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f(x)=2sin(x+a/2)cos(x+a/2)+2根号3 cos²(x+a/2)-根号3 如何化简?
人气:324 ℃ 时间:2019-08-21 17:53:56
解答
∵2sin(x+a/2)cos(x+a/2)=sin(2x+a)2 cos²(x+a/2)=1+cos(2x+a)∴f(x)=2sin(x+a/2)cos(x+a/2)+2根号3 cos²(x+a/2)-根号3=sin(2x+a)+√3[1+cos(2x+a)]-√3=sin(2x+a)+√3cos(2x+a)=2[1/2sin(2x+a)+√3/2*...太谢谢你了!还有一个问题,可以帮我算一下吗?若0≤a≤π时,求使函数f(x)为偶函数的a值解答题做法f(x)为偶函数f(-x)=f(x)∴sin[-2x+(a+π/3)]=sin[2x+(a+π/3)]∴sin(-2x)cos(a+π/3)+cos(-2x)sin(a+π/)=sin2xcos(a+π/3)+cos2xsin(a+π/)∴-sin2xcos(a+π/3)+cos2xsin(a+π/)=sin2xcos(a+π/3)+cos2xsin(a+π/)∴sin2xcos(a+π/3)=0∵sin2x是变量∴cos(a+π/3)=0∴a+π/3=kπ+π/2,k∈Z∵0≤a≤π∴取k=0,a=π/6
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