Bài 106575

Question

Chứng minh rằng trong mọi tam giác

$ABC$ ta luôn có:

$\frac{1}{sin A} + \frac{1}{sin B} + \frac{1}{sin C} = \frac{1}{2}\left( {\tan \frac{A}{2} + \tan \frac{B}{2} + \tan \frac{C}{2} + \cot \frac{A}{2}.\cot \frac{B}{2}.\cot \frac{C}{2}} \right)$

score 0 · Answer 1

Dễ chứng minh:

$\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \cot \frac{A}{2}.\cot \frac{B}{2}.\cot \frac{C}{2}(1)$ Do đó:

$\frac{1}{2}\left( {\tan \frac{A}{2} + \tan \frac{B}{2} + \tan \frac{C}{2} + \cot \frac{A}{2}.\cot \frac{B}{2}.\cot \frac{C}{2}} \right)$

$= \frac{1}{2}\left( {\tan \frac{A}{2} + \tan \frac{B}{2} + \tan \frac{C}{2} + \cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2}} \right)\\ = \frac{1}{2}\left( {\tan \frac{A}{2} + \cot \frac{A}{2} + \tan \frac{B}{2} + \cot \frac{B}{2} + \tan \frac{C}{2} + \cot \frac{C}{2}} \right)$

$\begin{array}{l} = \frac{1}{2}\left( {\frac{1}{{c{\rm{os}}\frac{A}{2}\sin \frac{A}{2}}} + \frac{1}{{c{\rm{os}}\frac{B}{2}\sin \frac{B}{2}}} + \frac{1}{{c{\rm{os}}\frac{C}{2}\sin \frac{C}{2}}}} \right)\\ = \frac{1}{2}\left( {\frac{2}{{\sin A}} + \frac{2}{{\sin B}} + \frac{2}{{\sin C}}} \right) = \frac{1}{{\sin A}} + \frac{1}{{\sin B}} + \frac{1}{{\sin C}} \end{array}$

Chứng minh (1):

$\cot(\frac{A}{2}+\frac{B}{2})=\frac{\cot\frac{A}{2}\cot\frac{B}{2}-1}{\cot\frac{A}{2}+\cot\frac{B}{2}}$

$\Leftrightarrow \frac{1}{\cot\frac{C}{2}}=\frac{\cot\frac{A}{2}\cot\frac{B}{2}-1}{\cot\frac{A}{2}+\cot\frac{B}{2}}$

$\Rightarrow$ (1)

Bài 106575

1 Đáp án

Lý thuyết liên quan

Bài 106575

1 Đáp án

Liên quan

Lý thuyết liên quan