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求函数y=cos^2x-sin^2-√3cos(3π/2+2x)+1的周期,单调减区间和最值
人气:349 ℃ 时间:2019-08-18 20:54:19
解答
y=cos^2x-sin^2-√3cos(3π/2+2x)+1
=cos2x-√3sin2x+1
=2cos(2x+π/3)+1
当2kπ+π/2≤2x+π/3≤2kπ+3π/2时,函数单调递减
2kπ+π/2≤2x+π/3≤2kπ+3π/2
2kπ+π/6≤2x≤2kπ+7π/6
kπ+π/12≤x≤kπ+7π/12
所以单调递减区间:[kπ+π/12,kπ+7π/12],k∈Z
最大值:ymax=3
最小值:ymin=-1
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