函数f(x)=sinx-cosx+sin2x,(x属于R)的最大值为_______ 题中确是X的正弦减余弦,不是乘

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/29 17:33:34
函数f(x)=sinx-cosx+sin2x,(x属于R)的最大值为_______ 题中确是X的正弦减余弦,不是乘

函数f(x)=sinx-cosx+sin2x,(x属于R)的最大值为_______ 题中确是X的正弦减余弦,不是乘
函数f(x)=sinx-cosx+sin2x,(x属于R)的最大值为_______ 题中确是X的正弦减余弦,不是乘

函数f(x)=sinx-cosx+sin2x,(x属于R)的最大值为_______ 题中确是X的正弦减余弦,不是乘
设t=sinx-cosx=根号2sin(x-Pai/4),故有:-根号2<=t<=根号2
t^2=1-2sinxcosx=1-sin2x
那么有f(t)=t+1-t^2=-(t-1/2)^2+5/4
故当t=1/2时有最大值是5/4,当t=-根号2时有最小值是-根号2+1-2=-根号2-1
即有最大值是:5/4

f(x)=sinx-cosx+sin2x=√2sin(x-π/4)+cos(2x-π/2)
=√2sin(x-π/4)+1-2sin^2(x-π/4)
=-2[sin^2(x-π/4)-√2/2sin(x-π/4)]+1
=-2[sin(x-π/4)-√2/4]^2+1+1/4
所以最大值时5/4
不懂可以追问,谢谢!...

全部展开

f(x)=sinx-cosx+sin2x=√2sin(x-π/4)+cos(2x-π/2)
=√2sin(x-π/4)+1-2sin^2(x-π/4)
=-2[sin^2(x-π/4)-√2/2sin(x-π/4)]+1
=-2[sin(x-π/4)-√2/4]^2+1+1/4
所以最大值时5/4
不懂可以追问,谢谢!

收起

f(x)=sinx-cosx+sin2x
=(sinx-cosx)+1-(sinx-cosx)²
=5/4-[1/2-(sinx-cosx)]²
=5/4-[1/2-√2(sin(x-π/4)]²
因此,最大值为5/4