求解:sinx=cosx-1 求证:sin2x/(1+sinx+cosx)=sinx+co...

\u6c42\u8bc1sinx-cosx+1/sinx+cosx-1=1+sinx/cosx

(sinx-cosx+1)/(sinx+cosx-1)
=[2sin(x/2)cos(x/2)+2(sin(x/2))^2] / [ 2sin(x/2)cos(x/2)-2(sin(x/2))^2
=[sin(x/2) +cos(x/2)] /[cos(x/2) -sin(x/2)]

(1+sinx)/cosx=(sin(x/2)+cos(x/2))^2/[cos(x/2))^2-sin(x/2)^2]
=[sin(x/2)+cos(x/2)]/[cos(x/2)-sin(x/2)]
\u5de6=\u53f3

\u8bc1\u660e\uff1a\u56e0\u4e3a sin2x=2sinxcosx
\u6240\u4ee5 sin2x/(1+sinx+cosx)
=2sinxcosx / (1+sinx+cosx)
=\uff082sinxcosx+1-1\uff09/ (1+sinx+cosx)
=\uff08sin^2 x +cos^2 x +2sinxcosx+1\uff09/ (1+sinx+cosx)
=(1+sinx+cosx) (sinx+cosx-1)\uff09/ (1+sinx+cosx)
=sinx+cosx-1

首先化为:cosx-sinx=1 ,然后转为:根号下2*(sin45度cosx-cos45度sinx)=1 ,最后化解一下,即可求出。答案为:0度或360度。

答案既然能以角度表示,,,0,270,360

cosx-sinx=2^0.5(cos(x+45))=1

270+45=360-45

应该用不到根号的
cosx-sinx=1………………(1)
cosx^2+sinx^2=1…………(2)
联立2个方程
将(1)平方
cosx^2+sinx^2-2sinx*cosx=1……(3)
(2)代入(3)
得2sinx*cosx=0
sin2x=0 (0<x<360)
得2x=0或者2x=360,或者2x=360,2x=720;
所以x=0或者x=180,或者x=360

sinx=cosx-1
移项sinx-cosx=-1
平方得sinx平方+cosx平方-2sinxcosx=1
化简sin2x=0即为所求
解得x=0 ,180 ,360

sinx=cosx-1
sinx^2=cosx^2-2cosx+1
sinx^2+cosx^2=1
cosx^2-2cosx+1+cosx^2=1
2cosx^2-2cosx=0
cosx(cosx-1)=0
cosx=0 x=90,270,
cosx=1 x=0,180,360

0度或360度

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