第1个回答 2013-02-11
f(x)=2√3sinxcosx-2cos(x+π/4)cos(x-π/4)
=√3sin2x+2sin(x+π/4-π/2)cos(x-π/4)
=√3sin2x+sin(2x-π/2)
=√3sin2x-cos2x
=2sin(2x-π/6)
T=2π/2=π
2x-π/6=π/2+kπ,k是整数
对称轴x=π/3+k/2π,k是整数
x∈【-π/12,π/2】
2x-π/6∈[-π/3,5/6π]
f(x)属于【-√3,2】
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第2个回答 2013-02-11
解:(1)f(x)=2√3sinxcosx-2cos(x+π/4)cos(x-π/4)=2sin(2x-π/6)
T=2π/w=π
对称轴2x-π/6=π/2+kπ(k∈Z)x=π/3+kπ/2(k∈Z)
(2)令2x-π/6=t,x∈【-π/12,π/2】 t∈【-π/3,5π/6】
f(x) ∈【-√3/2,1】
第3个回答 2013-02-11
1、=√3sin2x+2sin(x+π/4-π/2)cos(x-π/4)
=√3sin2x+sin(2x-π/2)
=√3sin2x-cos2x
=2sin(2x-π/6)
对称轴2x-π/6=π/2+kπ(k∈Z)x=π/3+kπ/2(k∈Z)
2、令2x-π/6=t,x∈【-π/12,π/2】 t∈【-π/3,5π/6】
f(x) ∈【-√3/2,1】