三角函数极值间周期
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发布时间:2022-05-01 00:38
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时间:2022-06-21 11:00
解析:
f(x)=sin(2x+ 3分之π)+sin(2x- 3分之π)+2cos²x-1
=sin2x*cos(3分之π) +cos2x*sin(3分之π)+sin2x*cos(3分之π) -cos2x*sin(3分之π)+cos2x
=2sin2x*cos(3分之π) +cos2x
=sin2x + cos2x
=√2*(sin2x*√2/2 + cos2x*√2/2)
=√2*sin(2x+ π/4)
则可知函数f(x)的最小正周期T=2π/2=π
若x∈[-π/4,π/4],即2x∈[-π/2,π/2],那么:2x+ π/4∈[-π/4,3π/4]
所以当2x+ π/4=π/2即x=π/8时,函数f(x)有最大值为√2;
当2x+ π/4=-π/4即x=-π/4时,函数f(x)有最小值为√2*(-√2/2)=-1.