化簡sin(nπ+2π/3)×cos(nπ+4π/3) n屬于Z
化簡sin(nπ+2π/3)×cos(nπ+4π/3) n屬于Z
數(shù)學(xué)人氣：697 ℃時間：2020-09-12 10:13:10
優(yōu)質(zhì)解答
①當(dāng)n=2k,k∈Z時, 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當(dāng)n=2k+1,k∈Z時, 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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