已知函数f(x)=sin(2x-π/6)＋1/2,三角形ABC中,A B C对应的边为a b c,若f(A/2＋π/12)＋f(B/2＋π/12)=2√6sinAsinB＋1,C=60°,c=3,求三角形的面积.

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已知函数f(x)=sin(2x-π/6)＋1/2,三角形ABC中,ABC对应的边为abc,若f(A/2＋π/12)＋f(B/2＋π/12)=2√6sinAsinB＋1,C=60°,c=3,求三角形的面

已知函数f(x)=sin(2x-π/6)＋1/2,三角形ABC中,A B C对应的边为a b c,若f(A/2＋π/12)＋f(B/2＋π/12)=2√6sinAsinB＋1,C=60°,c=3,求三角形的面积.
已知函数f(x)=sin(2x-π/6)＋1/2,三角形ABC中,A B C对应的边为a b c,若f(A/2＋π/12)＋f(B/2＋π/12)=2√6sinAsinB＋1,C=60°,c=3,求三角形的面积.

已知函数f(x)=sin(2x-π/6)＋1/2,三角形ABC中,A B C对应的边为a b c,若f(A/2＋π/12)＋f(B/2＋π/12)=2√6sinAsinB＋1,C=60°,c=3,求三角形的面积.
a/sinA=b/sinB=c/sinC=3/sin60度=3/(√3/2)=2√3
f(A/2＋π/12)＋f(B/2＋π/12)=2√6sinAsinB＋1
sin(A+π/6-π/6)+1/2+sin(B+π/6-π/6)+1/2=2√6sinAsinB+1
sinA+sinB=2√6sinAsinB
a/(2√3)+b/(2√3)=2√6*a/(2√3)*b/(2√3)
a+b=√2ab
cosC=(a²+b²-c²)/(2ab),a²+b²-9=2abcos60度
a²+2ab+b²-9=2ab+2ab*1/2,(a+b)²-9=3ab,2(ab)²-3(ab)-9=0
(2ab+3)(ab-3)=0
a>0,b>0,ab>0
所以 ab=3
S=1/2absinC=√3ab=3√3

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