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
2.
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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