已知函数f(x)=ocs^4x-2sinxcosx-sin^4x,如何化简,并求最小正周期
问题描述:
已知函数f(x)=ocs^4x-2sinxcosx-sin^4x,如何化简,并求最小正周期
答
=(cos^2x+sin^2x)(cos^2x-sin^2x)-sin2x
=cos2x-sin2x=√2sin(-2x+π/4)
最小正周期π
答
f(x)=ocs^4x-2sinxcosx-sin^4x
=(cos²x+sin²x)(cos²x-sin²x)-sin2x
=cos²x-sin²x-sin2x
=cos2x-sin2x
=√2sin(2x+3π/4)
∴最小正周期是2π/2=π