If `Delta` stands for the area of a triangle ABC , then `a^2 sin 2B+b^2 sin2A=`

Options

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

Answer

To solve the equation \( a^2 \sin 2B + b^2 \sin 2A \) where \( \Delta \) stands for the area of triangle \( ABC \), we can follow these steps:

### Step 1: Use the double angle identity for sine
Recall that:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
Applying this to \( \sin 2B \) and \( \sin 2A \):
\[ \sin 2B = 2 \sin B \cos B \quad \text{and} \quad \sin 2A = 2 \sin A \cos A \]
Thus, we can rewrite the expression:
\[ a^2 \sin 2B + b^2 \sin 2A = a^2 (2 \sin B \cos B) + b^2 (2 \sin A \cos A) \]
This simplifies to:
\[ 2a^2 \sin B \cos B + 2b^2 \sin A \cos A \]

### Step 2: Substitute the area of the triangle
The area \( \Delta \) of triangle \( ABC \) can be expressed using the formula:
\[ \Delta = \frac{1}{2}ab \sin C \]
Using the sine rule, we can express \( \sin A \) and \( \sin B \) in terms of the area:
\[ \sin A = \frac{2\Delta}{bc} \quad \text{and} \quad \sin B = \frac{2\Delta}{ac} \]
Substituting these into our expression gives:
\[ 2a^2 \left(\frac{2\Delta}{ac}\right) \cos B + 2b^2 \left(\frac{2\Delta}{bc}\right) \cos A \]
This simplifies to:
\[ \frac{4a\Delta}{c} \cos B + \frac{4b\Delta}{c} \cos A \]

### Step 3: Factor out common terms
Now, we can factor out \( \frac{4\Delta}{c} \):
\[ \frac{4\Delta}{c} \left(a \cos B + b \cos A\right) \]

### Step 4: Use the projection formula
From the projection formula in triangles, we know:
\[ a \cos B + b \cos A = c \]
Thus, we can substitute this back into our expression:
\[ \frac{4\Delta}{c} \cdot c = 4\Delta \]

### Final Result
Therefore, we have:
\[ a^2 \sin 2B + b^2 \sin 2A = 4\Delta \]