已知函数fx=cos²ωx一√3sinωxcosωx-1 2
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fx=-√3cos2x-sin2x=-2sin(2x+π/3)所以最小正周期为πf'x=-4cos(2x+π/3),f'x>0时递增x在(π/12,π/3)上递增f'x=0,x=π/12.极小值f(π
问题不完整啊再问:已知函数fx=sin(x/2)cos(x/2)+cosx/2-2再答:是化简吗再问:嗯嗯是的,打字慢,再问:再答:f(x)=1/2sinx+1/2cosx-3/2=sin(x+π/4
f(x_=(cosx+sinx)(cosx-sinx)=cos²x-sin²x=cos2x所以T=2π/2=πf(α/2)=cosα=1/3sin²α+cos²
若cosα=3/5.α属于(3π/2,2π),sinα=-4/5把f(2α+π/3)代入fx=√2cos(x-π/12),化简原式=cos2α-sin2αcos2α-sin2α怎么化简的就不用我说了吧
f(x)=sin²x+√3sinxcosx+2cos²x,=√3sinxcosx+cos²x+1=√3/2sin2x+1/2(1+cos2x)+1=√3/2sin2x+1
令t=sinx则f=(1-t^2)+2t=-t^2+2t+1=-(t-1)^2+2因为|t|
先化简f(x)=2根号3sinxcosx+2cos^2x-1=根号3sin2x+cos2x=2(根号3/2sin2x+1/2cos2x)=2sin(2x+π/6)则T=2π/ω=2π/2=πy=sin
1,=1/2sinwxcoswx+(1+cos2wx)/2=1/2sin2wx+1/2cos2wx+1/2=根号2/2sin(2wx+π/4)+1/22π/2w=π解得w=12,根号2/2sin(2x
答:f(x)=2cos²(x/2)-sinx=cosx+1-sinx=-√2*[(√2/2)*sinx-(√2/2)*cosx]+1=-√2*(sinxcosπ/4-cosxsinπ/4)+
(1)f(x)=[cos(x-π/6)]^2-(sinx)^2f(π/12)=(cos(π/12))^2-(sin(π/12))^2=cos(π/6)=√3/2(2)f(x)=[cos(x-π/6)]
f(x)=cos(2x-π/3)-cos2x=1/2cos2x+√3/2sin2x-cos2x=√3/2sin2x-1/2cos2x=sin(2x-π/6)最小正周期T=2π/2=π(2)0
你好f(x)=cos²x+sinx+a=(1-sin²x)+sinx+a=-sin²x+sinx+a+1=-(sinx-1/2)²+1/4a+1当f(x)=0有
(1)f(x)=cos²x=(1/2)+(1/2)cos2x,对称轴2x0=kπ,sin2x0=0;所以g(2x0)=1+(1/2)sin2x0=1;(2)h(x)=f(x)+g(x)=(1
f(x)=2sin(x-π/6)cosx+2cos²x=(2sinxcosπ/6-2cosxsinπ/6)cosx+2cos²x=√3sinxcosx-cos²x+2co
已知函数fx=√3sinxcosx+(cos∧2)x+a(1)求fx的最小正周期及单调递减区间(2)若fx在区间[~π/6,π/3]上的最大值与最小值的和为3/2,求a的值.(1)解析:f(x)=√3
f(x)=√3sin2x+cos2x=2sin(2x+π/6)∴f(x0)=2sin(2x0+π/6)=6/5∴sin(2x0+π/6)=3/5∵x0∈[π/4,π/2]∴2x0+π/6∈[2π/3,
解f(x)=2cos^2x+2√3sinxcosx-1=√3sin2x+cos2x=2sin(2x+π/6)∴最小正周期为:2π/2=π再答:不懂追问再问:在三角形ABC中,角ABC所对的边分别是ab