If A + B + C = 180^ , then prove that: ^2 A2 + ^2 B2 + ^2 C2 = 2 + 2 A2 B2 C2

We are given that $A+B+C=180^{\circ }$. We need to prove: $${\cos }^2\frac{A}{2}+{\cos }^2\frac{B}{2}+{\cos }^2\frac{C}{2}=2+2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}.$$

Step 1: Use the identity for ${\cos }^2\frac{x}{2}$

Recall the half-angle identity: $${\cos }^2\frac{x}{2}=\frac{1+\cos x}{2}.$$ Applying this to each term: $${\cos }^2\frac{A}{2}+{\cos }^2\frac{B}{2}+{\cos }^2\frac{C}{2}=\frac{1+\cos A}{2}+\frac{1+\cos B}{2}+\frac{1+\cos C}{2}.$$

Step 2: Simplify the expression

Combine the terms: $$=\frac{3+(\cos A+\cos B+\cos C)}{2}.$$

Step 3: Use the trigonometric identity for angles of a triangle

For $A+B+C=180^{\circ }$, we know: $$\cos A+\cos B+\cos C=1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}.$$ Substitute this into the expression: $$\frac{3+(1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2})}{2}=\frac{4+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}{2}.$$

Step 4: Simplify further

$$=2+2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}.$$

Thus, we have proven: $${\cos }^2\frac{A}{2}+{\cos }^2\frac{B}{2}+{\cos }^2\frac{C}{2}=2+2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}.$$