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求函数y=sinx%Cosx+sinxCosx的值域

先说下方法: 令t=sinx+cosx,因t^2=1+2sinxcosx, 所以y=t+2t^-2,注意t的范围是[-√2,√2]。 下面就变成一个初中就学过的二次函数,可以根据图像求出值域。

令t=sinx+cosx 则 t=√2sin(x+45°)∈[-√2,√2] 而sinxcosx =[(sinx+cosx)^2-(sinx)^2-(cosx)^2]/2 =(t^2-1)/2 ∴原式=t+(t^2-1)/2 =[(t+1)^2-2]/2 当t=-1时,有最小值是:-1,当t=根号2时,有最大值是:(1+2根号2)/2。 所以,y∈[-1,(1+2√2)/2]

假设sinx+cosx=√2sin(x+45)=t -√2

令sinx+cosx=x 2sinx*cosx=(sinx+cosx)^2-1=x^2-1 y=sinx+cosx+sinx*cosx=(x^2-1)/2+x=1/2(x+1)^2-1 x=sinx+cosx=√2sinx(x+π/2) ∴x∈[-√2,√2] 所x=-1时,y有最小值-1 x=√2时,y有最大值2+√2 y=sinx+cosx+sinx*cosx的值域为 -1

解: y=sinx+cosx+sinxcosx =½sin2x+√2sin(x+π/4) =-½cos(2x+π/2)-√2sin(x+π/4) =-½[1-2sin²(x+π/4)]-√2sin(x+π/4) =sin²(x+π/4)-√2sin(x+π/4)-½ =[sin(x+π/4) -√2/2]²-1 sin(x+π/4)=√2/2时,y取得最小值,y...

先说下方法: 令t=sinx+cosx,因t^2=1+2sinxcosx, 所以y=t+2t^-2,注意t的范围是[-√2,√2]。 下面就变成一个初中就学过的二次函数,可以根据图像求出值域。

y=sinx+cosx =√2(√2/2*sinx+√2/2*cosx) =√2sin(x+π/4) ∵-1≤sin(x+π/4)≤1 ∴-√2≤y≤√2 即值域为:[-√2,√2] 行家正解,不明白可以追问!祝您学习进步 满意请点击下面的【选为满意回答】按钮,O(∩_∩)O谢谢

题目是y=sinx-cosx+sinxcosx是吧。 解: y=sinx-cosx+sinxcosx =√2sin(x-π/4)+(1/2)sin(2x) =√2sin(x-π/4)+(1/2)cos(π/2-2x) =√2sin(x-π/4)+(1/2)cos[2(x-π/4)] =√2sin(x-π/4)+(1/2)[1-2sin²(x-π/4)] =-sin²(x-π/4)+√2sin(x-π/4)+1/2 ...

函数y=sinx+cosx的值域为:[-√2,√2] y=sinx+cosx =√2(√2/2*sinx+√2/2cosx) =√2(sinxcosπ/4+cosxsinπ/4) =√2sin(x+π/4) 值域[-√2,√2]

y=sinxcosx+sinx+cosx =1/2(2sinxcosx+1-1)+sinx+cosx =1/2(sinx+cosx)^2-1/2+(sinx+cosx) =1/2[(sinx+cosx)^2+2(sinx+cosx)+1-2] =1/2(sinx+cosx+1)^2-1 =1/2{(√2)*[sin(x+π/4)]+1}^2-1 所以值域为[-1,√2+1/2]

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