Chứng minh rằng:    $\frac{ a \sin \frac{ B-C}{ 2} }{ \sin \frac{ A}{ 2} } +\frac{ b \sin \frac{ C-A}{ 2} }{ \sin \frac{ B}{ 2} } +\frac{c \sin \frac{A-B }{ 2}  }{\sin \frac{C }{2 }    } =0    (1)$
Chứng minh:
Ta có:            $\frac{a \sin \frac{B-C }{ 2}  }{\sin \frac{ A}{ 2}  } =\frac{ 2R \sin A \sin \frac{ B-C}{ 2} }{\sin \frac{ A}{ 2}  } =\frac{ 4R \sin \frac{ A}{ 2} \cos \frac{ A}{ 2} \sin \frac{B-C }{ 2} }{ \sin \frac{ A}{ 2} } $
                                               $ =4R \sin \frac{B+C}{2 } \sin \frac{ B-C}{ 2} = 2R(\cos C-\cos B)       (2)$
Tương tự:       $\frac{ b \sin \frac{C-A }{ 2} }{\sin \frac{ B}{ 2}  } =2R(\cos A-\cos C)                                                                                                          (3)$     
                        $\frac{ b \sin \frac{ C-A}{ 2} }{ \sin \frac{ B}{ 2} } =2R(\cos A-\cos C)                                                        (4)$
Cộng vế với vế $(2) , (3) , (4)$ với nhau ta được :
$VT (1) = 2R(\cos C-\cos B+\cos A-\cos C+\cos B-\cos A)=0=VP  (1)$
Đẳng thức $(1)$ được chứng minh.
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