(1-tanx)[1+√2sin(2x+π/4)]
问题描述：
(1-tanx)[1+√2sin(2x+π/4)]
化简：(1-tanx)[1+√2sin(2x+π/4)]
答
(1-tanx)[1+√2sin(2x+π/4)]
=(1-tanx)[1+sin(2x)+cos(2x)]
=(1-tanx)[1+2sinxcosx+2cos²x-1]
=2cosx(1-tanx)(sinx+cosx)
=2(cos²x-sin²x)
=2cos2x