Question

在△ABC中，角A，B，C所对的边分别为a,b,c，且满足cos(B-C)-2sinBsinC=

且满足cos(B-C)-2sinBsinC=1/2.求角A(拜托拜托详细过程！)

Answer 1

cos(B-C)-2sinBsinC
=cosBcosC+sinBsinC-2sinBsinC
=cosBcosC-sinBsinC
=cos(B+C)
=-cosA
=½
cosA=-½
A=⅔π

Answer 2

考查要点，三角形余弦定理和正弦定理的应用，以及和差化积和积化和差公式的运用
cos(B-C)=cosBcosC+sinBsinC

cos(B-C)-2sinBsinC=cosBcosC-sinBsinC=cos(B+C)
A+B+C =pi
1/2 =cos(B+C)=cos(pi-A)=-cos(A),cos(A)=-1/2 ,A=2pi/3

Answer 3

依题有：cos(B-C)-2sinBsinC
=cosBcosC+sinBsinC-2sinBsinC
=cosBcosC-sinBsinC
=cos(B+C)=1/2
cosA=cos(180⁰-(B+C))=-cos(B+C)=-1/2
因此角A=120⁰

Answer 4

解：cos(B-C)-2sinBsinC=1/2.
cosBcosC+sinBsinC-2sinBsinC=1/2
cosBcosC-sinBsinC=1/2
cos（B+C）=1/2
cosA=-1/2
A∈（0，∏）所以A=2∏/3