已知函数f(x)=根号3sin^2+sinxcosx-二分之根号3 x属于R
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已知函数f(x)=根号3sin^2+sinxcosx-二分之根号3 x属于R
1,求f(π/4) 2,若x属于(0,π),求f(x)的最大值 3,在三角形ABC中,若A
1,求f(π/4) 2,若x属于(0,π),求f(x)的最大值 3,在三角形ABC中,若A
![已知函数f(x)=根号3sin^2+sinxcosx-二分之根号3 x属于R](/uploads/image/z/3710245-13-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D%E6%A0%B9%E5%8F%B73sin%5E2%2Bsinxcosx-%E4%BA%8C%E5%88%86%E4%B9%8B%E6%A0%B9%E5%8F%B73+x%E5%B1%9E%E4%BA%8ER)
f(x)=√3sin^2 x+sinxcosx-(√3/2)(x∈R)
=√3*[(1-cos2x)/2]+(1/2)sin2x-(√3/2)
=(√3/2)-(√3/2)cos2x+(1/2)sin2x-(√3/2)
=(1/2)sin2x-(√3/2)cos2x
=sin[2x-(π/3)]
(1)f(π/4)=sin[(π/2)-(π/3)]=sin(π/6)=1/2
(2)当x∈(0,π)时,2x∈(0,2π)
2x-(π/3)∈(-π/3,5π/3)
所以,f(x)的最大值为1
(1)A<B,且f(A)=f(B)=1/2
===> sin[2A-(π/3)]=sin[2B-(π/3)]=1/2
===> 2A-(π/3)=π/6;2B-(π/3)=5π/6
===> 2A=π/2;2B=7π/6
===> A=π/4,B=7π/12
所以,C=π-(π/4)-(7π/12)=π/6
在△ABC中,BC=a,AB=c
所以,由正弦定理有:BC/AB=a/c=sinA/sinC=√2.
=√3*[(1-cos2x)/2]+(1/2)sin2x-(√3/2)
=(√3/2)-(√3/2)cos2x+(1/2)sin2x-(√3/2)
=(1/2)sin2x-(√3/2)cos2x
=sin[2x-(π/3)]
(1)f(π/4)=sin[(π/2)-(π/3)]=sin(π/6)=1/2
(2)当x∈(0,π)时,2x∈(0,2π)
2x-(π/3)∈(-π/3,5π/3)
所以,f(x)的最大值为1
(1)A<B,且f(A)=f(B)=1/2
===> sin[2A-(π/3)]=sin[2B-(π/3)]=1/2
===> 2A-(π/3)=π/6;2B-(π/3)=5π/6
===> 2A=π/2;2B=7π/6
===> A=π/4,B=7π/12
所以,C=π-(π/4)-(7π/12)=π/6
在△ABC中,BC=a,AB=c
所以,由正弦定理有:BC/AB=a/c=sinA/sinC=√2.
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