a: Xét ΔABC có
\(cosC=\dfrac{CA^2+CB^2-AB^2}{2\cdot CA\cdot CB}\)
\(cosB=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)
\(b\cdot cosC+c\cdot cosB\)
\(=\dfrac{AC\left(CA^2+CB^2-AB^2\right)}{2\cdot CA\cdot CB}+\dfrac{AB\cdot\left(BA^2+BC^2-AC^2\right)}{2\cdot BA\cdot BC}\)
\(=\dfrac{CA^2+CB^2-AB^2}{2CB}+\dfrac{BA^2+BC^2-AC^2}{2BC}\)
\(=\dfrac{CA^2+CB^2-AB^2+BA^2+BC^2-AC^2}{2BC}=\dfrac{2BC^2}{2BC}=BC=a\)
b: Xét ΔABC có
\(\dfrac{AB}{sinC}=\dfrac{AC}{sinB}=\dfrac{BC}{sinA}=2R\)
=>\(sinC=\dfrac{AB}{2R};sinB=\dfrac{AC}{2R};sinA=\dfrac{BC}{2R}\)
\(sinB\cdot cosC+sinC\cdot cosB=sin\left(B+C\right)\)
\(=sin\left(180^0-\widehat{B}-\widehat{C}\right)=sinA\)
