已知向量m=（cosx,sinx）,n=（2√2+sinx,2√2-cosx）,函数fx=m向量*n向量,x∈R1.求函数fx的最大值 2.若x∈（-3π/2,-π）,且fx=1,求cos（x+5π/12）的值
问题描述：
已知向量m=（cosx,sinx）,n=（2√2+sinx,2√2-cosx）,函数fx=m向量*n向量,x∈R
1.求函数fx的最大值 2.若x∈（-3π/2,-π）,且fx=1,求cos（x+5π/12）的值
答
f(x)=cosx(2√2+sinx)+sinx(2√2-cosx)
=2√2(cosx+sinx)
=4(sinπ/4cosx+cosπ/4sinx)
=4sin(x+π/4)
所以当x=π/4+2kπ时,f(x)有最大值4
当f(x)=1时,
sin(x+π/4)=1/4
sin(x+π/4+π/2)=sin(x+3π/4)=cos(x+π/4)
因为x∈（-3π/2,-π）,所以x+π/4∈（-5π/4,-3π/4）,
cos(x+π/4)0,
所以sin(x+3π/4)=cos(x+π/4)=-√1-sin^2(x+π/4)=-√15/4,cos(x+3π/4)=1/4
cos（x+5π/12）=cos((x+3π/4)-π/3)
=cos(x+3π/4)/2+√3sin(x+3π/4)/2
=1/8-√45/8
=(1-3√5)/8