若sinx+cosx=1/5,求cosx,sinx,sin2x,cos2x
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若sinx+cosx=1/5,求cosx,sinx,sin2x,cos2x
![若sinx+cosx=1/5,求cosx,sinx,sin2x,cos2x](/uploads/image/z/8003077-61-7.jpg?t=%E8%8B%A5sinx%2Bcosx%3D1%2F5%2C%E6%B1%82cosx%2Csinx%2Csin2x%2Ccos2x)
与(sinx)^2+(cosx)^2=1 联立,解得
(1)sinx=4/5 cosx=-3/5 sin2x=2sinxcosx=-24/25 cos2x=(cosx)^2-(sinx)^2=-7/25
(2)sinx=-3/5 cosx=4/5 sin2x=2sinxcosx=-24/25 cos2x=(cosx)^2-(sinx)^2=7/25
(1)sinx=4/5 cosx=-3/5 sin2x=2sinxcosx=-24/25 cos2x=(cosx)^2-(sinx)^2=-7/25
(2)sinx=-3/5 cosx=4/5 sin2x=2sinxcosx=-24/25 cos2x=(cosx)^2-(sinx)^2=7/25
若sinx+cosx=1/5,求cosx,sinx,sin2x,cos2x
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解方程 sin2x/sinx=cos2x/cosx