y=cos²x-sinx求值域
展开全部
y=cos²x-sinx
=-sin²x-sinx+1
=-(sin²x+sinx+1/4)+5/4
= -(sinx+1/2)²+5/4
∵-1<=sinx<=1
∴-1/2<=sinx+1/2<=3/2
∴0<=(sinx+1/2)²<=9/4
-9/4<=-(sinx+1/2)²<=0
∴-1<= -(sinx+1/2)²+5/4<=5/4
∴值域是[-1，5/4}
=-sin²x-sinx+1
=-(sin²x+sinx+1/4)+5/4
= -(sinx+1/2)²+5/4
∵-1<=sinx<=1
∴-1/2<=sinx+1/2<=3/2
∴0<=(sinx+1/2)²<=9/4
-9/4<=-(sinx+1/2)²<=0
∴-1<= -(sinx+1/2)²+5/4<=5/4
∴值域是[-1，5/4}
本回答由提问者推荐
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
