求解这道题目第一问，谢谢？
1个回答
展开全部
f(x)=sinx + sinxcos(π/6) + cosxsin(π/6) - cos[π+(x + π/3)]
=sinx + (√3/2)sinx + (1/2)cosx - [-cos(x + π/3)]
=sinx + (√3/2)sinx + (1/2)cosx + cos(x + π/3)
=sinx + (√3/2)sinx + (1/2)cosx + cosxcos(π/3) - sinxsin(π/3)
=sinx + (√3/2)sinx + (1/2)cosx + (1/2)cosx - (√3/2)sinx
=sinx + cosx
=√2sin(x + π/4)
∴T=2π/1=2π
单调递减区间是：
2kπ + π/2≤x + π/4≤2kπ + 3π/2
∴2kπ + π/4≤x≤2kπ + 5π/4，(k∈Z)
=sinx + (√3/2)sinx + (1/2)cosx - [-cos(x + π/3)]
=sinx + (√3/2)sinx + (1/2)cosx + cos(x + π/3)
=sinx + (√3/2)sinx + (1/2)cosx + cosxcos(π/3) - sinxsin(π/3)
=sinx + (√3/2)sinx + (1/2)cosx + (1/2)cosx - (√3/2)sinx
=sinx + cosx
=√2sin(x + π/4)
∴T=2π/1=2π
单调递减区间是：
2kπ + π/2≤x + π/4≤2kπ + 3π/2
∴2kπ + π/4≤x≤2kπ + 5π/4，(k∈Z)
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
