怎样从-[1-2sin²(x-π/12)]得到-cos(2x-π/6)?

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怎样从-[1-2sin²(x-π/12)]得到-cos(2x-π/6)?怎样从-[1-2sin²(x-π/12)]得到-cos(2x-π/6)?怎样从-[1-2sin²(

怎样从-[1-2sin²(x-π/12)]得到-cos(2x-π/6)?
怎样从-[1-2sin²(x-π/12)]得到-cos(2x-π/6)?

怎样从-[1-2sin²(x-π/12)]得到-cos(2x-π/6)?
cos2x=(cosx)^2-(sinx)^2=1-2(sinx)^2
∴-[1-2sin(x-π/12)]=-cos(2x-π/6)