Ta có: \(u = a - b;v = a + b\).
Suy ra \(u + v = 2a \to a = \frac{{u + v}}{2}\)
\(u - v = 2b \to b = \frac{{u - v}}{2}\)
Ta có: \(\cos u + \cos v = 2\cos \frac{{u + v}}{2}\cos \frac{{u - v}}{2}\)
\(\cos u - \cos v = - 2\sin \frac{{u + v}}{2}\sin \frac{{u - v}}{2}\)
\(\sin u + \sin v = 2\sin \frac{{u + v}}{2}\cos \frac{{u - v}}{2}\)
\(\sin u - \sin v = 2\cos \frac{{u + v}}{2}\sin \frac{{u - v}}{2}\)