Question

三角形的三内角A,B,C所对应的边分别为a,b,c,设向量m=(c-a,b-a),n=(a+b,c),

Answer 1

1

m∥n,即：(c-a)/(a+b)=(b-a)/c

即：b^2-a^2=c^2-ac

即：a^2+c^2-b^2=ac

故：cosB=(a^2+c^2-b^2)/(2ac)=1/2

即：B=π/3

2

sinA^2+sinC^2=(1-cos(2A))/2+(1-cos(2C))/2

=1-(cos(2A)+cos(2C))/2

=1-cos(A+C)cos(A-C)

=1+cosBcos(A-C)=1+cos(A-C)/2

=1+cos(2A-2π/3)/2

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