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求函数y=sin方x+sinxcos(30度-x)的周期和单调递增区间
人气:189 ℃ 时间:2020-09-23 01:49:57
解答
y=sin^2 x+sinxcos(30°-x)
=sin^2 x+(√3sinxcosx)/2+(sin^2 x)/2
=√3/4(sin2x-√3cos2x)+3/4
=√3/2sin(2x-π/3)+3/4.
T=π.
2kπ-π/2≤2x-π/3≤2kπ+π/2,
kπ-π/12≤x≤kπ+5π/12.
单调递增区间[kπ-π/12, kπ+5π/12],k∈Z
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