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求函数y=sinx%Cosx+sinxCosx的值域

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

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]

y=sinx+cosx+sinxcosx 令sinx+cosx=T,(1) 由同角三角函数关系sinxcosx=[(sinx+cosx)^2-(sinx^2+cosx^2)]/2 把(1)式代入,得sinxcosx=(T^2-1)/2 所以y=T+(T^2-1)/2 整理得,y=1/2(T+1)^2-1 而sinx+cosx=√2sin(x+π/4)∈[-√2,√2] 所以y在T[∈-√2,√...

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

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]

令t=sinx+cosx t^2=1+2sinxcosx sinxcosx=(1/2)(t^2-1) y=t+(1/2)(t^2-1) =(1/2)t^2+t-(1/2) t=sinx+cosx=√2sin(x+π/4) -√2≤t≤√2 抛物线y(t) 开口向上,对称轴为: t=-1 函数y(t)在[-√2,√2]上的单调性是先减后增,且减区间短,增区间长; 所以,...

利用(sinx+cosx)^2=1+2sinxcosx把sinxcosx换掉再做,注意范围

y=sinx+cosx+sinxcosx 令sinx-cosx=T,(1) 由同角三角函数关系sinxcosx=[(sinx+cosx)^2-(sinx^2+cosx^2)]/2 把(1)式代入,得sinxcosx=(1-T^2)/2 所以y=T+(1-T^2)/2 整理得,y=-1/2(T-1)^2+3/4 而因为x∈[0,π]故sinx-cosx=√2sin(x-π/4)∈[-1,√2] ...

y=cos²x+sinxcosx =(1+cos2x)/2+sin2x/2 =(1/2)+(1/2)(sin2x+cos2x) =(1/2)+(1/2)(根号2)sin(2x+45) 因为-1=

因为x是三角形的最小内角,所以0

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