求函数f(x)=2cosxsin(x+π/3)-根号3/2的最小正周期
人气:156 ℃ 时间:2019-11-18 06:24:00
解答
公式:2sinAcosB = sin(A+B) +sin(A-B)
f(x)=2cosxsin(x+π/3)-√3/2 = sin(x+π/3+x) + sin(x+π/3-x) - √3/2
= sin(2x+π/3) + √3/2 - √3/2
= sin(2x+π/3)
最小正周期:2π/2 = π
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