已知cosx+cosy=1/2,sinx-siny=1/3,则cos(x+y)=
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已知cosx+cosy=1/2,sinx-siny=1/3,则cos(x+y)=
解
cosx+cosy=1/2
两边平方得:
cos²x+cos²y+2cosxcosy=1/4 ①
sinx-siny=1/3
两边平方得:
sin²x-2sinxsiny+sin²y=1/9 ②
①+②得:1+1+2cosxcosy-2sinxsiny=1/4+1/9=13/36
∴2(cosxcosy-sinxsiny)=13/36-2=-59/36
∴cosxcosy-sinxsiny=-59/72
∴cos(x+y)
=cosxcosy-sinxsiny
=-59/72
cosx+cosy=1/2
两边平方得:
cos²x+cos²y+2cosxcosy=1/4 ①
sinx-siny=1/3
两边平方得:
sin²x-2sinxsiny+sin²y=1/9 ②
①+②得:1+1+2cosxcosy-2sinxsiny=1/4+1/9=13/36
∴2(cosxcosy-sinxsiny)=13/36-2=-59/36
∴cosxcosy-sinxsiny=-59/72
∴cos(x+y)
=cosxcosy-sinxsiny
=-59/72
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