Question

函数f(x)=cos方x+√3sinxcosx在区间（-π／4，π／3）上的最大值为

Answer 1

f(x)=(1+cos2x)/2+(√3/2)sin2x
=(√3/2)sin2x+(1/2)cos2x+1/2
=√[(√3/2)²+(1/2)²]*sin(2x+z)+1/2
=sin(2x+z)+1/2
其中 tanz=(1/2)/(√3/2)=√3/3
所以 z=π/6
f(x)=sin(2x+π/6)+1/2

-π/4<x<π/3
-π/3<=2x+π/6<=5π/6
因为sinx在 (-π/3,π/2)是减函数
所以sin(-π/3)<sin(2x+π/6)<=sin(π/2)
-√3/2<=sin(2x+π/6)<=1
所以最大值=1+1/2=3/2

Answer 2

cos方x+√3sinxcosx=cos2x/2+√3sin2x/2+1/2=sinπ/6cos2x+cosπ/6sin2x+1/2=
sin(2x+π/6)+1/2 x=π/6 时f(x)=cos方x+√3sinxcosx的最大值为1+1/2=3/2

Answer 3

上式可得1/2cos2x+1/2+2根号3sin2x=0
sin(2x+30度）+1/2=o
在当x处于该区间时
后面的你会吧