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f(x)=sin(x+3兀/2)sin(x-2π) 求函数f(x)的最直和最小正周期 计算f(π/6)+f(π/12)
人气:250 ℃ 时间:2020-05-24 02:47:26
解答
f(x) = sin(x+3π/2)sin(x-2π)
= -cosxsinx
= -1/2sin2x
最大值1/2
最小值-1/2
最小正周期2π/2=π
f(π/6)=-1/2sinπ/3=-√3/4
f(π/12)=-1/2sinπ/6=-1/4
f(π/6)+f(π/6)=-(√3+1)/4
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