生成{Sin[1],Sin[2],Sin[3]}A.a={1,2,3};Sin[a]B.Table[Sin[k],{k,3}]C.Table[Sin(k),{k,3}]D.Table[{Sin[k]},{k,3}]
生成{Sin[1],Sin[2],Sin[3]}
A.a={1,2,3};Sin[a]
B.Table[Sin[k],{k,3}]
C.Table[Sin(k),{k,3}]
D.Table[{Sin[k]},{k,3}]
参考答案和解析
a={1,2,3};Sin[a];Table[Sin[k],{k,3}]
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