4个回答
展开全部
分子=(1/2)[4cos^4x-4cos²x+]=(1/2)(2cos²x-1)²=(1/2)cos²2x
分母=2sin²(x+π/4)×[sin(π/4-x)/cos(π/4-x)]
. =2sin²(x+π/4)×[cos(π/4+x)/sin(π/4+x)]
. =2sin(x+π/4)cos(π/4+x)
. =sin(π/2+2x)
. =cos2x
答案是:(1/2)cos2x
分母=2sin²(x+π/4)×[sin(π/4-x)/cos(π/4-x)]
. =2sin²(x+π/4)×[cos(π/4+x)/sin(π/4+x)]
. =2sin(x+π/4)cos(π/4+x)
. =sin(π/2+2x)
. =cos2x
答案是:(1/2)cos2x
本回答被网友采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
=(cos^4x-cos^2x+1/4)/[sin^2(x+π/4)cot(x+π/4)]
=(cos^2x-1/2)^2/[sin(x+π/4)cos(x+π/4)]
=4(2cos^2x-1)^2/[1/2sin(2x+π/2)]
=8(cos2x)^2/cos(2x)
=8cos2x
=(cos^2x-1/2)^2/[sin(x+π/4)cos(x+π/4)]
=4(2cos^2x-1)^2/[1/2sin(2x+π/2)]
=8(cos2x)^2/cos(2x)
=8cos2x
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
2cos⁴x-2cos²+½=2(cos⁴x-cos²+¼)=2(cos²x-½)²
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询