请教真苗大侠!!!

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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.

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