解由(1+2sinxcosx+cos^2x-sin^2x)+1=(cos^2x+sin^2x+2sinxcosx+cos^2x-sin^2x)+1=(cos^2x+2sinxcosx+sin^2x+cos^2x-sin^2x)+1=[(cosx+sinx)²+cos^2x-sin^2x]+1=[(cosx+sinx)²+（cosx-sinx）（cosx+sinx）]...倒数第二步[(cosx+sinx)²+（cosx-sinx）（cosx+sinx）]+1是怎么到[(cosx+sinx)（cosx+sinx+cosx-sinx）]+1？[(cosx+sinx)²+（cosx-sinx）（cosx+sinx）]+1是怎么到[(cosx+sinx)（cosx+sinx+cosx-sinx）]+1？在中括弧内取公因式（cosx+sinx）