已知函数f(x)=2sinxcosx+2cosx² (x∈R) 1,求f(x) 的值 2,当x∈[0,π/2]时f(x)的值域
问题描述:
已知函数f(x)=2sinxcosx+2cosx² (x∈R) 1,求f(x) 的值 2,当x∈[0,π/2]时f(x)的值域
答
(1)f(x)=sin2x+cos2x+1
=√2(sin2x *√/2+cos2x *√2)+1
=√2(sin2x cosπ/4+cos2x cosπ/2)+1
=√2sin(2x+π/4)+1
(2) 当x∈[0,π/2]
2x+π/4∈[π/4,3π/4]
sin(2x+π/4)∈[√2/2,1]
所以 y∈【2,√2+1】