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z sinx siny sin x y

sinx+siny+sinz-sin(x+y+z)=4sin[(x+y)/2]sin[(x+z)/2]sin[(y+z)/2] sinx+siny+sinz-sin(x+y+z) =2sin[(x+y)/2]cos[(x-y)/2]+sinz-sin(x+y)cosz-sinzcos(x+y) =2sin[(x+y)/2]cos[(x-y)/2]+sinz[1-cos(x+y)]-sin(x+y)cosz =2sin[(x+y)/2]cos[(x-y...

z对x的偏导=cosx+cos(x+y)=0时,cosx=-cos(x+y)=cos(pi-x-y),所以x=pi-x-y.同理z对y的偏导=0时,有y=pi-x-y.所以x=y=pi/3。此时z=3根号3/2. 当x=0时,z=2siny 最大2,最小0。当y=0时一样。当x=pi/2时,z=1+根号2sin(y+pi/4) 最大值1+根号2,最小...

这已经是最简了,没法化简

解法一:根据已知:sin(y+x)=sinycosx+cosysinx=siny+sinx化简得:siny(1-cosx)+sinx(1-cosy)=0即2siny2cosy2×2sin 2x2+2sinx2cosx2×2sin 2y2=0即4sinx2siny2×(cosy2×sinx2+cosx2×siny2)=0即4sinx2siny2sinx+y2=0上式成立,所以必有x2...

和差化积 sinX+sinY=2sin[(X+Y)/2]cos[(X-Y)/2]; sinX-sinY=2cos[(X+Y)/2]sin[(X-Y)/2] cosX+cosY=2cos[(X+Y)/2]cos[(X-Y)/2] cosX-cosY=-2sin[X+Y)/2]sin[(X-Y)/2] 积化和差sinXcosY=1/2[sin(X+Y)+sin(X-Y)] cosXsinY=1/2[sin(X+Y)-sin(X-Y)]

积化和差:sinxsiny=1/2(cos(x-y)-cos(x+y))

题中x应不等于y,否则不等式就不成立了。 由于不等式涉及到sinx-siny和x-y,很容易想到利用和差化积公式: |sinx-siny|=|2cos[(x+y)/2]sin[(x-y)/2]|≤2|sin[(x-y)/2]|

Plot3D[Sin[x] Sin[y], {x, -0.3, 0.3}, {y, -0.3, 0.3}] (*呵呵,范围选大一点图形大不一样:*) Plot3D[Sin[x] Sin[y], {x, -0.3, 0.3}, {y, -5, 5}]

解: sinx+siny=1/3,cosx-cosy=1/5 两式分别平方得 sin²x+sin²y+2sinxsiny=1/9 cos²x+cos²y-2cosxcosy=1/25 相加得 1+1+2sinxsiny-2cosxcosy=1/9+1/25=34/225 2cosxcosy-2sinxsiny=2-34/225 cosxcosy-sinxsiny=1-17/225=208...

这是个公式 cos(x+y)=cosxcosy-sinxsiny cos(x-y)=cosxcosy+sinxsiny 现在 sinxsin(x+y)+cosxcos(x+y)=cos[x-(x+y)]=cosy

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