已知向量a=(2cosx,sinx),向量b=(0,√3cosx),f(x)=|向量a+向量b|1.求f(π/6)的值2.当x∈(0,π)时,求f(x)的值域有谁有温州二模的答案啊啊
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![已知向量a=(2cosx,sinx),向量b=(0,√3cosx),f(x)=|向量a+向量b|1.求f(π/6)的值2.当x∈(0,π)时,求f(x)的值域有谁有温州二模的答案啊啊](/uploads/image/z/3743689-49-9.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%282cosx%2Csinx%29%2C%E5%90%91%E9%87%8Fb%3D%280%2C%E2%88%9A3cosx%29%2Cf%28x%29%3D%7C%E5%90%91%E9%87%8Fa%2B%E5%90%91%E9%87%8Fb%7C1.%E6%B1%82f%28%CF%80%2F6%29%E7%9A%84%E5%80%BC2.%E5%BD%93x%E2%88%88%280%2C%CF%80%EF%BC%89%E6%97%B6%2C%E6%B1%82f%28x%29%E7%9A%84%E5%80%BC%E5%9F%9F%E6%9C%89%E8%B0%81%E6%9C%89%E6%B8%A9%E5%B7%9E%E4%BA%8C%E6%A8%A1%E7%9A%84%E7%AD%94%E6%A1%88%E5%95%8A%E5%95%8A)
已知向量a=(2cosx,sinx),向量b=(0,√3cosx),f(x)=|向量a+向量b|1.求f(π/6)的值2.当x∈(0,π)时,求f(x)的值域有谁有温州二模的答案啊啊
已知向量a=(2cosx,sinx),向量b=(0,√3cosx),f(x)=|向量a+向量b|
1.求f(π/6)的值2.当x∈(0,π)时,求f(x)的值域
有谁有温州二模的答案啊啊
已知向量a=(2cosx,sinx),向量b=(0,√3cosx),f(x)=|向量a+向量b|1.求f(π/6)的值2.当x∈(0,π)时,求f(x)的值域有谁有温州二模的答案啊啊
(1)a+b=(2cosx,sinx+√3cosx)
得到f(x)=|向量a+向量b|=√(4cosxcosx+sinxsinx+3cosxcosx+2√3sinxcosx)
=√(6cosxcosx+2√3sinxcosx+1)
=√(√3sin2x+3cos2x+4)
=√[2√3sin(2x+π/3)+4]
所以f(π/6)=√7
(2)当x∈(0,π)时,2x+π/3属于[π/3,7π/3]
得到sin(2x+π/3)属于[-1,1]
所以2√3sin(2x+π/3)+4属于[-2√3+4,2√3+4]
得到f(x)的值域是[√3-1,√3+1]
a+b
=(2cosx,sinx+√3cosx)
|a+b|^2
=(2cosx)^2+(sinx+√3cosx)^2
=6(cosx)^2+2√3sinxcosx +1
=3(cos2x-1) + √3sin2x +1
=3cos2x+√3sin2x -2
=(2√3)sin(2x+π/3) -2
f(π/6) = (2√3)sin...
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a+b
=(2cosx,sinx+√3cosx)
|a+b|^2
=(2cosx)^2+(sinx+√3cosx)^2
=6(cosx)^2+2√3sinxcosx +1
=3(cos2x-1) + √3sin2x +1
=3cos2x+√3sin2x -2
=(2√3)sin(2x+π/3) -2
f(π/6) = (2√3)sin(2π/3) -2
= 3-2
=1
f(x)= (2√3)sin(2x+π/3) -2
x∈(0,π)
max f(x) = 2√3 -2
min f(x) = f(π) = -3 -2 =-5
值域 [-5,2√3 -2]
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