This math problem will get the help of the great god! Senior one semester

1+cos2A=2cos？ A

cos2B+cos2C = 2 cos(B+C)cos(B-C)= 2 cos( 180-A)cos(B-C)=-2 cos cos(B-C)

1+cos2A-cos2B-cos2C

=2cos？ a+2 cos(B-C)

=2cosA*[cosA+cos(B-C)]

=2cosA*[cos(B-C)-cos(B+C)]

=2cosA*2sinBsinC

=4cosAsinBsinC

So 4cosAsinBsinC=2sinBsinC

that is

2cosA= 1

cosA= 1/2

A=60

f(B)=sin？ B+ crime? C

=( 1-cos2B)/2+( 1-cos2C)/2

=(2-cos2B-cos2C)/2

= 1-(cos2B+cos2C)/2

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

= 1-cos 120 cos(B-C)

= 1+cos(B-C)/2

Therefore, if and only if b = c = 60, f(B) takes 3/2 of the maximum value.

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