Question

高一数学、三角恒等变形！！！！急求！！

若cos(x+y)=4/5,cos(x-y)=-4/5,且3π/2<x+y<2π,π/2<x-y<π，求cos2x,cos2y
要过程！！！

Answer 1

因为3π/2<x+y<2π,π/2<x-y<π
所以sin(x+y)=-3/5,sin(x-y)=3/5
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

Answer 2

∵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