已知函数f(x)=2根号3sinwx*coswx+2cos^2wx(w>0),且函数f(x)的图像的相邻两条对称轴的距离为派,若f(x/2)...

Question

已知函数f(x)=2根号3sinwx*coswx+2cos^2wx(w>0),且函数f(x)的图像的相邻两条对称轴的距离为派,若f(x/2)=2,t=(2派/3)-x
1.求cost的值

Answer 1

f(x)=√3sin(2wx)＋cos(2wx)＋1=2sin(2wx＋π/6)＋1，此函数两对称轴之间的距离是半个周期，则其周期是2π，则：T=2π/|2w|=2π，得：w=1/2
f(x)=2sin(x＋π/6)＋1，因f(x/2)=2，则：2sin(x/2＋π/6)＋1=2，sin(x/2＋π/6)=1/2
cost=cos(2π/3－x)=－cos(x＋π/3)=－cos[2(π/6＋x/2)]=－[1－2sin²(x/2＋π/6)]=－1/2

Answer 2

f(x)=4(根号3/2*2sinwxcoswx+1/2cos^2wx)
=4sin (2wx+π/6)
4sin(x/2)=2
sinx/2=1/2
sinx=√3/2
cosx=1/2
t=2π/3-x.
cost=cos(2π/3-x)=cos2π/3cosx+sin2π/3sinx
cost=√3/2×1/2-1/2×√3/2
cost=1

Answer 3

f(x)=2√3sinwx*coswx+2cos^2wx=√3/sin(2wx)+cos2wx+1=2sin(2wx+π/6)+1
所以f(x)的图像的相邻两条对称轴的距离为周期的一半T/2=π/(2w)=π，所以w=1/2
所以f(x)=2sin(x+π/6)+1
f(x/2)=2sin(x/2+π/6)+1=2则sin(x/2+π/6)=1/2
而cost=cos[(2π/3)-x]=-cos[π-(2π/3-x)]=-cos(x+π/3)
=-cos[2(x/2+π/6)]=-[1-2sin²(x/2+π/6)]=-(1-1/2)=-1/2