Question

函数f(x)=sin^2 x +根号3sinxcosx在区间[π/4,π/2]上的最大值

f(x)=(1-cos2x)/2+√3/2*sin2x
=√3/2sin2x-1/2*cos2x+1/2
=sin2xcosπ/6-cos2xsinπ/6+1/2
=sin(2x-π/6)+1/2

π/4<=x<=π/2
π/2<=2x<=π
π/3<=2x-π/6<=5π/6
所以2x-π/6=π/2时
原式最大=1+1/2=3/2（为什么一定当2x-π/6=π/2时，f(x)的值最大呢？不用求对称轴什么的吗）

Answer 1

f(x)=(1-cos2x)/2+√3/简嫌2*sin2x
=√3/2sin2x-1/2*cos2x+1/2
=sin2xcosπ/6-cos2xsinπ/6+1/2
=sin(2x-π/6)+1/2

π/4<=x<=π/2
π/2<=2x<=π
π/3<=2x-π/6<=5π/6
所以2x-π/6=π/2时
原式最大宽咐哪=1+1/2=3/2
求采纳慎码为满意回答。

追问

为什么一定当2x-π/6=π/2时，f(x)的值最大呢？不用求单调递增区间什么的吗