Răspuns:
Explicație pas cu pas:
[tex]bcosB+ccosC=2RsinBcosB+2RsinCcosC=\\ =2R(sinBcosB+sinCcosC)=\\ \\ =2R(\frac{sin(B+B)+sin(B-B)}{2} +\frac{sin(C+C)+sin(C-C)}{2} )=\\ \\ =2R(\frac{sin2B+1}{2} +\frac{sin2C+1}{2} )=\\ \\ =R(sin2B+sin2C+2)=\\ \\ =R(2sin\frac{2B+2C}{2} cos\frac{2B-2C}{2} +2)=\\ \\ =R[2sin(B+C)cos(B-C)+2]=\\ \\ =2R[sin(\pi -A)cos(B-C)+1]=\\ \\ =\frac{a}{sinA} [-sinAcos(B-C)+1]=...[/tex]
