2、由(1)若何推出(2)
(1)[kπ,kπ+π/3]∪[kπ+π/2,kπ+(5π)/6],k∈z
(2)[(kπ)/2,(kπ)/2+π/3],k∈z
[kπ,kπ+π/3]∪[kπ+π/2,kπ+(5π)/6]=[2k(π/2),2k(π/2)+π/3]∪[2k(π/2)+π/2,2k(π/2)+π/2+π/3]]=[2k(π/2),2k(π/2)+π/3]∪[(2k+1)(π/2),(2k+1)(π/2)+π/3]=[(kπ)/2,(kπ)/2+π/3],k∈z
上式中前一部门为π/2的偶数倍，后一部门为π/2的奇数倍。
3.由(1)若何推出(2)
(1){kπ}∪{kπ+π/2},k∈z
(2){(kπ)/2},k∈z
[kπ]∪[kπ+π/2]=[2k(π/2)]∪[[(2k+1)(π/2)]=[(kπ)/2],k∈z
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