化简sin(kπ+π/3)cos(3kπ+2π/3)(k属于Z)
2个回答
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sin(kπ+π/3)cos(3kπ+2π/3)
=sin(kπ+π/3)cos(kπ+2π/3)
=sin(kπ+π/3)(-cos(kπ+π/3))
=-sin(kπ+π/3)cos(kπ+π/3)
=-1/2sin(2kπ-2π/3)
=1/2*sin(π2/3)
=√3/4
=sin(kπ+π/3)cos(kπ+2π/3)
=sin(kπ+π/3)(-cos(kπ+π/3))
=-sin(kπ+π/3)cos(kπ+π/3)
=-1/2sin(2kπ-2π/3)
=1/2*sin(π2/3)
=√3/4
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展开全部
sin(kπ+π/3)cos(3kπ+2π/3)
=sin(kπ+π/3)cos(kπ+2π/3)
={sin[(kπ+π/3)+(kπ+2π/3)]+sin[(kπ+π/3)-(kπ+2π/3)]}/2
=[sin(2kπ+π)+sin(-π/3)]/2
=[sin(π)-sin(π/3)]/2
=－√3/4
=sin(kπ+π/3)cos(kπ+2π/3)
={sin[(kπ+π/3)+(kπ+2π/3)]+sin[(kπ+π/3)-(kπ+2π/3)]}/2
=[sin(2kπ+π)+sin(-π/3)]/2
=[sin(π)-sin(π/3)]/2
=－√3/4
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