余弦函数的单调性

Question

余弦函数的单调性

Answer 1

有y=cosx的图像可知[0+2kπ,π+2kπ]k属于z为单调递减区间,[-π+2kπ,0+2kπ]k属于z为单调递减区间,当函数为y=Acos(wx+B）时,（注：w为偶米噶,B为西格玛,都打不来）,求单调性,比如求单调递减区间要令t=wx+B,∵y=Acost的单调递减区间为[0+2kπ,π+2kπ]k属于z,∴0+2kπ≤wx+B≤π+2kπ,再化为-B/w+2kπ/w≤x≤（π-B）/w+2kπ/w,得函数的单调递减为[-B/w+2kπ/w,π-B）/w+2kπ/w]k属于z