已知函数f(x)=2sinxcosx-1+2sinx2,(1)求f(x)的最小正周期和最大值
已知函数f(x)=2sinxcosx-1+2(sinx)2,(1)求f(x)的最小正周期和最大值(2)若f(α/2+π/8)=(3√2)/5,α是第二象限角,求sin(π...
已知函数f(x)=2sinxcosx-1+2(sinx)2,(1)求f(x)的最小正周期和最大值(2)若f(α/2+π/8)=(3√2)/5,α是第二象限角,求sin(π/3-α)和cos(π/3+α)的值
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2012-11-09 · 知道合伙人教育行家
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解
(1)
f(x)=2sinxcosx-1+2sin²x
=sin2x-cos2x
=√2sin2xcosπ/4-√2cos2xsinπ/4
=√2sin(2x-π/4)
T=2π/ω=2π/2=π
f(x)max=√2
(2)若f(α/2+π/8)=(3√2)/5,α是第二象限角
则√2sin[2(α/2+π/8)-π/4]=√2sinα=3√2/5
sinα=3/5
则cosα=-4/5
sin(π/3-α)=sinπ/3*cosα-cosπ/3*sinα=√3/2*(-4/5)-1/2*3/5=-(2√3)/5-3/10
cos(π/3+α)=cosπ/3*cosα-sinπ/3*sinα=1/2*(-4/5)-√3/2*3/5=-2/5-(3√3)/10
数学辅导团为您解答,不理解请追问,理解请及时选为满意回答!(*^__^*)
(1)
f(x)=2sinxcosx-1+2sin²x
=sin2x-cos2x
=√2sin2xcosπ/4-√2cos2xsinπ/4
=√2sin(2x-π/4)
T=2π/ω=2π/2=π
f(x)max=√2
(2)若f(α/2+π/8)=(3√2)/5,α是第二象限角
则√2sin[2(α/2+π/8)-π/4]=√2sinα=3√2/5
sinα=3/5
则cosα=-4/5
sin(π/3-α)=sinπ/3*cosα-cosπ/3*sinα=√3/2*(-4/5)-1/2*3/5=-(2√3)/5-3/10
cos(π/3+α)=cosπ/3*cosα-sinπ/3*sinα=1/2*(-4/5)-√3/2*3/5=-2/5-(3√3)/10
数学辅导团为您解答,不理解请追问,理解请及时选为满意回答!(*^__^*)
追问
√2sin[2(α/2+π/8)-π/4]=√2sinα=3√2/5
这里为什么等于]√2sinα呢?
追答
√2前面本来就有的,后面化简就是a,你选另一个为最佳吧
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已知函数f(x)=2sinxcosx-1+2sin^2x
=sin2x-cos2x
=√2sin(2x-π/4)
,(1)f(x)的最小正周期T=2π/2=π
最大值=√2
(2)若f(α/2+π/8)=√2sin(α+π/4-π/4)=√2sinα=(3√2)/5
所以 sinα=3/5 α是第二象限角,cosα=-4/5
sin(π/3-α)
=sinπ/3cosα-cosπ/3sinα
=√3/2*(-4/5)-1/2*3/5
=-(4√3+3)/10
cos(π/3+α)
=cosπ/3cosα-cosπ/3cosα
=1/2*(-4/5)-√3/2*3/5
=-(4+3√3)/10
=sin2x-cos2x
=√2sin(2x-π/4)
,(1)f(x)的最小正周期T=2π/2=π
最大值=√2
(2)若f(α/2+π/8)=√2sin(α+π/4-π/4)=√2sinα=(3√2)/5
所以 sinα=3/5 α是第二象限角,cosα=-4/5
sin(π/3-α)
=sinπ/3cosα-cosπ/3sinα
=√3/2*(-4/5)-1/2*3/5
=-(4√3+3)/10
cos(π/3+α)
=cosπ/3cosα-cosπ/3cosα
=1/2*(-4/5)-√3/2*3/5
=-(4+3√3)/10
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