Cho tam giác ABC. Chứng minh rằng \(1 + \frac{r}{R} = \cos A + \cos B + \cos C\).

Ta có

VP = cosA + cosB + cosC

=\(2\cos \frac{{A + B}}{2}.\cos \frac{{A - B}}{2} + \cos C\)

\( = 2\cos \frac{{180^\circ - C}}{2}.\cos \frac{{A - B}}{2} + \cos C\)

\[ = 2\cos \left( {90^\circ - \frac{C}{2}} \right).\cos \frac{{A - B}}{2} + \cos \left( {2.\frac{C}{2}} \right)\]

\[ = 2\sin \frac{C}{2}.\cos \frac{{A - B}}{2} + 1 - 2{\sin ^2}\frac{C}{2}\]

\[ = 2\sin \frac{C}{2}\left( {cos\frac{{A - B}}{2} - \sin \frac{C}{2}} \right) + 1\]

\[ = 2\sin \frac{C}{2}\left[ {cos\frac{{A - B}}{2} - \cos \left( {90^\circ - \frac{C}{2}} \right)} \right] + 1\]

\[ = 2\sin \frac{C}{2}\left[ {cos\frac{{A - B}}{2} - \cos \frac{{180^\circ - C}}{2}} \right] + 1\]

\[ = 2\sin \frac{C}{2}\left[ {cos\frac{{A - B}}{2} - \cos \frac{{A + B}}{2}} \right] + 1\]

\[ = 2\sin \frac{C}{2}\left( {2\sin \frac{A}{2}.\sin \frac{B}{2}} \right) + 1\]

\[ = 4\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2} + 1\]               (1)

Ta có \(S = \frac{{abc}}{{4R}}\) và \(p = \frac{{a + b + c}}{2}\)

Suy ra \(r = \frac{S}{p} = \frac{{abc}}{{4{\rm{R}}{\rm{.}}\frac{{a + b + c}}{2}}}\)

\( = \frac{{\sin {\rm{A}}.2{\rm{R}}.\sin B.2{\rm{R}}.\sin C.2{\rm{R}}}}{{2{\rm{R}}\left( {\sin {\rm{A}}2{\rm{R}} + \sin B.2{\rm{R + }}\sin C.2{\rm{R}}} \right)}}\)

\( = \frac{{2{\rm{R}}\sin A.\sin B.\sin C}}{{\sin {\rm{A}} + \sin B + \sin C}}\)

Ta có: \(\sin A + \sin B + \sin C = 2\sin \frac{{A + B}}{2}.c{\rm{os}}\frac{{A - B}}{2} + \sin C\)

\( = 2\sin \frac{{180^\circ - C}}{2}.c{\rm{os}}\frac{{A - B}}{2} + 2\sin \frac{C}{2}.\cos \frac{C}{2}\)

\( = 2\sin \left( {90^\circ - \frac{C}{2}} \right).c{\rm{os}}\frac{{A - B}}{2} + 2\sin \frac{C}{2}.\cos \frac{C}{2}\)

\( = 2\cos \frac{C}{2}.c{\rm{os}}\frac{{A - B}}{2} + 2\sin \frac{C}{2}.\cos \frac{C}{2}\)

\( = 2\cos \frac{C}{2}.\left( {c{\rm{os}}\frac{{A - B}}{2} + \sin \frac{C}{2}} \right)\)

\( = 2\cos \frac{C}{2}.\left[ {c{\rm{os}}\frac{{A - B}}{2} + \cos \left( {90^\circ - \frac{C}{2}} \right)} \right]\)

\( = 2\cos \frac{C}{2}.\left[ {c{\rm{os}}\frac{{A - B}}{2} + \cos \frac{{180^\circ - C}}{2}} \right]\)

\( = 2\cos \frac{C}{2}.\left[ {c{\rm{os}}\frac{{A - B}}{2} + \cos \frac{{A + B}}{2}} \right]\)

\( = 2\cos \frac{C}{2}.\left( {2\cos \frac{A}{2}.\cos \frac{B}{2}} \right)\)

\( = 4\cos \frac{A}{2}.\cos \frac{B}{2}.\cos \frac{C}{2}\)

Suy ra

\(r = \frac{{2{\rm{R}}\sin A.\sin B.\sin C}}{{4\cos \frac{A}{2}.\cos \frac{B}{2}.\cos \frac{C}{2}}}\)

\(r = \frac{{2{\rm{R}}{\rm{.}}\left( {{\rm{2}}{\rm{.}}\sin \frac{A}{2}.\cos \frac{A}{2}} \right){\rm{.}}\left( {{\rm{2}}{\rm{.}}\sin \frac{B}{2}.\cos \frac{B}{2}} \right){\rm{.}}\left( {{\rm{2}}{\rm{.}}\sin \frac{C}{2}.\cos \frac{C}{2}} \right)}}{{4\cos \frac{A}{2}.\cos \frac{B}{2}.\cos \frac{C}{2}}}\)

\(r = 4R.\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}\)

Suy ra \(1 + \frac{r}{R} = 1 + \frac{{4R.\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}}}{R} = 1 + 4\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}\)       (2)

Từ (1) và (2) suy ra \(1 + \frac{r}{R} = \cos A + \cos B + \cos C\)

Vậy \(1 + \frac{r}{R} = \cos A + \cos B + \cos C\).