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解:∵ a+c=2b
由正弦定理可知sinA+sinC=2sinB ①
由 积化和差公式 知
sinA+sinC=2* sin[(A+C)/2]* cos[(A-C)/2]
∵A+B+C=π,A-C=π/3
∴ sinA+sinC=2* sin[(A+C)/2]* cos[(A-C)/2]
=2*sin(π/2-B/2)*cosπ/6
=√3cos(B/2) ②
由①②两式得
2sinB=√3cos(B/2)
而sinB=2sin(B/2)*cos(B/2)
∴4sin(B/2)*cos(B/2)=√3cos(B/2)
得sin(B/2)=√3/4
∵B/2一定是锐角,
∴cos(B/2)=√13/4
∴sinB=2sin(B/2)*cos(B/2)=√39/8
由正弦定理可知sinA+sinC=2sinB ①
由 积化和差公式 知
sinA+sinC=2* sin[(A+C)/2]* cos[(A-C)/2]
∵A+B+C=π,A-C=π/3
∴ sinA+sinC=2* sin[(A+C)/2]* cos[(A-C)/2]
=2*sin(π/2-B/2)*cosπ/6
=√3cos(B/2) ②
由①②两式得
2sinB=√3cos(B/2)
而sinB=2sin(B/2)*cos(B/2)
∴4sin(B/2)*cos(B/2)=√3cos(B/2)
得sin(B/2)=√3/4
∵B/2一定是锐角,
∴cos(B/2)=√13/4
∴sinB=2sin(B/2)*cos(B/2)=√39/8
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