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求函数y=sinx+根号3cosx在区间[-π/2,π/2]上的最大值与最小值
人气:307 ℃ 时间:2019-08-17 19:04:18
解答
y=2(sinxcosπ/3+cosxsinπ/3)
=2sin(x+π/3)
x=π/6属于[-π/2,π/2],ymax=2
x=-π/2属于[-π/2,π/2],ymin=sin(-π/6)=-1/2
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