sinx=1/3,sin(x+y)=1,求sin(2y+x)=?

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sinx=1/3,sin(x+y)=1,求sin(2y+x)=?

sinx=1/3,sin(x+y)=1,求sin(2y+x)=?
sinx=1/3,sin(x+y)=1,求sin(2y+x)=?

sinx=1/3,sin(x+y)=1,求sin(2y+x)=?
sin(x+y)=1,cos(x+y)=0
sin(2x+2y)=sin[2*(x+y)]=2sin(x+y)cos(x+y)=2*1*0=0
cos(2x+2y)=[ cos(x+y)]^2-[sin(x+y)]^2=0-1=-1
sin(2y+x)=sin[(2x+2y)-x]=sin(2x+2y)cosx-cos(2x+2y)sinx=0*cosx-(-1)*1/3=1/3

sin(x+y)=1 ==>x+y=2kπ+π/2 (k∈Z)
sin(2y+x)=sin[2(x+y)-x]=sin(4kπ+π-x)
=sin(π-x)=sinx=1/3