y=sin^2(x+π/2)-sin^2(x-π/4)的最小正周期和值域

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/06 20:06:40
y=sin^2(x+π/2)-sin^2(x-π/4)的最小正周期和值域

y=sin^2(x+π/2)-sin^2(x-π/4)的最小正周期和值域
y=sin^2(x+π/2)-sin^2(x-π/4)的最小正周期和值域

y=sin^2(x+π/2)-sin^2(x-π/4)的最小正周期和值域
y=sin²(x+π/2) - sin²(x-π/4)
=cos²x + [1-2sin²(x-π/4)-1]/2
=(2cos²x-1+1)/2 + [cos2(x-π/4)-1]/2
=1/2cos2x+1/2 + 1/2cos(2x-π/2)-1/2
=1/2cos2x + 1/2sin2x
=√2/2(√2/2cos2x+√2/2sin2x)
=√2/2sin(2x+π/4)
所以最小正周期T=2π/W=2π/2=π
因为-1≤sin(2x+π/4)≤1
所以-√2/2≤√2/2sin(2x+π/4)≤√2/2
即值域为[-√2/2,√2/2]

y=sin^2(x+π/2)-sin^2(x-π/4)=cos^2(x)-(sinxcosπ/4-sinπ/4cosx)^2=cos^2x-(0.5sin^2x+0.5cos^2x-sinxcosx)=0.5cos^2x-0.5sin^2x+sinxcosx=0.5cos2x+0.5sin2x=(√2/2)sin(2x+π/4)
最小正周期是π,值域是[-√2/2,√2/2]。