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已知函数f(x)=sin(3pai/2-x)cosx-sinxcos(pai+x)

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已知函数f(x)=sin(3pai/2-x)cosx-sinxcos(pai+x)
(1)求函数的单调递增区间.(2)三角形ABC的三个内角A.B.C成等差数列,若A为锐角,f(A)=0,BC=2,求AC的长
已知函数f(x)=sin(3pai/2-x)cosx-sinxcos(pai+x)
f(x)=sin(3pai/2-x)cosx-sinxcos(pai+x)
=-cosx*cosx-sinx*(-cosx)
=-(1+cos2x)/2+(1/2)sin2x
=(1/2)sin2x-(1/2)cos2x-1/2
=(√2/2)*[sin2x*cos(π/4)-cos2x*sin(π/4)]-1/2
=(√2/2)sin(2x-π/4)-1/2
(1)增区间
2kπ-π/2≤2x-π/4≤2kπ+π/2
2kπ-π/4≤2x≤2kπ+3π/4
即 kπ-π/8≤x≤kπ+3π/8
∴ 增区间为[ kπ-π/8,kπ+3π/8],k∈Z
(2)三角形ABC的三个内角A.B.C成等差数列,
则A+C=2B
∴ 3B=A+B+C=π
∴ B=π/3
f(A)=0,∴ (√2/2)sin(2A-π/4)-1/2=0
∴ sin(2A-π/4)=√2/2
∴ A=π/4
利用正弦定理AC/sinB=BC/sinA
∴ AC=BCsinB/sinA=2*(√3/2)/(√2/2)=√6