已知sinx+siny=√2,cosx+cosy=2√3/3,求tanx*tany
2个回答
这种题目就是平方相加或相减
sinx+siny=√2,①
cosx+cosy=2√3/3 ②
①平方
sin²x+sin²y+2sinxsiny=2 ③
②平方
cos²x+cos²y+2cosxcosy=4/3 ④
③+④
2+2(cosxcosy+sinxsiny)=10/3
∴ cosxcosy+sinxsiny=2/3 ⑤
即 cos(x-y)=2/3
④-③
cos2x+cos2y+2(cosxcosy-sinxsiny)=-2/3
2cos(x+y)cos(x-y)+2cos(x+y)=-2/3
∴ 2*cos(x+y)*(2/3)+2cos(x+y)=-2/3
∴ cos(x+y)=-1/5
即 cosxcosy-sinxsiny=-1/5 ⑥
∴ ⑤*3+⑥*10
∴ 3(cosxcosy+sinxsiny)+10(cosxcosy-sinxsiny)=0
∴ 13cosxcosy=7sinxsiny
∴ tanx*tany=(sinxsiny)/(cosxcosy)=13/7
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