设cos^2(x-y)-cos^2(x+y)=1/2,(1+cos2x)*(1+cos2y)=1/3,则tanxtany

52悠悠我心2022-10-04 11:39:542条回答

设cos^2(x-y)-cos^2(x+y)=1/2,(1+cos2x)*(1+cos2y)=1/3,则tanxtany=?

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xln127 共回答了20个问题 | 采纳率95%: ∵cos^2(x-y)-cos^2(x+y)=1/2
==>[cos(x-y)+cos(x+y)][cos(x-y)-cos(x+y)]=1/2
==>(2cosxcosy)(2sinxsiny)=1/2 (应用和差角公式)
==>cosxcosysinxsiny=1/8.(1)
(1+cos2x)*(1+cos2y)=1/3
==>(2cos²x)(2cos²y)=1/3 (应用倍角公式)
==>cos²xcos²y=1/12.(2)
∴由(1)式和()式,得tanxtany=(sinx/cosx)(siny/cosy)
=(sinxsiny)/(cosxcosy)
=(cosxcosysinxsiny)/(cos²xcos²y) (分子分母同乘cosxcosy)
=(1/8)/(1/12)
=3/2; 1年前

鱼香腊肉丝共回答了16个问题 | 采纳率: =(sin2x+cos2x)/2+3/2 =(√2/2)*sin(2x+π/4)+3/2 所以T=2π2、（1）(a+c)^2=15cos^2 x+1+15sin^2 y+1+8(cosxcosy-sinx; 1年前

sinx+siny=根号3 cosx+cosy=2根号3/3 求tanxtany sinx+siny=根号3 cosx+cosy=2根号3/3 求tanxtany 错了，第一个我等于根号2 包包1951年前2: hushanzhenmian 共回答了18个问题 | 采纳率88.9%

sinx+siny=根号2 (1)
cosx+cosy=2根号3/3 (2)
(1)平方+(2)平方
2+2sinxsiny+2cosxcosy=2+4/3
cosxcosy+sinxsiny=2/3 (3)
cos (x-y)=2/3 (4)
(2)平方-(1)平方
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

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