函数f(x)=sin눀(x+π⼀4)-sin눀(x-π⼀4)的最小正周期为

2024-12-30 23:22:02
推荐回答(2个)
回答1:

cos2x=1-2sⅰn²ⅹ,
sⅰn²ⅹ=(1-cos2ⅹ)/2,
f(x)=[1-cos(4x-π/2)]/2
=(1-sⅰn4ⅹ)/2
∴T=2π/ω
=π/2

回答2:

f(x)=sin²(x+π/4)+cos²(x-π/4)-1
=(sinx+cosx)²*(√2/2)²+(cosx+sinx)²(√2/2)²-1
=(sinx+cosx)²-1
=2sinxcosx
=sin2x
f(-x)=sin(-2x)=-sin2x=-f(x)
f(x)为奇函数
正周期为:2π/2=π