函数f(x)=cos(-x/2)+sin(π-x/2),x属于R,(1)求f(x)的周期,(2)求f(x)在[0,π]上的减区间为什么=√2sin(x+π/4),不是=√2sin(x/2+π/4)吗?
网友回答
(1)f(x)=cos(-x/2)+sin(π-x/2)
=cos(x/2)+sin(x/2)
=√2[(√2/2)cos(x/2)+(√2/2)sin(x/2)]
=√2[sin(π/4)cos(x/2)+cos(π/4)sin(x/2)]
=√2sin(x+π/4)
则：f(x)的周期
T=2π/1=2π
(2)设t=x+π/4
则当t属于[2kπ+π/2,2kπ+3π/2]时
f(x)单调递减
即：2kπ+π/2=
======以下答案可供参考======
供参考答案1:
f(x)=cos(-x/2)+sin(π-x/2),
=cos(x/2)+sin(x/2)
=√2sin[(x/2)+π/4],
周期为4ππ/2π/4π/2