

3. **Substituting for C**:
From the equation \( A + B + C = 180^\circ \), we can express \( C \) in terms of \( A \) and \( B \):
\[ C = 180^\circ - A - B \]

4. **Substituting C in the AP Condition**:
Now, substituting \( C \) in the AP condition:
\[ 2B = A + (180^\circ - A - B) \]
Simplifying this gives:
\[ 2B = 180^\circ - B \implies 3B = 180^\circ \implies B = 60^\circ \]

5. **Finding A and C**:
Since \( B = 60^\circ \), we can find \( A \) and \( C \):
\[ A + C = 2B = 120^\circ \]
Also, since \( A + B + C = 180^\circ \):
\[ A + 60^\circ + C = 180^\circ \implies A + C = 120^\circ \]
This confirms our earlier calculation.

6. **Using Sine Addition**:
Now we can express \( \sin A + 2 \sin B + \sin C \):
\[ \sin A + 2 \sin B + \sin C = \sin A + 2 \sin(60^\circ) + \sin C \]
We know \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \), so:
\[ = \sin A + 2 \cdot \frac{\sqrt{3}}{2} + \sin C = \sin A + \sqrt{3} + \sin C \]

7. **Using the Sine Addition Formula**:
Since \( A + C = 120^\circ \), we can use the sine addition formula:
\[ \sin(A + C) = \sin(120^\circ) = \frac{\sqrt{3}}{2} \]
Therefore, we can express \( \sin A + \sin C \) using the sine addition formula:
\[ \sin A + \sin C = 2 \sin\left(\frac{A + C}{2}\right) \cos\left(\frac{A - C}{2}\right) \]
Here, \( \frac{A + C}{2} = 60^\circ \) and \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \).

8. **Final Calculation**:
Thus, we have:
\[ \sin A + \sin C = 2 \cdot \frac{\sqrt{3}}{2} \cdot \cos\left(\frac{A - C}{2}\right) = \sqrt{3} \cdot \cos\left(\frac{A - C}{2}\right) \]
Therefore:
\[ \sin A + 2 \sin B + \sin C = \sqrt{3} \cdot \cos\left(\frac{A - C}{2}\right) + \sqrt{3} \]

9. **Conclusion**:
Since \( \cos\left(\frac{A - C}{2}\right) \) can vary depending on the specific values of \( A \) and \( C \), we can conclude that:
\[ \sin A + 2 \sin B + \sin C = 2\sqrt{3} \]
when \( A = C \).