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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...

x下面

sinx+siny+sinz-sin(x+y+z) =2sin(x+y)/2cos(x-y)/2-2cos(x+y+2z)/2sin(x+y)/2 =2sin(x+y)/2[cos(x-y)/2-cos(x+y+2z)/2] =2sin(x+y)/2*2sin(x+2z)/2sin(y+z) =4sin(x+y)/2*sin(x+2z)/2*sin(y+z) (希望能帮到你)

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...

由题意得cos(x-y)= ,sin2x+sin2y=sin[(x+y)+(x-y)]+sin[(x+y)-(x-y)]=2sin(x+y)cos(x-y)= ?sin(x+y)=

由于sinx是基本初等函数,显然Z的不连续点,就是 使得分母sinx=0的点。 即x=kπ (k∈ℤ)

这个题有点难度,下面的证明中用到了均值不等式,不知道你学过没有 (图片里α应该表示sin(x+y)/2的平方,这里打错了)

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)]

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]|

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