作业帮已知函数f(x)=1+sinxcosx,g(x)=[cos(x+(π/12))]^2
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![已知函数f(x)=1+sinxcosx,g(x)=[cos(x+(π/12))]^2](/uploads/image/z/6890777-17-7.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D1%2Bsinxcosx%2Cg%28x%29%3D%5Bcos%28x%2B%28%CF%80%2F12%29%29%5D%5E2)
已知函数f(x)=1+sinxcosx,g(x)=[cos(x+(π/12))]^2
已知函数f(x)=1+sinxcosx,g(x)=[cos(x+(π/12))]^2
已知函数f(x)=1+sinxcosx,g(x)=[cos(x+(π/12))]^2
(1)设X=Xo是函数y=f(x)图像的一条对称轴,求g(x).\x0d因为f(x)=1+(1/2)sin2x,对于正弦函数来说,当x=xo为对称轴时函数f(x)取得最大值或者最小值.即：sin2x=1,或者sin2x=-1\x0d所以,2x=2xo=kπ+(π/2)（k∈Z）\x0d那么,g(x)=[cos(2x+(π/6))+1]/2=[cos(kπ+(π/2)+(π/6))+1]/2\x0d=[cos(kπ+(2π/3))+1]/2\x0d当k为偶数时,g(x)=[cos(2π/3)+1]/2=[(-1/2)+1]/2=1/4\x0d当k为奇数时,g(x)=[cos(5π/3)+1]/2=[(1/2)+1]/2=3/4\x0d(2)求h(x)=f(wx/2)+g(wx/2)(w0)在区间[-2π/3,π/3]上是增函数的w的最大值.\x0d由前面知,f(x)=1+(1/2)sin2x,g(x)=[cos(2x+(π/6))+1]/2\x0d所以,f(wx/2)=1+(1/2)sin(2*wx/2)=1+(1/2)sin(wz)\x0dg(wx/2)=[cos(2*wx/2+(π/6))+1]/2=[cos(wx+(π/6))+1]/2\x0d所以：f(wx/2)+g(wx/2)=1+(1/2)sin(wx)+(1/2)cos(wx+(π/6))+(1/2)\x0d=(3/2)+(1/2)[sin(wx)+cos(wx+(π/6))]\x0d=(3/2)+(1/2)[sin(wx)+cos(wx)*cos(π/6)-sin(wx)*sin(π/6)]\x0d=(3/2)+(1/2)[sin(wx)+cos(wx)*(√3/2)]-sin(wx)*(1/2)]\x0d=(3/2)+(1/2)[(1/2)sin(wx)+(√3/2)cos(wx)]\x0d=(3/2)+(1/2)sin[(wx)+(π/3)]\x0d=(3/2)+(1/2)sin[w(x+(π/3w))]\x0d则其周期为T=2π/w\x0d区间[-2π/3,π/3]的长度为(π/3)-(-2π/3)=π\x0d要保证其在[-2π/3,π/3]上为增函数,则：\x0dπ/3w≥2π/3,且T/4=π/(2w)≥π\x0d即,w的最大值为1/2