9.已知函数f(x)=sinx*2+2√3sinxcos+3cos*2x (2)已知f(a)=3,且a∈(0,π),求a
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9.f(x)=sin^2x+2√3sinxcosx+3cos^2x
=2cos^2x+√3sin2x+1
=cos2x+√3sin2x+2
=sin(2x+π/6)+2
所以f(a)=sin(2a+π/6)+2=3
即sin(2a+π/6)=1
则有2a+π/6=2kπ+π/2(k∈N,a∈(0,π))
可得a=π/6
10.f(x)=sin(π/2+x)cosx-sinxcos(π-x)
=cosxcosx-sinx(-cosx)
=(1+cos2x)/2+(sin2x)/2
=√2/2sin(2x+π/4)+1/2
因f(A)=1且A为锐角,易知A=π/4
由正弦定理可求得AC长
11.｜OP｜=√(1/4+cos⁴θ),｜OQ｜=√(sin⁴θ+1)
sin(α+β)=sinαcosβ+cosαsinβ
=[(sin²θ*cos²θ)+1/2]/√(1/4+cos⁴θ)√(1+sin⁴θ)
=1/4sin²2θ+1/2/√[1/4+1/4(1+cos2θ)²]√[1+1/4(1-cos2θ)²]
=1/4+1/2/√(1/2)√(5/4)
=(3/4 )/ √5/8
=3√10/10
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