`2 abc.cos ""A/2 cos "" B/2 ""cos""C/2=`

Options

A. `aS`
B. `bS`
C. `cS`
D. `2sS`

Answer

To solve the equation \( 2abc \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \), we can use the properties of triangles and some trigonometric identities.

### Step-by-Step Solution:

1. **Understanding the Triangle Properties**:
- Let \( A, B, C \) be the angles of triangle \( ABC \).
- The semi-perimeter \( s \) of triangle \( ABC \) is given by \( s = \frac{a + b + c}{2} \), where \( a, b, c \) are the lengths of the sides opposite to angles \( A, B, C \) respectively.

**Hint**: Remember that the semi-perimeter is half the sum of the lengths of the sides of the triangle.

2. **Using the Cosine of Half Angles**:
- The cosine of half angles can be expressed in terms of the sides of the triangle:
\[ \cos \frac{A}{2} = \sqrt{\frac{s(s-a)}{bc}}, \quad \cos \frac{B}{2} = \sqrt{\frac{s(s-b)}{ac}}, \quad \cos \frac{C}{2} = \sqrt{\frac{s(s-c)}{ab}} \]

**Hint**: Recall the formulas for the cosine of half angles in terms of the sides of the triangle.

3. **Substituting the Half Angle Formulas**:
- Substitute the expressions for \( \cos \frac{A}{2}, \cos \frac{B}{2}, \cos \frac{C}{2} \) into the equation:
\[ 2abc \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} = 2abc \left(\sqrt{\frac{s(s-a)}{bc}} \cdot \sqrt{\frac{s(s-b)}{ac}} \cdot \sqrt{\frac{s(s-c)}{ab}}\right) \]

**Hint**: When substituting, ensure to multiply the square roots correctly.

4. **Simplifying the Expression**:
- The right-hand side simplifies as follows:
\[ = 2abc \cdot \frac{s(s-a)s(s-b)s(s-c)}{abc} = 2s^3 \cdot \sqrt{(s-a)(s-b)(s-c)} \]

**Hint**: Notice how the \( abc \) terms cancel out.

5. **Final Expression**:
- The final expression can be simplified to:
\[ 2abc \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} = \sqrt{s(s-a)(s-b)(s-c)} \]

**Hint**: The final result is a well-known identity in triangle geometry.

### Final Result:
Thus, we have shown that:
\[ 2abc \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} = \sqrt{s(s-a)(s-b)(s-c)} \]