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函数f(x)=√3cos(-x/2)+sin(x/2-丌),x€R.求f(x)的最小正周期
人气:155 ℃ 时间:2020-07-17 19:51:30
解答
f(x)=√3cos(-x/2)+sin(x/2-丌)
=√3cos(x/2)-sin(x/2)
=2[cos(x/2)cos(π/3)-sin(x/2)sin(π/3)]
=2cos(x/2+π/3)
T=2π/(1/2)=4π
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