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y=tan(2x-π/4)+tan(2x+π/4)的最小正周期是?
最好用切割化弦做一下。
人气:299 ℃ 时间:2020-07-02 23:25:52
解答
y=sin(2x-π/4)/cos(2x-π/4)+sin(2x+π/4)/cos(2x+π/4)
=[sin(2x-π/4)cos(2x+π/4)+cos(2x-π/4)sin(2x+π/4)] / cos(2x+π/4)cos(2x-π/4)
=2sin4x/cos4x
=2tan4x
最小正周期是T=π/4
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