化简sin(nπ+2π/3)×cos(nπ+4π/3) n属于Z
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①当n=2k,k∈Z时, sin(nπ+2π/3)·cos(nπ+4π/3) = sin(2π/3)·cos(4π/3) = sin(π-π/3)·cos(π+π/3) = sin(π/3)·[-cos(π/3)] =-√3/4 ②当n=2k+1,k∈Z时, sin(nπ+2π/3)·cos(nπ+4π/3) = sin(π+2π/3)·cos(π+4π/3) = sin(2π/3)·cos(4π/3) = sin(π-π/3)·cos(π+π/3) = sin(π/3)·[-cos(π/3)] =-√3/4 故：sin(nπ+2π/3)·cos(nπ+4π/3) =-√3/4
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