求一道数学题 一道简单的数学题。。。
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\u5219\u5973\u6027\u804c\u5de5\u5c31\u662f11\u4eba
\u5c06\u8fd9\u4e24\u4e2a\u6570\u5b57\u5e26\u5165\u9898\u76ee\u6761\u4ef6\u5c31\u53ef\u5f97\u51fa\u7537\u6027\u8fd1\u89c6\u67092\u4eba\uff0c\u5973\u6027\u8fd1\u89c6\u67093\u4eba
\u5219\u7537\u6027\u4e0d\u8fd1\u89c617\u4eba\uff0c\u5973\u6027\u4e0d\u8fd1\u89c68\u4eba\uff0c\u7b54\u6848\u5c31\u662f9\u4eba
\u7ea6\u4e3a\u957f14.76cm\uff0c\u5bbd7.38cm
【参考答案】f(x)=2cos(2x+ π/3)+4√3sinxcosx
=2[(1/2)cos2x-(√3/2)sin2x]+4√3sinxcosx
=cos2x-√3sin2x+2√3sin2x
=cos2x+√3sin2x
=2[(1/2)cos2x+(√3/2)sin2x]
=2sin(2x+ π/6)
①当0≤x≤π/2时，π/6≤2x+ π/6≤7π/6
此时，sin2x∈[-1/2，1]
∴函数f(x)在[0，π/2]上的值域是[-1，2]
②f(B)=1即2sin(2B+π/6)=1
sin(2B+ π/6)=1/2
则2B+π/6=5π/6或π/6（舍去）
∴B=π/3, A+C=2π/3
∵sinA、sinB、sinC成等比数列
∴sin²B=sinAsinC=(√3/2)²=3/4
根据积化和差公式得
3/4=-[cos(A+C)-cos(A-C)]/2
3/4=-[(-1/2)-cos(A-C)]/2
cos(A-C)=1
∴A=C
又∵A+C=2π/3
∴A=C=π/3
∴△ABC是等边三角形
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