Đề bài

Chứng minh rằng trong tam giác \(ABC\), ta có: a) \(\sin B = \sin \left( {A + C} \right)\)                                   b) \(\cos C =  - \cos \left( {A + B + 2C} \right)\) c) \(\sin \frac{A}{2} = \cos \frac{{B + C}}{2}\)                                d) \(\tan \frac{{A + B - 2C}}{2} = \cot \frac{{3C}}{2}\)

Lời giải chi tiết

Trong tam giác \(ABC\), ta có \(A + B + C = \pi \).a) Do \(A + B + C = \pi  \Rightarrow A + C = \pi  - B \Rightarrow \sin \left( {A + C} \right) = \sin \left( {\pi  - B} \right) = \sin B\).b) Do \(A + B + C = \pi  \Rightarrow A + B + 2C = \pi  + C\)\( \Rightarrow \cos \left( {A + B + 2C} \right) = \cos \left( {\pi  + C} \right) =  - \cos C\)c) Do \(A + B + C = \pi  \Rightarrow \frac{{A + B + C}}{2} = \frac{\pi }{2} \Rightarrow \frac{{B + C}}{2} = \frac{\pi }{2} - \frac{A}{2}\)\( \Rightarrow \sin \frac{A}{2} = \cos \left( {\frac{\pi }{2} - \frac{A}{2}} \right) = \cos \frac{{B + C}}{2}\)d)Do \(A + B + C = \pi  \Rightarrow \frac{{A + B + C}}{2} = \frac{\pi }{2} \Rightarrow \frac{{A + B - 2C}}{2} = \frac{{A + B + C - 3C}}{2} = \frac{\pi }{2} - \frac{{3C}}{2}\)\( \Rightarrow \tan \frac{{A + B - 2C}}{2} = \tan \left( {\frac{\pi }{2} - \frac{{3C}}{2}} \right) = \cot \frac{{3C}}{2}\).