=½[1+cos(2x-π/6)]+½[1-cos(2x+π/6)]-1
=½[cos(2x-π/6)-cos(2x+π/6)]
=-½ * 2 sin[(2x-π/6+2x+π/6)/2] sin[(2x-π/6-2x-π/6)/2]
=sin2x sin(π/6)
=½sin2x
=½sin(2x+2kπ)
=½sin2(x+kπ) k为整数
最小正周期为π
以上回答你满意么？
=1/2+1/2cos(2x-π/6)+1/2-1/2cos(2x+π/6)-1
=1/2[cos(2x)cos(π/6)+sin(2x)sin(π/6)]-1/2[cos(2x)cos(π/6)-sin(2x)sin(π/6)]
=1/2sin(2x)
T=2π/2=π本回答被提问者采纳
函数y=cos²(x-π\/12)+sin²(x+π\/12)-1的最小正周期
sin²x=(1-cos2x)\/2得来的。根据以上公式，所以 cos²(x-π\/12)=[1+cos(2x-π\/6)]\/2=1\/2+cos(2x-π\/6)\/2 =1\/2+1\/2cos(2x-π\/6)同理，sin²(x+π\/12)=[1-cos(2x+π\/6)]\/2=1\/2-cos(2x+π\/6)\/2 =1\/2-1\/2cos(2x+π\/6)
函数y=cos²(x-派\/12)+sin²(x+派\/12)-1的最小正周期为?
y=cos²(x-π\/12)+sin²(x+π\/12)-1 =1\/2+1\/2cos(2x-π\/6)+1\/2-1\/2cos(2x+π\/6)-1 =1\/2[cos(2x)cos(π\/6)+sin(2x)sin(π\/6)]-1\/2[cos(2x)cos(π\/6)-sin(2x)sin(π\/6)]=1\/2sin(2x)T=2π\/2=π 打字不易，如满意，望采纳。
函数y=cos^(x-π\/12)+sin^2(x+π\/12)-1的最小正周期为( )
y=cos²(x-π\/12)+sin²(x+π\/12)-1 =½[1+cos(2x-π\/6)]+½[1-cos(2x+π\/6)]-1 =½[cos(2x-π\/6)-cos(2x+π\/6)]=-½ * 2 sin[(2x-π\/6+2x+π\/6)\/2] sin[(2x-π\/6-2x-π\/6)\/2]=sin2x sin(π\/6)=½sin2x =...
函数y=cos²(x- π\/12) + sin²(x+ π\/12)的周期,单调递增区间
y=cos²(x- π\/12) + sin²(x+ π\/12)=[1+cos(2x-π\/6)]\/2+[1-cos(2x+π\/6)]\/2 =1+[cos(2x-π\/6)-cos(2x+π\/6)]\/2 =1+[cos(2x)cos(π\/6)+sin(2x)sin(π\/6)-cos(2x)cos(π\/6)+sin(2x)sin(π\/6)]\/2 =1+sin(2x)sin(π\/6)=1+(1\/2)sin...
为什么函数y=cos^2(x-∏\/12)+sin^2(x+∏\/12)-1是 奇函数
y=cos²(x-π\/12) + sin²(x＋π\/12)－1 =(1+cos(2x-π\/6))\/2+(1-cos(2x+π\/6))\/2－1 = cos(2x-π\/6)\/2-cos(2x+π\/6)\/2 =1\/2[cos(2x-π\/6)- cos(2x+π\/6)]=1\/2[cos2x cosπ\/6+sin2x sinπ\/6-( cos2x cosπ\/6-sin2x sinπ\/6)]=1\/2[2...
求解: 函数y=sin(x-π\/12)cos(x+π\/12)+cos(x-π\/12)sin(x+π
求解: 函数y=sin(x-π\/12)cos(x+π\/12)+cos(x-π\/12)sin(x+π 求解:函数y=sin(x-π\/12)cos(x+π\/12)+cos(x-π\/12)sin(x+π\/12)的最小正周期?请给出详细的解题过程,一定采纳!谢谢。... 求解: 函数y=sin(x-π\/12)cos(x+π\/12)+cos(x-π\/12)sin(x+π\/12)的最小正周期? 请...
求函数f(x)=cos^2(x-pai\/12)+sin^2(x+pai\/12)-1的周期
f(x)=cos^2(x-pai\/12)+sin^2(x+pai\/12)-1 =[1+cos(2x-π\/6)]\/2+[1-cos(2x+π\/6)]\/2-1 =[cos(2x-π\/6)-cos(2x+π\/6)]\/2 =(sin2x)\/2 所以最小正周期为π 这里主要公式是二倍角公式和两角和差公式，都是常规公式，书上有 参考资料：团队：我最爱数学！
化简cos²(x-π\/12)+sin²(x+π\/12)-1等于
化简cos²(x-π\/12)+sin²(x+π\/12)-1等于 我来答 1个回答 #热议# 生活中有哪些成瘾食物?匿名用户 2014-12-05 展开全部 已赞过 已踩过< 你对这个回答的评价是? 评论 收起 为你推荐: 特别推荐 关键时刻可以救命的18种婴幼儿急救常识! 小米,防癌、降糖,还助眠。 凌烟阁二十四功臣结局如何?
函数fx=sin(x-π\/12)xsin(x+5π\/12),求最小正周期
回答：π\/12)后面是x,还是乘号?
然后得 Y=cos²(x-π\/12)-cos²(x+π\/12) 用平方差公式可得 Y=4si...
Y=cos²（x-π\/12）-cos²（x+π\/12）=[cos(x-π\/12)+cos(x+π\/12)][cos(x-π\/12)-cos(x+π\/12)]=[cosxcosπ\/12+sinxsinπ\/12+cosxcosπ\/12-sinxsinπ\/12][cosxcosπ\/12+sinxsinπ\/12-cosxcosπ\/12+sinxsinπ\/12]=[2cosxcosπ\/12][2]sinxsinπ\/12 =4s...
