已知函数f(x)=cos(x-π/3)-sin(π/2-x).1.求函数的最小值. 2.若a∈(0,2/π),且f(a+

f(x)=cos(x-π/3)-sin(π/2-x)
=cos(x-π/3)-cosx
=cosxcos(π/3)+sinxsin(π/3)-cosx
=1/2*cosx+√3/2*sinx-cosx
=√3/2*sinx-1/2*cosx
=sinxcos(π/6)-cosxsin(π/6)
=sin(x-π/6)
1、当x-π/6=-π/2+2kπ,k∈Z时,f(x)有最小值-1
即当x=-π/3+2kπ,k∈Z时,f(x)有最小值-1
2、∵f(a+π/6)=sin(a+π/6-π/6)=sina=3/5
又∵a∈(0,π/2)
∴cosa=√(1-sin²a)=4/5
∴f(2a)=sin(2a-π/6)
=sin2acos(π/6)-cos2asin(π/6)
=2sinacosa*√3/2-(2cos²a-1)*1/2
=2*(3/5)*(4/5)*√3/2-[2*(4/5)²-1]*1/2
=12√3/25-7/50
=(24√3-7)/50
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