Question

In any triangle ABC prove that: (a-b)^2 cos^2(c/2)+(a+b)^2 sin^2(c/2) = c^2?

Answer 1

Simplify the terms of the LHS as follows:
\[(a-b)^2 \cos^2{\frac{C}{2}}=(a^2-2ab+b^2)\cos^2{\frac{C}{2}}\]
\[(a+b)^2 \sin^2{\frac{C}{2}}=(a^2+2ab+b^2) \sin^2{\frac{C}{2}}\]
After you add these two, group the terms like this:
\[a^2\left(\cos^2{\frac{C}{2}}+\sin^2{\frac{C}{2}}\right)-2ab\left(\cos^2{\frac{C}{2}}-\sin^2{\frac{C}{2}}\right)+b^2\left(\cos^2{\frac{C}{2}}+\sin^2{\frac{C}{2}}\right)\]
Two Pythagorean identities, a double-angle identity, and the Law of Cosines later, it comes out as \(c^2\).