f(x)=(cosx)^2-(sinx)^2-根号3cos(π/2+2x)求周期,
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f(x)=(cosx)^2-(sinx)^2-根号3cos(π/2+2x)求周期,
f(x)=(cosx)^2-(sinx)^2-根号3cos(π/2+2x)
求周期,
f(x)=(cosx)^2-(sinx)^2-根号3cos(π/2+2x)求周期,
由倍角与半角的关系得:(cosx)^2-(sinx)^2=cos2x
由诱导公式得:cos(π/2+2x)=-sin2x
所以原式=cos2x-3sin2x=√10(1/√10cos2x-3/√10sin2x) (1)
令sina=1/√10,则cosa=3//√10,所以a=arcsin(1/√10),
所以(1)式=√10cos(2x-a)
因此T=2π/2=π
如图