设α是第一象限角,且f(3/2α+π/2)=23/26,求sin(α+π/4）/cos(4π+2α）.f(x)=sin(2/3x+π/6）+1/2,fx图像任意两相邻对称轴的间距为3π/2

Question

设α是第一象限角,且f(3/2α+π/2)=23/26,求sin(α+π/4）/cos(4π+2α）.f(x)=sin(2/3x+π/6）+1/2,fx图像任意两相邻对称轴的间距为3π/2

Answer 1

①∵向量m⊥向量n∴向量m与向量n的内积为0
-f(x)+(coswx+√3sinwx)coswx=0 f(x)=(1+cos2wx)/2+√3(sin2wx)/2=sin(2wx+π/6)+1/2
又函数f（x）的图像上任意两相邻对称轴的间距为3/2π
∴函数f（x）的周期为2×3/2π=3π 从而2π/(2w)=3π,w=1/3
②由①知f(x)=sin(2/3x+π/6)+1/2,由条件f（3/2α+π/2）=23/26有
sin[2/3（3/2α+π/2）+π/6]+1/2=23/26,整理得sin(α+π/2)=5/13,cosα=5/13
又α是第一象限角,所以sinα=√(1-cos^2α)=12/13
sin（α+π/4）/cos（4π+2α）=[sinαcos(π/4)+cosαsin(π/4)]/(cos^2α-sin^2α)=1/[√2(cosα-sinα)]
=1/[√2(5/13-12/13)]=-13√2/14