Cho tam giác ABC. Khi đó: a) a = b cos C + c cos B b) sin A = sin B cos C + sin C cos B

a) Đúng

b) Đúng

c) Đúng

d) Đúng

a) \(\cos B = \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}};\cos C = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\). \(VP = b\cos C + c\cos B = b \cdot \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} + c \cdot \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\) \( = \frac{{{a^2} + {b^2} - {c^2}}}{{2a}} + \frac{{{a^2} + {c^2} - {b^2}}}{{2a}} = \frac{{2{a^2}}}{{2a}} = a = VP(dpcm)\).

b) \(\sin B = \frac{b}{{2R}};\sin C = \frac{c}{{2R}}\) \(VP = \sin B\cos C + \sin C\cos B = \frac{b}{{2R}} \cdot \cos C + \frac{c}{{2R}} \cdot \cos B = \frac{1}{{2R}}(b\cos C + c\cos B)\)

(mà \(a = b\cos C + c\cos B\), chứng minh câu \(a). = \frac{1}{{2R}} \cdot a = VP(dpcm)\).

c) \(S = \frac{1}{2}a \cdot {h_a} = \frac{1}{2}bc\sin A\) \( \Rightarrow {h_a} = \frac{{bc\sin A}}{a} = \frac{{2R\sin B \cdot 2R\sin C \cdot \sin A}}{{2R\sin A}} = 2R\sin B\sin C = VP(dp\;cm)\).

d) \(a(b\cos C - c\cos B) = ab \cdot \cos C - ac \cdot \cos B = ab \cdot \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} - ac \cdot \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\) \( = \frac{{{a^2} + {b^2} - {c^2}}}{2} - \frac{{{a^2} + {c^2} - {b^2}}}{2} = \frac{{{a^2} + {b^2} - {c^2} - {a^2} - {c^2} + {b^2}}}{2} = {b^2} - {c^2}\).