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求函数f(x)=cos^2x+2√3sinxcosx-sin^2x的周期和单调区间
人气:477 ℃ 时间:2019-08-20 19:50:30
解答
f(x)=cos^2x+2√3sinxcosx-sin^2x
=cos^2x-sin^2x+2√3sinxcosx
=cos2x+2√3sinxcosx
=cos2x+√3sin2x
=2[(1/2)*cos2x+(√3/2)*sin2x]
=2*sin(2x+π/6)
=2*sin[2(x+π/12)]
周期为2π/2=π
单调区间为:
[kπ-π/3,kπ+π/6)上单调增;
[kπ+π/6,kπ+2π/3)上单调减.
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