求函数f(x)=1/2sin^2x+cos^2x+根号3/4sin2x的最大值
1个回答
展开全部
f(x)=sin^2x/2+cos^2x+√3/4sin2x
=1/2+cos^2x/2+√3/4sin2x
=1/2+(cos2x+1)/4+√3/4sin2x
=1/2+1/4*cos2x+1/4+√3/4sin2x
=1/4*cos2x+√3/4sin2x+3/4
=1/2*(1/2*cos2x+√3/2sin2x)+3/4
=1/2*(sinπ/6*cos2x+cosπ/6sin2x)+3/4
=1/2*sin(π/6+2x)+3/4
-1<=sin(π/6+2x)<=1
-1/2<=1/2sin(π/6+2x)<=1/2
1/4<=1/2sin(π/6+2x)+3/4<=5/4
所以函数f(x)=sin^2x/2+cos^2x+√3/4sin2x的最大值为：5/4
=1/2+cos^2x/2+√3/4sin2x
=1/2+(cos2x+1)/4+√3/4sin2x
=1/2+1/4*cos2x+1/4+√3/4sin2x
=1/4*cos2x+√3/4sin2x+3/4
=1/2*(1/2*cos2x+√3/2sin2x)+3/4
=1/2*(sinπ/6*cos2x+cosπ/6sin2x)+3/4
=1/2*sin(π/6+2x)+3/4
-1<=sin(π/6+2x)<=1
-1/2<=1/2sin(π/6+2x)<=1/2
1/4<=1/2sin(π/6+2x)+3/4<=5/4
所以函数f(x)=sin^2x/2+cos^2x+√3/4sin2x的最大值为：5/4
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
