In `DeltaABC`, prove that:
`a(cos^(2)C/2-cos^(2)B/2)=(b-c).cos^(2)A/2`

Answer

To prove that in triangle \( \Delta ABC \):

\[ a \left( \cos^2 \frac{C}{2} - \cos^2 \frac{B}{2} \right) = (b - c) \cos^2 \frac{A}{2} \]

we will start with the left-hand side (LHS) and manipulate it to show that it equals the right-hand side (RHS).

### Step 1: Write down the LHS
We start with the left-hand side:

\[ \text{LHS} = a \left( \cos^2 \frac{C}{2} - \cos^2 \frac{B}{2} \right) \]

### Step 2: Use the half-angle formula for cosine
Recall the half-angle formula for cosine:

\[ \cos \frac{A}{2} = \sqrt{\frac{s(s-a)}{bc}}, \quad \cos \frac{B}{2} = \sqrt{\frac{s(s-b)}{ac}}, \quad \cos \frac{C}{2} = \sqrt{\frac{s(s-c)}{ab}} \]

where \( s = \frac{a + b + c}{2} \) is the semi-perimeter of the triangle.

### Step 3: Substitute the half-angle formulas into the LHS
Substituting the half-angle formulas into the LHS:

\[ \text{LHS} = a \left( \frac{s(s-c)}{ab} - \frac{s(s-b)}{ac} \right) \]

### Step 4: Simplify the expression
Now, we can simplify the expression:

\[ \text{LHS} = a \left( \frac{s(s-c) \cdot c - s(s-b) \cdot b}{abc} \right) \]

This can be rewritten as:

\[ \text{LHS} = \frac{a s \left( c(s-c) - b(s-b) \right)}{abc} \]

### Step 5: Factor out common terms
Now, we can factor out \( s \):

\[ \text{LHS} = \frac{s \left( ac - ab + bc - b^2 \right)}{bc} \]

### Step 6: Use the semi-perimeter definition
Since \( s = \frac{a + b + c}{2} \), we can express the terms in the numerator in terms of \( s \):

\[ \text{LHS} = \frac{s(b - c)(s - a)}{bc} \]

### Step 7: Relate to the RHS
Now, we can relate this to the right-hand side:

\[ \text{RHS} = (b - c) \cos^2 \frac{A}{2} \]

Using the half-angle formula for \( \cos^2 \frac{A}{2} \):

\[ \cos^2 \frac{A}{2} = \frac{s(s-a)}{bc} \]

Thus, we can write:

\[ \text{RHS} = (b - c) \cdot \frac{s(s-a)}{bc} \]

### Step 8: Conclude the proof
Since both sides are equal:

\[ \text{LHS} = \text{RHS} \]

Thus, we have proved that:

\[ a \left( \cos^2 \frac{C}{2} - \cos^2 \frac{B}{2} \right) = (b - c) \cos^2 \frac{A}{2} \]