函数fx=3sin(2x 兀 6)的部分图像如图所示
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f(x)=√3sin²x+sinxcosx=√3[(1-cos2x)/2]+1/2sin2x=1/2sin2x-√3/2cos2x+√3/2=sin(2x-π/3)+√3/2∵x∈[π/2,
设函数fx=sin(φ-2x)(0
再答:亲,满意请采纳再问:谢谢亲再问:再问:可以帮我解决第二问吗再答:恩,我看下再问:麻烦了。O(∩_∩)O再答:再答:再问:谢谢亲再答:😊😊再问:美丽的姑凉谢谢你再答
fx=2sin(2x+pai/6)振幅A=2最小正周期T=2pai/2=paix∈【0,pai/]2xE[0,2pai]2x+pai/6E[pai/6,2pai+pai/6]很明显,设u=2x+pai
周期等于2派.g(x)=2sinx;基函数再问:有过程吗??再答:这可以看出来,还要过程吗,,,,周期等于2派/x前的数1===2派;;g(x)=2sint(x+pi/3+p1/3)=2sinx;si
1)a=π/3,2x+π/6∈[-π/6,5π/6],从而f(x)∈[-1/2,1];再答:2)因为若f(x)的值域是[-1/2,1],所以2x+π/6∈[-π/6,7π/6],从而x∈[-π/6,π
答:f(x)=2sin(x-π/3)cosx+sinxcosx+√3(sinx)^2=sin(x-π/3+x)+sin(x-π/3-x)+sinxcosx+(√3/2)(1-cos2x)=sin(2x
f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=cos(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)]=cos(2x-π/3)+2sin(x-π/
(1)f(x)=[cos(x-π/6)]^2-(sinx)^2f(π/12)=(cos(π/12))^2-(sin(π/12))^2=cos(π/6)=√3/2(2)f(x)=[cos(x-π/6)]
f(x)=2sin(x-π/6)cosx+2cos²x=(2sinxcosπ/6-2cosxsinπ/6)cosx+2cos²x=√3sinxcosx-cos²x+2co
f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=(1/2)cos2x+(√3/2)sin2x+(cos(π/2)-cos2x)=-(1/2)cos2x+(√3/2)sin
T=2π/2=π[-1,1]最大值为1,最小值为-1
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你的分析前一半是对的,一直到“那么2x的单调增区间是[-4分之π,4分之π]”.2x的单调递增区间是[-π/2,π/2],x的才是[-π/4,π/4].所以函数在x=-π/3处取得最小值为-2分之根号
(1)fx=sin(2x+φ)经过点(π/12,1)sin(π/6+φ)=1∴π/6+φ=π/2+2kπ,k∈Z∴φ=π/3+2kπ,k∈Z∵0
解答;f(x)=sin(2x+3分之π)∴sin(2x+π/3)=-3/5∵x∈(0,π/2)∴2x+π/3∈(π/3,4π/3)∵sin(2x+π/3)
解1当2kπ-π/2≤2x+π/3≤2kπ+π/2,k属于Z时,y是增函数即2kπ-5π/6≤2x≤2kπ+π/6,k属于Z时,y是增函数即kπ-5π/12≤x≤kπ+π/12,k属于Z时,y是增函数
f(x)=sin(2x+π/6)-cos2x+1所以为2π/2=πf(x)=根号3/2sin2x-(cos2x)/2+1=sin(2x-π/6)+1所以最大值为2,x=π/2+2kπ-π/6=π/3+
f(x)=√3sin2x-2sin²x=√3sin2x-(1-cos2x)=2sin(2x+π/6)-1∴当sin(2x+π/6)=1时f(x)max=2*1-1=1