已知f(x)=sin(x+π/6)+sin(x-π/6)+acosx+b,(a,b∈R,且均为常数).(1)求函数f(x)的最小正周期;(2)若f(x)在区间[-π/3,0]上单调递增,且恰好能够取到f(x)的最小值2,试求a,b的值.
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![已知f(x)=sin(x+π/6)+sin(x-π/6)+acosx+b,(a,b∈R,且均为常数).(1)求函数f(x)的最小正周期;(2)若f(x)在区间[-π/3,0]上单调递增,且恰好能够取到f(x)的最小值2,试求a,b的值.](/uploads/image/z/2553724-28-4.jpg?t=%E5%B7%B2%E7%9F%A5f%28x%29%3Dsin%28x%2B%CF%80%2F6%29%2Bsin%28x-%CF%80%2F6%29%2Bacosx%2Bb%2C%28a%2Cb%E2%88%88R%2C%E4%B8%94%E5%9D%87%E4%B8%BA%E5%B8%B8%E6%95%B0%29.%EF%BC%881%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%EF%BC%9B%EF%BC%882%EF%BC%89%E8%8B%A5f%28x%29%E5%9C%A8%E5%8C%BA%E9%97%B4%5B-%CF%80%2F3%2C0%5D%E4%B8%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%2C%E4%B8%94%E6%81%B0%E5%A5%BD%E8%83%BD%E5%A4%9F%E5%8F%96%E5%88%B0f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC2%2C%E8%AF%95%E6%B1%82a%2Cb%E7%9A%84%E5%80%BC.)
已知f(x)=sin(x+π/6)+sin(x-π/6)+acosx+b,(a,b∈R,且均为常数).(1)求函数f(x)的最小正周期;(2)若f(x)在区间[-π/3,0]上单调递增,且恰好能够取到f(x)的最小值2,试求a,b的值.
已知f(x)=sin(x+π/6)+sin(x-π/6)+acosx+b,(a,b∈R,且均为常数).(1)求函数f(x)的最小正周期;
(2)若f(x)在区间[-π/3,0]上单调递增,且恰好能够取到f(x)的最小值2,试求a,b的值.
已知f(x)=sin(x+π/6)+sin(x-π/6)+acosx+b,(a,b∈R,且均为常数).(1)求函数f(x)的最小正周期;(2)若f(x)在区间[-π/3,0]上单调递增,且恰好能够取到f(x)的最小值2,试求a,b的值.
f(x)=sin(x+ π/6) +sin(x-π/6) +acosx +b(a,b∈r,且均为常数)
=2sinxcosπ/6+acosx+b
=√3sinx+acosx+b
=√(3+a^2)sin(x+θ)+b (sinθ=a/√(3+a^2),cosθ=√3/√(3+a^2),
f(x)在区间[-π/3,0]上单调递增,且恰好能够取到f(x)的最小值2
∴-π/3+θ=-π/2,b-√(3+a^2)=2
θ=-π/6
a=-1,b=2+2=4
+sin(x-π/6) +acosx +b(a,b∈r,且均为常数)??? f(x)=sin(x+ π/6) +sin(x-π/6) +acosx +b(a,b∈r,且均为常数) =2sinx
f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a
=2sinxcos(π/6)+cosx+a
=√3sinx+cosx+a
=2sin(x+π/6)+a,
1.最小正周期是2π.
2.x∈[-π/2,π/2],
x+π/6∈[-π/3,2π/3],
f(x)的最大值是2+a,最小值是a-√3,
依题意2+a+a-√3=a,
∴a=√3-2.