Question

已知x+y=3-cos4x,x-y=4sin2x,求证x的2分之1次幂+y的2分之1次幂

Answer 1

x+y=3-cos4x=3-(1-2sin²2x)=2sin²2x+2

x-y=4sin2x

x=sin²2x+2sin2x+1=(sin2x+1)²

y=sin²2x-2sin2x+1=(sin2x-1)²

所以√x+√y

=|sin2x+1|+|sin2x-1|

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