已知函数f(x)=cos(2π-x) cos(π/2-x)-sin^2x (1)求函数f(x)的最小正周期(2)当x∈[-π/8,3/8π]时,求函数f(x)的值域
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![已知函数f(x)=cos(2π-x) cos(π/2-x)-sin^2x (1)求函数f(x)的最小正周期(2)当x∈[-π/8,3/8π]时,求函数f(x)的值域](/uploads/image/z/6897484-28-4.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dcos%282%CF%80-x%29+cos%28%CF%80%2F2-x%29-sin%5E2x+%281%29%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%882%EF%BC%89%E5%BD%93x%E2%88%88%5B-%CF%80%2F8%2C3%2F8%CF%80%5D%E6%97%B6%2C%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E5%80%BC%E5%9F%9F)
已知函数f(x)=cos(2π-x) cos(π/2-x)-sin^2x (1)求函数f(x)的最小正周期(2)当x∈[-π/8,3/8π]时,求函数f(x)的值域
已知函数f(x)=cos(2π-x) cos(π/2-x)-sin^2x (1)求函数f(x)的最小正周期
(2)当x∈[-π/8,3/8π]时,求函数f(x)的值域
已知函数f(x)=cos(2π-x) cos(π/2-x)-sin^2x (1)求函数f(x)的最小正周期(2)当x∈[-π/8,3/8π]时,求函数f(x)的值域
1、f(x)=cos(2π-x) cos(π/2-x)-sin^2x
=-cosxsinx-sin^2x
=-½sin2x-(1-cos2x)/2
=-1/2sin2x+1/2cos2x-1/2
= - √2/2sin(2x-π/4)-1/2
所以函数f(x)的最小正周期为π
2、当x∈[-π/8,3π/8]时,-π/2≤2x-π/4≤π/2 ,f(x)最小值为x=3π/8时,f(x)=-1/2-√2/2
x=π/8时f(x)最大值为√2/2-1/2
所以当x∈[-π/8,3/8π]时,函数f(x)的值域为[-1/2-√2/2,√2/2-1/2]
1、f(x)=cos(2π-x) cos(π/2-x)-sin^2x
=cosx*sinx+1/2(1-2sin^2x )-1/2
=1/2sin2x+1/2cos2x-1/2
=√2/2(sin2xcosπ/4+cos2xsinπ/4)-1/2
=√2/2sin(2x+π/4)-1/2
所以函数f(x)的最小正周期...
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1、f(x)=cos(2π-x) cos(π/2-x)-sin^2x
=cosx*sinx+1/2(1-2sin^2x )-1/2
=1/2sin2x+1/2cos2x-1/2
=√2/2(sin2xcosπ/4+cos2xsinπ/4)-1/2
=√2/2sin(2x+π/4)-1/2
所以函数f(x)的最小正周期为π
2、当x∈[-π/8,3/8π]时,f(x)最小值为x=-π/8或3/8π时,f(x)=-1/2
x=π/8时f(x)最大值为√2/2-1/2
所以当x∈[-π/8,3/8π]时,函数f(x)的值域为[-1/2,√2/2-1/2]
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