Question

已知sinx+siny=√2,cosx+cosy=2√3/3，求tanx*tany

Answer 1

解答：
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

追问

厉害，谢谢

Answer 2

(sinx siny)^2==sinx^2 siny^2 2sinxsiny=2 (1)
(cosx cosy)^2==cosx^2 cosy^2 2cosxcosy=4/3 (2)
(1) (2)
sinx^2 siny^2 2sinxsiny cosx^2 cosy^2 2cosxcosy
=2sinxsiny 2cosxcosy=2 4/3-2=4/3
=sinxsiny cosxcosy=2/3 (3)
(2)-(1)
2cosxcosy-2sinxsiny=-2/3
cosxcosy-sinxsiny=-1/3 (4)
(3) (4)
cosxcosy=1/6 (5)
(3)-(4)
sinxsiny=1/2 (6)
(6)/(5)
tanxtany=1/2*6=3

麻烦点

sinx siny=√2 (1)
cosx cosy=2√3/3 (2)
(1)^2 (2)^2
2 2sinxsiny 2cosxcosy=2 4/3
cosxcosy sinxsiny=2/3 (3)
cos (x-y)=2/3 (4)
(2)^2-(1)^2
cos2x cos2y 2(cosxcosy-sinxsiny)=-2/3
2cos(x y)cos(x-y) 2cos(x y)=-2/3
把(4)代入,得到cos(x y)=-1/5
所以cosxcosy-sinxsiny=-1/5 (5)
由(3)和(5)解得cosxcosy=7/30,sinxsiny=13/30
所以tanxtany=sinxsiny/cosxcosy=13/7

祝您。 学习愉快