In \(\triangle ABC\), prove that \(a^2 = b^2 + c^2 - 2bc\cos A\)

Answer: Proof of the cosine rule using coordinates.

Take \(A\) as the origin and the line \(AB\) as the positive X-axis, with \(C\) above the X-axis. Then \(A \equiv (0, 0)\), \(B \equiv (c, 0)\), and since \(AC = b\) makes angle \(A\) with the X-axis, \(C \equiv (b\cos A, b\sin A)\). (This holds whether \(A\) is acute, right or obtuse.)

By the distance formula,

\(a^2 = BC^2 = (b\cos A - c)^2 + (b\sin A - 0)^2\)

\(= b^2\cos^2 A - 2bc\cos A + c^2 + b^2\sin^2 A\)

\(= b^2(\cos^2 A + \sin^2 A) + c^2 - 2bc\cos A\)

\(= b^2 + c^2 - 2bc\cos A\).

Hence \(a^2 = b^2 + c^2 - 2bc\cos A\).