![](https://iknow-base.cdn.bcebos.com/lxb/notice.png)
已知函数y=sinxcosx+sinx+cosx求值域
2个回答
展开全部
y=sinxcosx+sinx+cosx
=1/2(2sinxcosx+1-1)+sinx+cosx
=1/2(sinx+cosx)^2-1/2+(sinx+cosx)
=1/2[(sinx+cosx)^2+2(sinx+cosx)+1-2]
=1/2(sinx+cosx+1)^2-1
=1/2{(√2)*[sin(x+π/4)]+1}^2-1
所以值域为[-1,√2+1/2]
=1/2(2sinxcosx+1-1)+sinx+cosx
=1/2(sinx+cosx)^2-1/2+(sinx+cosx)
=1/2[(sinx+cosx)^2+2(sinx+cosx)+1-2]
=1/2(sinx+cosx+1)^2-1
=1/2{(√2)*[sin(x+π/4)]+1}^2-1
所以值域为[-1,√2+1/2]
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