f(x)=sin2xcos(π/3)+cos2xsin(π/3)+sin2xcos(π/3)-cos2xsin(π/3)+cos2x
=2sin2xcos(π/3)+cos2x
=sin2x+cos2x
=√2sin(2x+π/4)
f(max)=√2
f(min)= - √2
求函数f(x) 在区间[-π/4,π/4]的最大值和最小值
求函数f(x) 在区间[-π/4,π/4]的最大值和最小值
f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos²x-1,x∈R
f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos²x-1,x∈R
数学人气:275 ℃时间:2019-08-17 10:58:18
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