设函数f(x)=2cosxsin(x+π/6)+2sinxcos(x+π/6)⑴当x属于0到2分之π的闭区间求f(x)的值域⑵设三角形ABC的三个内角ABC所对的三边依次为abc已知f(A)=1,a=根号7,三角形ABC面积为2分之3倍根号3求b+c
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![设函数f(x)=2cosxsin(x+π/6)+2sinxcos(x+π/6)⑴当x属于0到2分之π的闭区间求f(x)的值域⑵设三角形ABC的三个内角ABC所对的三边依次为abc已知f(A)=1,a=根号7,三角形ABC面积为2分之3倍根号3求b+c](/uploads/image/z/11642027-59-7.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D2cosxsin%28x%2B%CF%80%EF%BC%8F6%29%2B2sinxcos%28x%2B%CF%80%EF%BC%8F6%29%E2%91%B4%E5%BD%93x%E5%B1%9E%E4%BA%8E0%E5%88%B02%E5%88%86%E4%B9%8B%CF%80%E7%9A%84%E9%97%AD%E5%8C%BA%E9%97%B4%E6%B1%82f%EF%BC%88x%EF%BC%89%E7%9A%84%E5%80%BC%E5%9F%9F%E2%91%B5%E8%AE%BE%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E4%B8%89%E4%B8%AA%E5%86%85%E8%A7%92ABC%E6%89%80%E5%AF%B9%E7%9A%84%E4%B8%89%E8%BE%B9%E4%BE%9D%E6%AC%A1%E4%B8%BAabc%E5%B7%B2%E7%9F%A5f%EF%BC%88A%EF%BC%89%3D1%2Ca%3D%E6%A0%B9%E5%8F%B77%2C%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E9%9D%A2%E7%A7%AF%E4%B8%BA2%E5%88%86%E4%B9%8B3%E5%80%8D%E6%A0%B9%E5%8F%B73%E6%B1%82b%2Bc)
设函数f(x)=2cosxsin(x+π/6)+2sinxcos(x+π/6)⑴当x属于0到2分之π的闭区间求f(x)的值域⑵设三角形ABC的三个内角ABC所对的三边依次为abc已知f(A)=1,a=根号7,三角形ABC面积为2分之3倍根号3求b+c
设函数f(x)=2cosxsin(x+π/6)+2sinxcos(x+π/6)⑴当x属于0到2分之π的闭区间求f(x)的值域⑵设三角形ABC的三个内角ABC所对的三边依次为abc已知f(A)=1,a=根号7,三角形ABC面积为2分之3倍根号3求b+c
设函数f(x)=2cosxsin(x+π/6)+2sinxcos(x+π/6)⑴当x属于0到2分之π的闭区间求f(x)的值域⑵设三角形ABC的三个内角ABC所对的三边依次为abc已知f(A)=1,a=根号7,三角形ABC面积为2分之3倍根号3求b+c
f(x)=2cosxsin(x+π/6)+2sinxcos(x+π/6)
=2sin(2x+π/6),
(1)x∈[0,π/6],
∴2x+π/6∈[π/6,π/2],
∴f(x)的值域是[1,2].
(2)f(A)=2sin(2A+π/6)=1,
∴sin(2A+π/6)=1/2,
2A+π/6=5π/6,
A=π/3,
由余弦定理,7=b^+c^-bc,
(1/2)bc*√3/2=(3/2)√3,bc=6,
∴(b+c)^=25,
b+c=5.
(1)f(x)=2sin(2x+π/6),
x属于0到2分之π的闭区间
值域[-1,2]
(2)f(A)=2sin(2A+π/6)=1,
A=π/3或π/2或11π/12
S=1/2bcsinA,
得到bc=6,3根号3,3根号3/(根号6—根号2)
余弦定理:a2=(b+c)2-2bc-2bccosA
可得=(b+c)