Question

sinx +siny=1/3 cosx+cosy=1/4 求tan(x+y)和sin(x+y)

Answer 1

sinx+siny=sin((x+y)/2+(x-y)/2)+sin((x+y)/2-(x-y)/2)=1/3
cosx+cosy=cos((x+y)/2+(x-y)/2)+cos((x+y)/2-(x-y)/2)=1/4
根据三角函数公式化简得
2cos[(x-y)/2]sin[(x+y)/2]=1/3
2cos[(x+y)/2]cos[(x-y)/2]=1/4
tan[(x+y)/2]=4/3
cos[(x+y)/2]=+-3/5
sin[(x+y)/2]=+-4/5(由于tan>0,sin,cos 必须同号）

sin(x+y)=2(12)/25 =24/25 （cos[(x+y)/2]，sin[(x+y)/2]同号，乘积为正）
cos(x+y)=2(cos[(x+y)/2])^2-1=-7/25
tan(x+y)=sin(x+y)/cos(x+y)=-24/7

Answer 2

sinx+siny=1/3......(1)
cosx+cosy=1/4......(2)
(1)^2+(2)^2:1+1+2(cosxcosy+sinxsiny)=1/9+1/16=25/144
--->cos(x-y)=-263/288.
(1):2sin[(x+y)/2]cos[(x-y)/2]=1/3......(3)
(2):2cos[(x+y)/2]cos[(x-y)/2]=1/4......(4)
(3)/(4):tan[(x+y)/2]=4/3
利用倍角公式tan2α＝2tanα/（1－tanαtanα）得
tan(x+y)=tan(2((x+y)/2))=(2*3/4)÷(1-(3/4)^2)=24/7
cos(x+y)=cos[2(x+y)/2]
={1-[tan(x/2+y/2)]^2}/{1+[tan(x/2+y/2)]^2}
=[1-(4/3)^2]/[1+(4/3)^2]
=(9-16)/(9+16)
=-7/25
sin(x+y)=cos(x+y)*tan(x+y)
=(-7/25)*(24/7)=-24/25