题目内容
当k∈Z时,在①sin(kπ+A.①和② B.③和④ C.①和④ D.②和③
解析:(1)当k=2n时,sin(kπ+
)=sin(2nπ+
)=sin
.
当k=2n+1时,sin(kπ+
)=sin[(2n+1)π+
]=sin(2nπ+π+
)=sin(π+
)=-sin
.
(2)sin(2kπ±
)=sin(±
)=±sin
.
(3)当k=2n时,sin[kπ+(-1)k·
]=sin[2nπ+(-1)2n·
]=sin
.
当k=2n+1时,sin[kπ+(-1)k·
]=sin[2nπ+π-
]=sin
.
(4)cos[2kπ+(-1)k·
]=cos[(-1)k·
].
当k=2n时,原式=cos
=sin
.
当k=2n+1时,原式=cos[(-1)2n+1·
]=cos
=sin
.故选B.
答案:B
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