In a `triangleABC, a^(2) sin 2C+c^(2) sin 2A=`

Options

A. `Delta`
B. `2Delta`
C. `3Delta`
D. `4Delta`

Answer

To solve the problem \( a^2 \sin 2C + c^2 \sin 2A \) in triangle \( ABC \), we will use trigonometric identities and properties of triangles.

### Step-by-Step Solution:

1. **Use the double angle identity for sine**:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
Therefore, we can rewrite \( \sin 2C \) and \( \sin 2A \):
\[ \sin 2C = 2 \sin C \cos C \quad \text{and} \quad \sin 2A = 2 \sin A \cos A \]
Substituting these into the original expression gives:
\[ a^2 \sin 2C + c^2 \sin 2A = a^2 (2 \sin C \cos C) + c^2 (2 \sin A \cos A) \]
Simplifying this, we have:
\[ = 2a^2 \sin C \cos C + 2c^2 \sin A \cos A \]

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

3. **Use the sine rule**:
According to the sine rule:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
where \( R \) is the circumradius of the triangle. Thus, we can express \( a \) and \( c \) in terms of \( R \):
\[ a = 2R \sin A \quad \text{and} \quad c = 2R \sin C \]

4. **Substituting back into the expression**:
Substitute \( a \) and \( c \) back into the expression:
\[ = 2((2R \sin A)^2 \sin C \cos C + (2R \sin C)^2 \sin A \cos A) \]
Simplifying this gives:
\[ = 2(4R^2 \sin^2 A \sin C \cos C + 4R^2 \sin^2 C \sin A \cos A) \]
\[ = 8R^2 (\sin^2 A \sin C \cos C + \sin^2 C \sin A \cos A) \]

5. **Using the area of triangle**:
The area \( \Delta \) of triangle \( ABC \) can be expressed as:
\[ \Delta = \frac{1}{2}bc \sin A \]
Thus, we can relate \( \sin A \) and \( \sin C \) to the area:
\[ \Delta = \frac{1}{2} \cdot 2R \sin A \cdot 2R \sin C \cdot \sin A \]
This gives us:
\[ = 2R^2 \sin A \sin C \]

6. **Final Expression**:
Therefore, we can conclude:
\[ a^2 \sin 2C + c^2 \sin 2A = 8R^2 \Delta \]

### Conclusion:
The value of \( a^2 \sin 2C + c^2 \sin 2A \) can be expressed in terms of the area of triangle \( ABC \) and the circumradius \( R \).