
①sin(nπ+) ②cos(2nπ+
) ③sin(2nπ+
) ④cos[(2n+1)π-
] ⑤sin[(2n+1)π-
](以上n∈Z
A.①② B.①③④ C.②③⑤ D.①③⑤
①sin(nπ+) ②cos(2nπ+
) ③sin(2nπ+
) ④cos[(2n+1)π-
] ⑤sin[(2n+1)π-
](以上n∈Z
A.①② B.①③④ C.②③⑤ D.①③⑤
解析:
②cos(2nπ+③sin(2nπ+)=sin
.
⑤sin[(2n+1)π-]=sin[2nπ+(π-
)]=sin(π-
)=sin
.
答案:
C