sinxcosy=1/3 求sinycosx取值范围
\u5df2\u77e5cosx+cosy=1\uff0c\u5219sinx-siny\u7684\u53d6\u503c\u8303\u56f4\u662f______\u2235\uff08sinx-siny\uff092+\uff08cosx+cosy\uff092=\uff08sin2x+cos2x\uff09+\uff08sin2y+cos2y\uff09+2\uff08cosxcosy-sinxsiny\uff09=2+2cos\uff08x+y\uff09\uff0c\u53c8\u2235cosx+cosy=1\uff0c\u2234\uff08sinx-siny\uff092=1+2cos\uff08x+y\uff09\u22643\uff0c\u2234-3\u2264sinx-siny\u22643\uff0c\u5219sinx-siny\u7684\u53d6\u503c\u8303\u56f4\u662f[-3\uff0c3]\uff0e\u6545\u7b54\u6848\u4e3a\uff1a[-3\uff0c3]
\uff08sinx+siny\uff09^2+(cosx+cosy)^2
=2+2(sinxsiny+cosxcosy)
=2+2cos(x-y)
2cos(x-y)\u503c\u57df[-2\uff0c2]
\uff08sinx+siny\uff09^2=1/9
(cosx+cosy)^2\u5c5e\u4e8e[-1/9,35/9]\u4e14(cosx+cosy)^2>=0\u5f97\u89e3
我认为这道题的范围很容易求错,现在我来求一求:
令t=sinycosx,平方得到t^2=sin^2ycos^2x=(1-cos^2y)(1-sin^2x)=1-sin^2x-cos^2y+(sinxcosy)^2=10/9-(sin^2x+1/9sin^2x)≤10/9-2/3=4/9[当且仅当sin^2=1/3取等号)所以解得 -2/3≤t≤2/3
sinxcosy+sinycosx=sin(x+y)，而sin(x+y)大于等于－1，小于等于1
所以sinycosx大于等于－4/3,小于等于2/3
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