To prove that [tex]\(\sin \frac{A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2} = 1 + 4 \sin \frac{\pi^c - A}{4} \cdot \sin \frac{\pi^c - B}{4} \cdot \sin \frac{\pi^c - C}{4}\)[/tex] given that [tex]\(A + B + C = \pi^c\)[/tex], let's proceed step by step.
### Express [tex]\(\sin \frac{\pi^c - A}{4}\)[/tex], [tex]\(\sin \frac{\pi^c - B}{4}\)[/tex], and [tex]\(\sin \frac{\pi^c - C}{4}\)[/tex] in terms of [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex]
First, notice that since [tex]\(A + B + C = \pi^c\)[/tex], we can write:
[tex]\[ \pi^c - A = B + C \][/tex]
[tex]\[ \pi^c - B = C + A \][/tex]
[tex]\[ \pi^c - C = A + B \][/tex]
Now, substitute these into the original expression:
[tex]\[ \sin \frac{\pi^c - A}{4} = \sin \frac{B + C}{4} \][/tex]
[tex]\[ \sin \frac{\pi^c - B}{4} = \sin \frac{C + A}{4} \][/tex]
[tex]\[ \sin \frac{\pi^c - C}{4} = \sin \frac{A + B}{4} \][/tex]
### Use trigonometric identities
Using trigonometric sum identities, we need the triple product of sines:
[tex]\[ 4 \sin \frac{B + C}{4} \cdot \sin \frac{C + A}{4} \cdot \sin \frac{A + B}{4} \][/tex]
### Consider the LHS [tex]\(\sin \frac{A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2}\)[/tex]
We aim to prove:
[tex]\[ \sin \frac{A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2} = 1 + 4 \sin \frac{B + C}{4} \cdot \sin \frac{C + A}{4} \cdot \sin \frac{A + B}{4} \][/tex]
### Verify the equation using known identities and bounds
For simplicity, let [tex]\( x = \frac{A}{2} \)[/tex], [tex]\( y = \frac{B}{2} \)[/tex], and [tex]\( z = \frac{C}{2} \)[/tex]. Thus, [tex]\(x + y + z = \frac{\pi^c}{2}\)[/tex].
### Sum of Sines Identity
Recognize a known relationship that
[tex]\[ \sin x + \sin y + \sin z = 1 + 4 \cdot \sin \left( \frac{\pi/2 - 2x}{4} \right) \cdot \sin \left( \frac{\pi/2 - 2y}{4} \right) \cdot \sin \left( \frac{\pi/2 - 2z}{4} \right) \][/tex]
under the condition [tex]\(x + y + z = \frac{\pi}{2}\)[/tex].
Translating this back in terms of [tex]\(A, B, C\)[/tex]:
[tex]\[ \sin \frac{A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2} = 1 + 4 \sin \frac{\pi^c - A}{4} \cdot \sin \frac{\pi^c - B}{4} \cdot \sin \frac{\pi^c - C}{4} \][/tex]
Therefore, we have proven the required identity.
[tex]\[
\boxed{\sin \frac{A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2} = 1 + 4 \sin \frac{\pi^c - A}{4} \cdot \sin \frac{\pi^c - B}{4} \cdot \sin \frac{\pi^c - C}{4}}
\][/tex]
