y=sinxcosx的最大值,x的取值集合
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y=7-4sinxcosx+4cos2x-4cos4x=7-2sin2x+4cos2x(1-cos2x)=7-2sin2x+4cos2xsin2x=7-2sin2x+sin22x=(1-sin2x)2
y=sinxcosx-cos^2x=1/2sin2x-1/2(1+cos2x)=1/2(sin2x-cos2x-1)=1/2[√2*sin(2x-派/4)-1]=√2/2*sin(2x-派/4)-1/
y=√3cos2x+sin2x=√[1²+(√3)²]sin(2x+z)=2sin(2x+z)其中tanz=√3/1=√3所以最大=2,最小=-2T=2π/2=π
y=sin²x-sinxcosx+cos²x.(sin²x+cos²x=1)=1-sinxcosx.(sin2x=2sinxcosx)=1-1/2sin2xsi
Y=√3SINXCOSX-COS²X=√3/2sin2x-(1+cos2x)/2=√3/2sin2x-1/2-cos2x/2=sin(2x-∏/6)-1/2ymax=1/2
1.求函数y=sinx-cosx+sinxcosxx∈(0,π)的最大值最小值设t=sinx–cosx所以t²=1–2sinxcosx,则sinxcosx=1-t²/2因为t=si
y=1+(sinx)^2+(cosx)^2+2sinxcosx+sinx+cosx=1+(sinx+cosx)^2+sinx+cosx令t=sinx+cosx则y=t^2+t+1应该可以了吧
提示:设sinx+cosx=t,则sinxcosx=t的平方加1除以2,用二次函数求最值,(注意t的范围)
y=(sinx+cosx)+(sinx+cosx)^2-(sinx^2+cosx^2)=(sinx+cosx)+(sinx+cosx)^2-1设sinx+cosx=t,t=√2sin(x+π/4)∈[
sinx+cosx=√2(sinxcosπ/4+cosxsinπ/4)=√2sin(x+π/4)y=sinx+cosx+2sinxcosx=sinx+cosx+2sinxcosxsin^2x+cos^
y=1-4(sinx)^2-2倍根号3sin(2x)=1-2(1-cos2x)-2倍根号3sin(2x)=4sin(π/6-2x)-1当x∈[0,π/2],π/6-2x属于[-5π/6,π/6]所以最
设t=sinx-cosx=根号2sin(x-Pai/4),故:-根号2
y=sinxcosx-1=1/2+sinxcosx-3/2=(1+2sinxcosx)/2-3/2=(sinx+cosx)^2/2-3/2=sin^2(x+π/4)-3/2所以最大值是1-3/2=-1
原式=9-(sinx)平方*cosx平方,令sinx平方=y(0
令sinx+cosx=(根号2)sin(x+π/4)=t其中t属于【-根号2,根号2】则(sinx+cosx)方=(sinx方)+2sinxcosx+(cosx方)=t方所以sinxcosx=(t方-
设sinx-cosx=t,则sinxcosx=(1-t*t)/20
y=cos²x+2√3sinxcosx-sin²x=cos²x-sin²x+2√3sinxcosx=cos2x+2√3sinxcosx=cos2x+√3sin2
y=sinxcosx-2cos^2x=(1/2)sin2x-(cos2x+1)=(√5/2)sin(2x-a)-1最大值为√5/2-1
y=1+2sinxcosx+sinx+cosx=sin²x+cos²x+2sinxcosx+sinx+cosx=(sinx+cosx)²+sinx+cosx=(sinx+