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幼苗
共回答了17个问题采纳率:94.1% 举报
答:
(1)
f(x)=√3(sinx)^2+sinxcosx-√3/2
=(√3/2)(1-cos2x)+(1/2)sin2x-√3/2
=sin2xcosπ/3-cos2xsinπ/3
=sin(2x-π/3)
所以:f(π/4)=sin(π/2-π/3)=sinπ/6=1/2
所以:f(π/4)=1/2
(2)02A-π/3>-π/3.
所以:f(B)=sin(2A-π/3)=1/2
所以:2B-π/3=5π/6,B=7π/12
所以:f(A)=sin(2A-π/3)=1/2
所以:2A-π/3=π/6,A=π/4
所以:C=π-7π/12-π/4=π/6
根据正弦定理得:
BC/sinA=AB/sinC=AC/sinB=2R
所以:
BC/AB=sinA/sinC
=sin(π/4)/sin(π/6)
=(√2/2)/(1/2)
=√2
所以:BC/AB=√2
1年前
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