设函数f(x)=sinxcosx+cosx^2,求f(x)的最小正周期,当x属于【0,π/2】时,求函数f(x)的最大值和最小值答案是f(x)=sinxcosx+cosx^2=½(sin2x+cos2x+1)=½[√2sin(2x+π/4)+1],即最小正周期为π.当x属于[0,π/2]时,2x∈[

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设函数f(x)=sinxcosx+cosx^2,求f(x)的最小正周期,当x属于【0,π/2】时,求函数f(x)的最大值和最小值答案是f(x)=sinxcosx+cosx^2=½(sin2x+cos2x+1)=½[√2sin(2x+π/4)+1],即最小正周期为π.当x属于[0,π/2]时,2x∈[

设函数f(x)=sinxcosx+cosx^2,求f(x)的最小正周期,当x属于【0,π/2】时,求函数f(x)的最大值和最小值答案是f(x)=sinxcosx+cosx^2=½(sin2x+cos2x+1)=½[√2sin(2x+π/4)+1],即最小正周期为π.当x属于[0,π/2]时,2x∈[
设函数f(x)=sinxcosx+cosx^2,求f(x)的最小正周期,当x属于【0,π/2】时,求函数f(x)的最大值和最小值
答案是f(x)=sinxcosx+cosx^2=½(sin2x+cos2x+1)=½[√2sin(2x+π/4)+1],即最小正周期为π.
当x属于[0,π/2]时,2x∈[0,π],2x+π/4∈[π/4,5π/4],sin(2x+π/4)∈[-√2/2,1],f(x)∈[0,(√2+1)/2],故f(x)最大值为(√2+1)/2,最小值为0.
我想知道答案里的sinπ/4为什么会等于-√2/2?

设函数f(x)=sinxcosx+cosx^2,求f(x)的最小正周期,当x属于【0,π/2】时,求函数f(x)的最大值和最小值答案是f(x)=sinxcosx+cosx^2=½(sin2x+cos2x+1)=½[√2sin(2x+π/4)+1],即最小正周期为π.当x属于[0,π/2]时,2x∈[
sin(2x+π/4)∈[-√2/2,1]这一步对吗?
sin π/2=1为最大
sin 5π/4=-√2/2为最小
两者构成该[-√2/2,1]
算 f(x)的区间 时并不一定要取x区间的端点

令a=2x+π/4∈[π/4,5π/4],
sina∈[-√2/2,1],
在a=5π/4时取到-√2/2

并非sinπ/4=-√2/2