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