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?
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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?
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?
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?
(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=
f(x)=cos(-x/2)+sin(π-x/2),
=cos(x/2)+sin(x/2)
=√2sin[(x/2)+π/4],
周期为4π
π/2<(x/2)+π/4<3π/2是减函数
π/4<(x/2)<5π/4
π/2<(x)<5π/2
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