请教真苗大侠！！！2、由(1)如何推出(2)(1)kπ,kπ+如果不给出(2),那么是很难想到如何把(1)化成(2)的.2.[kπ,kπ+π/3]∪[kπ+π/2,kπ+(5π)/6]=[2*(kπ/2),2*(kπ/2)+π/3]∪[2*(kπ/2)+π/2,2*(kπ/2)+(5π)
请教真苗大侠！！！2、由(1)如何推出(2)(1)kπ,kπ+
如果不给出(2),那么是很难想到如何把(1)化成(2)的.
2. [kπ,kπ+π/3]∪[kπ+π/2,kπ+(5π)/6]
=[2*(kπ/2),2*(kπ/2)+π/3]∪[2*(kπ/2)+π/2,2*(kπ/2)+(5π)/6]]
=[2kπ/2,2kπ/2+π/3]∪[(2k+1)π/2,(2k+1)π/2+π/3]
=[(nπ)/2,(nπ)/2+π/3].然后把n换成k就可以了.
3. {kπ}∪{kπ+π/2}
={2*(kπ/2)}∪{2*(kπ/2)+π/2}
={2kπ/2}∪{(2k+1)π/2}
={(nπ)/2}.同样把n换成k就可以了.
其实3应该是高中题.
2、由(1)如何推出(2)(1)kπ,kπ+π/3-∪kπ+?
2、由(1)如何推出(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
上式中前一部分为π/2的偶数倍，后一部分为π/2的奇数倍。
3.由(1)如何推出(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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