sinx-cosx=1//2,x∈（0，π），则sinx+cosx等于多少？
\u03c0\uff1cx<2\u03c0\uff0csinX+cosX=1/5\uff0ctanX\u7b49\u4e8e\uff1f\u89e3\u6cd5\u4e00\uff1a
sinX+cosX=1/5\uff08\u65b9\u7a0b\u4e00\uff09
(sinX+cosX)^2=1/25
1+2sinX*cosX=1/25
2sinX*cosX=-24/25
\u6240\u4ee5(sinX-cosX)^2=1-2sinX*cosX=49/25
sinX-cosX=\u00b17/5
\u56e0\u4e3a\u03c0\uff1cx<2\u03c0\u4e14sinX+cosX=1/5\uff1e0\uff0c\u6240\u4ee5\u03c03/2\uff1cx<2\u03c0\uff0c\u6240\u4ee5sinX-cosX\uff1c0\uff0c
\u6240\u4ee5sinX-cosX=-7/5\uff08\u65b9\u7a0b\u4e8c\uff09
\u8054\u7acb\u65b9\u7a0b\u4e00\u548c\u65b9\u7a0b\u4e8c\u5f97sinX=-3/5,cosX=4/5\uff0c\u6240\u4ee5tanX=sinX/cosX=-3/4
\u89e3\u6cd5\u4e8c\uff1a
sinX+cosX=1/5\uff08\u65b9\u7a0b\u4e00\uff09
sinX^2+cosX^2=1(\u65b9\u7a0b\u4e8c)
\u4e24\u65b9\u7a0b\u8054\u7acb\u89e3\u5f97
\u89e3\u5f97sinX= 4/5\uff0ccosX=-3/5
\u6216sinX=-3/5 cosX=4/5
\u53c8\u56e0\u4e3a\u03c0\uff1cx<2\u03c0\uff0csinx<0
\u6240\u4ee5sinX=-3/5 cosX=4/5
\u6240\u4ee5tanX=sinX/cosX=-3/4
cosx/\uff08sinx-1\uff09+\uff081+sinx\uff09/cosx=[(cosx)^2+(sinx)^2-1]/(sinx-1)cosx=0/(sinx-1)cosx=0
\uff081+sinx\uff09/cosx=-1/2
sinx-cosx=1/2
(sinx-cosx)²=1/4
sin²x+cos²x-2sinxcosx=1/4
1-2sinxcosx=1/4
∴2sinxcos=3/4
(sinx+cos)²
=sin²x+cos²x+2sinxcosx
=1+2sinxcosx
=1+3/4
=7/4
∴sinx+cos=√7/2
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