sinx+siny=根号2/2 cosx+cosy=根号6/2 则sin(x+y)=

问题描述:

sinx+siny=根号2/2 cosx+cosy=根号6/2 则sin(x+y)=

sinx+siny=√2/2→2sin[(x+y)/2]cos[(x-y)/2]=√2/2;
cosx+coy=√6/2→2cos[(x+y)/2]cos[(x-y)/2]=√6/2.
两式相除,得tan[(x+y)/2]=√3/3.
∴依万能公式,得sin(x+y)=2×(√3/3)/[1+(√3/3)²]=√3/2。

sinx+siny=根号2/2
sin^2x+sin^2y+2sinxsiny=1/2.......(1)
cosx+cosy=根号6/2
cos^2x+cos^2y+2cosxcosy=3/2....(2)
(1)+(2)得
2+2cos(x-y)=2
cos(x-y)=0
x=y
sinx=根号2/4,cosx=根号6/4
sin(x+y)=sin2x=2sinxcosx=2*根号2/4*根号6/4=根号3/4

sinx+siny=√2/2,cosx+cosy=√6/2; 则sin(x+y)=
由sinx+siny=√2/2得sin²x+2sinxsiny+sin²y=1/2.(1)
由cosx+cosy=√6/2得cos²x+2cosxcosy+cos²y=3/2.(2)
(1)+(2)得2+2(cosxcosy+sinxsiny)=2+2cos(x-y)=2,故cos(x-y)=0;
(2)-(1)得cos²x-sin²x+2(cosxcosy-sinxsiny)+cos²y-sin²y=1
即有cos2x+2cos(x+y)+cos2y=2cos(x+y)+cos2x+cos2y
=2cos(x+y)+2cos(x+y)cos(x-y)=2cos(x+y)=1;
故cos(x+y)=1/2,x+y=π/3+2kπ.
∴sin(x+y)=sin(2kπ+π/3)=sin(π/3)=√3/2.