问题补充:
已知函数f(x)=cos(x/2)平方-sin(x/2)平方+sinx,求函数f(x)的最小正周期,
答案:
f(x)=cos(x/2)^2-sin(x/2)^2+sinx
=cosx+sinx
=√2(√2/2*cosx+√2/2*sinx)
=√2(sinπ/4*cosx+sinxcos*π/4)
=√2sin(π/4+x)
所以:T=2π/1=2π
时间:2024-05-07 17:15:39
已知函数f(x)=cos(x/2)平方-sin(x/2)平方+sinx,求函数f(x)的最小正周期,
f(x)=cos(x/2)^2-sin(x/2)^2+sinx
=cosx+sinx
=√2(√2/2*cosx+√2/2*sinx)
=√2(sinπ/4*cosx+sinxcos*π/4)
=√2sin(π/4+x)
所以:T=2π/1=2π