在三角形ABC中，1+cos2A-cos2B-cos2C=2sinBsinC,设f(B)=sinB的平方+sinC的平方，求f(B)的最大值

Question

过程要详细点，谢谢

Answer 1

先说两点sinA=sin[180-(B+C)]=sin(A+B) cosA=cos[180-(B+C)]= -cos(B+C)
1+cos2A-cos2B-cos2C=2sinBsinC
1+(1-2sin^2 A)-(1-2sin^2 B)-(1-2sin^2 C)=2sinBsinC [sin^2 A指sinA的平方，余同。]
整理得，2sin^2 B+2sin^2 C-2sin^2 A=2sinBsinC
sin^2 B+sin^2 C-sin^2 A=sinBsinC [同除以2]
sin^2 B+sin^2 C-sin^2 (B+C)=sinBsinC
sin^2 B+sin^2 C-(sinBcosC+cosBsinC)^2 =sinBsinC
展开得，sin^2 B+sin^2 C-sin^2 B*cos^2 C-cos^2 B*sin^2 C-2sinBsinCcosBcosC=sinBsinC
sin^2 B(1-cos^2 C)+sin^2 C(1-cos^2 B)-2sinBsinCcosBcosC=sinBsinC
即sin^2 B*sin^2 C+sin^2 C*sin^2 B-2sinBsinCcosBcosC=sinBsinC
即2sin^2 C*sin^2 B-2sinBsinCcosBcosC=sinBsinC
两边同除以2sinBsinC得
sinBsinC-cosBcosC=1/2
cosBcosC-sinBsinC= -1/2
即cos(B+C)= -1/2= -cosA [如题前所指出]
所以cosA=1/2
A=60度