展开全部
原式=[sin²(x+π/4)+sin²(π/4-x)][sin²(x+π/4)-sin²(π/4-x)]
={sin²(x+π/4)+cos²[π/2-(π/4-x)]}{sin²(x+π/4)-cos²[π/2-(π/4-x)]}
=[sin²(x+π/4)+cos²(π/4+x)][sin²(x+π/4)-cos²(π/4+x)]
=1*{-cos[2(x+π/4)]}
=-cos(2x+π/2)
=sin2x
={sin²(x+π/4)+cos²[π/2-(π/4-x)]}{sin²(x+π/4)-cos²[π/2-(π/4-x)]}
=[sin²(x+π/4)+cos²(π/4+x)][sin²(x+π/4)-cos²(π/4+x)]
=1*{-cos[2(x+π/4)]}
=-cos(2x+π/2)
=sin2x
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
光点科技
2023-08-15 广告
通常情况下，我们会按照结构模型把系统产生的数据分为三种类型：结构化数据、半结构化数据和非结构化数据。结构化数据，即行数据，是存储在数据库里，可以用二维表结构来逻辑表达实现的数据。最常见的就是数字数据和文本数据，它们可以某种标准格式存在于文件...
点击进入详情页
本回答由光点科技提供
展开全部
sin^4(x+π/4)-sin^4(x-π/4)
= [sin^2(x+π/4)+sin^2(x-π/4)][sin^2(x+π/4)-sin^2(x-π/4)]
= [cos^2(x-π/4)+sin^2(x-π/4)][cos^2(x-π/4)-sin^2(x-π/4)]
= 1×cos2(x-π/4)
= cos(2x-π/2)
= cos(π/2-2x)
= sin2x
= [sin^2(x+π/4)+sin^2(x-π/4)][sin^2(x+π/4)-sin^2(x-π/4)]
= [cos^2(x-π/4)+sin^2(x-π/4)][cos^2(x-π/4)-sin^2(x-π/4)]
= 1×cos2(x-π/4)
= cos(2x-π/2)
= cos(π/2-2x)
= sin2x
本回答被提问者采纳
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
