已知函数f(x)=2cosx(sinx-cosx)+1,x属于R.(1)求函数f(x)的最小正周期;(2)求函数f(x)在区间〔π/8,3π/4]上的最小值和最大值过程详细点~~~最好解释下
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![已知函数f(x)=2cosx(sinx-cosx)+1,x属于R.(1)求函数f(x)的最小正周期;(2)求函数f(x)在区间〔π/8,3π/4]上的最小值和最大值过程详细点~~~最好解释下](/uploads/image/z/616005-45-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3D2cosx%28sinx-cosx%29%2B1%2Cx%E5%B1%9E%E4%BA%8ER.%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%3B%282%29%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%E5%8C%BA%E9%97%B4%E3%80%94%CF%80%2F8%2C3%CF%80%2F4%5D%E4%B8%8A%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E5%92%8C%E6%9C%80%E5%A4%A7%E5%80%BC%E8%BF%87%E7%A8%8B%E8%AF%A6%E7%BB%86%E7%82%B9%7E%7E%7E%E6%9C%80%E5%A5%BD%E8%A7%A3%E9%87%8A%E4%B8%8B)
已知函数f(x)=2cosx(sinx-cosx)+1,x属于R.(1)求函数f(x)的最小正周期;(2)求函数f(x)在区间〔π/8,3π/4]上的最小值和最大值过程详细点~~~最好解释下
已知函数f(x)=2cosx(sinx-cosx)+1,x属于R.(1)求函数f(x)的最小正周期;(2)求函数f(x)在区间〔π/8,3π/4]上的最小值和最大值
过程详细点~~~最好解释下
已知函数f(x)=2cosx(sinx-cosx)+1,x属于R.(1)求函数f(x)的最小正周期;(2)求函数f(x)在区间〔π/8,3π/4]上的最小值和最大值过程详细点~~~最好解释下
f(x)=2cosx(sinx-cosx)+1
=2sinxcosx-2(cosx)^2+1
=2sinxcosx-[2(cosx)^2-1]
=sin2x-cos2x
=√2(√2/2*sin2x-√2/2*cos2x)
=√2(sin2xcosπ/4-cos2xsinπ/4)
=√2sin(2x-π/4)
(1)
f(x)=√2sin(2x-π/4)
∴函数f(x)的最小正周期:
T=2π/2=π
(2)f(x)=2sinxcosx+1-2cosx^2
=sin2x-cos2x
=√2sin(2x-π/4)
π/8≤x≤3π/4
得 0≤2x-π/4≤5π/4
f(x)最小值是=√2sin5π/4=-1
f(x)最大值是=√2 2sinπ/2=√2
f(x)=2cosx(sinx-cosx)+1
=2sinxcosx-2(cosx)^2+1
=2sinxcosx-[2(cosx)^2-1]
=sin2x-cos2x
=√2(√2/2*sin2x-√2/2*cos2x)
=√2(sin2xcosπ/4-cos2xsinπ/4)
=√2sin(2x-π/4)
(1)
f(x)=...
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f(x)=2cosx(sinx-cosx)+1
=2sinxcosx-2(cosx)^2+1
=2sinxcosx-[2(cosx)^2-1]
=sin2x-cos2x
=√2(√2/2*sin2x-√2/2*cos2x)
=√2(sin2xcosπ/4-cos2xsinπ/4)
=√2sin(2x-π/4)
(1)
f(x)=√2sin(2x-π/4)
T=2π/2=π
(2)f(x)=2sinxcosx+1-2cosx^2
=sin2x-cos2x
=√2sin(2x-π/4)
π/8≤x≤3π/4
得 0≤2x-π/4≤5π/4
f(x)最小值是=√2sin5π/4=-1
f(x)最大值是=√2 2sinπ/2=√2
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我现在怀疑你是不是二十二中的,我也在写