求证(1-2cos^2(2a))/(2tan(2a-π/4)sin^2(π/4+2a))=1
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求证(1-2cos^2(2a))/(2tan(2a-π/4)sin^2(π/4+2a))=1
左边=(1-2cos²2a)/[2tan(2a-π/4)cos²(π/2-(π/4+2a))]
=(1-2cos²2a)/[2tan(2a-π/4)cos²(π/4-2a)]
=(1-2cos²2a)/[2sin(2a-π/4)/cos(2a-π/4)*cos²(2a-π/4)]
=-(2cos²2a-1)/[2sin(2a-π/4)cos(2a-π/4)]
=-cos4a/sin(4a-π/2)
=-cos4a/(-cos4a)
=1
=右边
命题得证
=(1-2cos²2a)/[2tan(2a-π/4)cos²(π/4-2a)]
=(1-2cos²2a)/[2sin(2a-π/4)/cos(2a-π/4)*cos²(2a-π/4)]
=-(2cos²2a-1)/[2sin(2a-π/4)cos(2a-π/4)]
=-cos4a/sin(4a-π/2)
=-cos4a/(-cos4a)
=1
=右边
命题得证
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