sinx+cosy=1/3 cosx+siny=1/6 则sin(x+y)=?
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sinx+cosy=1/3 cosx+siny=1/6 则sin(x+y)=?
RT
RT
sinx+cosy=1/3 ,cosx+siny=1/6
(sinx+cosy)^2 = 1/9,(cosx+siny)^2 = 1/36
sin^2x +cos^2y +2sinxcosy +cos^2x+sin^2y +2cosxsiny = 1/9+1/36 = 5/36
sin^2x +cos^2x+sin^2y +cos^2y +2sinxcosy +2cosxsiny = 5/36
1 + 1 +2sinxcosy +2cosxsiny = 5/36
1 +sinxcosy +cosxsiny = 5/72
sinxcosy +cosxsiny = -67/72
sin(x+y)= -67/72
(sinx+cosy)^2 = 1/9,(cosx+siny)^2 = 1/36
sin^2x +cos^2y +2sinxcosy +cos^2x+sin^2y +2cosxsiny = 1/9+1/36 = 5/36
sin^2x +cos^2x+sin^2y +cos^2y +2sinxcosy +2cosxsiny = 5/36
1 + 1 +2sinxcosy +2cosxsiny = 5/36
1 +sinxcosy +cosxsiny = 5/72
sinxcosy +cosxsiny = -67/72
sin(x+y)= -67/72
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