`(sin2A+sin2B+sin2C)/(sinA+sinB +sinC)` is equal to

Options

A. `8 sin((A)/(2))sin((B)/(2))sin((C)/(2))`
B. `8cos((A)/(2))cos((B)/(2))cos((C)/(2))`
C. `8tan((A)/(2))tan((B)/(2))tan((C)/(2))`
D. `8cot((A)/(2))cot((B)/(2))cot((C)/(2))`

Explanation

To solve the problem \((\sin 2A + \sin 2B + \sin 2C) / (\sin A + \sin B + \sin C)\), we will use trigonometric identities step by step.

### Step-by-Step Solution:

1. **Use the identity for sine of double angles**:
We know that \(\sin 2A = 2 \sin A \cos A\). Therefore, we can rewrite the numerator:
\[ \sin 2A + \sin 2B + \sin 2C = 2 \sin A \cos A + 2 \sin B \cos B + 2 \sin C \cos C \]

2. **Factor out the common factor of 2**:
\[ = 2(\sin A \cos A + \sin B \cos B + \sin C \cos C) \]

3. **Use the identity for the sum of sines**:
We also know that:
\[ \sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \]

4. **Substituting the identities into the expression**:
Now we can substitute these into our original expression:
\[ \frac{2(\sin A \cos A + \sin B \cos B + \sin C \cos C)}{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}} \]

5. **Use the identity for sine in terms of half angles**:
We can express \(\sin A\) as \(2 \sin \frac{A}{2} \cos \frac{A}{2}\):
\[ \sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}, \quad \sin B = 2 \sin \frac{B}{2} \cos \frac{B}{2}, \quad \sin C = 2 \sin \frac{C}{2} \cos \frac{C}{2} \]

6. **Substituting these into the numerator**:
\[ = 2(2 \sin \frac{A}{2} \cos \frac{A}{2} \cos A + 2 \sin \frac{B}{2} \cos \frac{B}{2} \cos B + 2 \sin \frac{C}{2} \cos \frac{C}{2} \cos C) \]

7. **Simplifying the expression**:
After substituting and simplifying, we will cancel out the common terms:
\[ = \frac{8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}} \]

8. **Final simplification**:
\[ = 2 \cdot \frac{2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}{\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}} = 8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \]

Thus, the final answer is:
\[ \frac{\sin 2A + \sin 2B + \sin 2C}{\sin A + \sin B + \sin C} = 8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \]