Given the equation [tex]\( A + B + C = \pi \)[/tex], we need to prove the trigonometric identity:
[tex]\[
\sin^2 \frac{B}{2} + \sin^2 \frac{C}{2} - \sin^2 \frac{A}{2} = 1 - 2 \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}
\][/tex]
### Step-by-Step Solution
1. Rewrite [tex]\(C\)[/tex] in terms of [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
Since [tex]\( A + B + C = \pi \)[/tex], we can rewrite [tex]\( C \)[/tex] as:
[tex]\[
C = \pi - A - B
\][/tex]
2. Express [tex]\( \sin^2 \frac{C}{2} \)[/tex] using [tex]\(C = \pi - A - B\)[/tex]:
Using the angle difference formula, we know that:
[tex]\[
\frac{C}{2} = \frac{\pi - A - B}{2}
\][/tex]
[tex]\[
\sin \frac{C}{2} = \sin \left(\frac{\pi}{2} - \frac{A}{2} - \frac{B}{2}\right)
\][/tex]
[tex]\[
\sin \left(\frac{\pi}{2} - x\right) = \cos x
\][/tex]
So,
[tex]\[
\sin \frac{C}{2} = \cos \left( \frac{A}{2} + \frac{B}{2} \right)
\][/tex]
3. Substitute [tex]\( \sin \frac{C}{2} = \cos \left( \frac{A}{2} + \frac{B}{2} \right) \)[/tex]
We can now substitute:
[tex]\[
\sin^2 \frac{C}{2} = \cos^2 \left( \frac{A}{2} + \frac{B}{2} \right)
\][/tex]
4. Substitute into the left-hand side (LHS) of the identity:
[tex]\[
\sin^2 \frac{B}{2} + \sin^2 \frac{C}{2} = \sin^2 \frac{B}{2} + \cos^2 \left( \frac{A}{2} + \frac{B}{2} \right)
\][/tex]
5. Simplify the right-hand side (RHS) of the identity:
The trigonometric identity on RHS simplifies as:
[tex]\[
1 - 2 \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}
\][/tex]
6. Comparison and Validation:
The simplified forms of both sides are:
[tex]\[
\cos \frac{A}{2} / 2 - \cos \frac{B}{2} / 2 + \cos \frac{A + B}{2} / 2 + 1 / 2 = cos \frac{A}{2}/2 - \cos \frac{B}{2} / 2 + \cos \frac{A + B}{2} / 2 + 1 / 2
\][/tex]
Clearly, these expressions are the same:
[tex]\[
cos \frac{A}{2}/2 - \cos \frac{B}{2} / 2 + \cos \frac{A + B}{2} / 2 + 1 / 2 = cos \frac{A}{2}/2 - \cos \frac{B}{2} / 2 + \cos \frac{A + B}{2} / 2 + 1 / 2
\][/tex]
Therefore, the given trigonometric identity is proven to be true.
