十八题的第二问是怎么做 ?
3个回答
展开全部
f(x)=2cosx(sinxcosπ/6+cosxsinπ/6)+1
=2cosxsinx·√3/2+2cos²x·1/2+1
=√3/2 sin2x+cos²x+1
=sin2xcosπ/6+1/2 (1+cos2x) +1
=sin2xcosπ/6 +cos2xsinπ/6+3/2
=sin(2x+π/6)+3/2
x∈【-π/6,π/3】
2x∈[-π/3,2π/3]
2x+π/6∈[-π/6,5π/6]
最小值=sin(-π/6)+3/2
=-1/2+3/2
=1
最大值=1+3/2=5/2
=2cosxsinx·√3/2+2cos²x·1/2+1
=√3/2 sin2x+cos²x+1
=sin2xcosπ/6+1/2 (1+cos2x) +1
=sin2xcosπ/6 +cos2xsinπ/6+3/2
=sin(2x+π/6)+3/2
x∈【-π/6,π/3】
2x∈[-π/3,2π/3]
2x+π/6∈[-π/6,5π/6]
最小值=sin(-π/6)+3/2
=-1/2+3/2
=1
最大值=1+3/2=5/2
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