若cos(x+y)=4/5,cos(x-y)=-4/5,且3π/2
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若cos(x+y)=4/5,cos(x-y)=-4/5,且3π/2若cos(x+y)=4/5,cos(x-y)=-4/5,且3π/2若cos(x+y)=4/5,cos(x-y)=-4/5,且3π/2因
若cos(x+y)=4/5,cos(x-y)=-4/5,且3π/2
若cos(x+y)=4/5,cos(x-y)=-4/5,且3π/2
若cos(x+y)=4/5,cos(x-y)=-4/5,且3π/2
因为3π/2
所以cos2x=cos(x+y+x-y)=cos(x+y)cos(x-y)-sin(x+y)sin(x-y)=-16/25+9/25=-7/25
cos2y=cos[(x+y)-(x-y)]=cos(x+y)cos(x-y)+sin(x+y)sin(x-y)=-16/25-9/25=-1
∵cos(x+y)=4/5
∴sin(x+y)=-3/5 (3/2π~2π)
∵cos(x-y)=-4/5
∴sin(x-y)=3/5
cos2x=cos[(x+y)+(x-y)]=cos(x+y)cos(x-y)-sin(x+y)sin(x-y)=-7/25
cos2y=cos[(x+y)-(x-y)]=cos(x+y)cos(x-y)+sin(x+y)sin(x-y)=-1
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