Question

已知sinθ+cosθ=2sinα,sinθ·cosθ=sin²β,求证：2cos2α=cos2β.

Answer 1

∵sinθ+cosθ=2sinα

∴(sinθ+cosθ)²=(2sinα)²

==>sin²θ+cos²θ+2sinθ*cosθ=4sin²α

==>1+2sinθ*cosθ=4sin²α

∵sinθ·cosθ=sin²β

∴1+2sin²β=4sin²α

==>1+1-cos(2β)=2(1-cos(2α))

==>2-cos(2β)=2-2cos(2α)

==>-cos(2β)=-2cos(2α)

==>2cos(2α)=cos(2β)

故原命题成立,证毕.