第6题求解过程
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B/2=60°-C/2
a/(b+c)+b/(a+c)
=sinA/(sinB+sinC)+sinB/(sinA+sinC)
=sin(B+C)/[2sin(B+C)/2cos(B-C)/2]+sin(A+C)/[2sin(A+C)/2cos(A-C)/2]
=[sin(B+C)/2]/[cos(B-C)/2]+[sin(A+C)/2]/[cos(A-C)/2]
=cos(B/2+60°)/cos(B/2-30°)+sin(90°-B/2)/cos(30°-B/2)
=[cos(B/2+60°)+cos(90°-B/2)]/cos(30°-B/2)
=2cos60°cos(30°-B/2)/cos(30°-B/2)
=1
a/(b+c)+b/(a+c)
=sinA/(sinB+sinC)+sinB/(sinA+sinC)
=sin(B+C)/[2sin(B+C)/2cos(B-C)/2]+sin(A+C)/[2sin(A+C)/2cos(A-C)/2]
=[sin(B+C)/2]/[cos(B-C)/2]+[sin(A+C)/2]/[cos(A-C)/2]
=cos(B/2+60°)/cos(B/2-30°)+sin(90°-B/2)/cos(30°-B/2)
=[cos(B/2+60°)+cos(90°-B/2)]/cos(30°-B/2)
=2cos60°cos(30°-B/2)/cos(30°-B/2)
=1
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