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设F(sinx)=Cos2x+1,求F(Cosx)

f(cosx)=f[sin(x+π/2)]=cos2(x+π/2)+1=cos(2x+π)+1=-cos2x+1

待续

解: (1) f(x)=cos²x-√3sinxcosx+½ =½[1+cos(2x)]-(√3/2)sin(2x)+½ =½cos(2x)-(√3/2)sin(2x)+1 =cos(2x+π/3)+1 最小正周期T=2π/2=π cos(2x+π/3)=1时,f(x)取得最大值f(x)max=1+1=2 cos(2x+π/3)=-1时,f(x)取得最小值f(...

f(sinx)=1-2(sinx)^2+1/(sinx)^2+1 设t=sinx f(t)=2-2t^2+1/t^2 f(x)=2 2x^2+1/x^2

f(x)=cos2x+2sinxcosx+1 =cos2x+sin2x+1 =√2(√2/2*cos2x+√2/2*sin2x)+1 =√2cos(2x-π/4)+1 ∵-1≤cos(2x-π/4)≤1 ∴1-√2≤√2cos(2x-π/4)+1≤1+√2 则f(x)值域为:[1-√2,1+√2] 行家解答,质量保证

你应该告诉我原题的,因为化简出来的结果未必是你想要的

(1)f(x)=12sin2x3+32(1+cos2x3)=12sin2x3+32cos2x3+32=sin(2x3+π3)+32,由sin(2x3+π3)=0,得2x3+π3=kπ(k∈Z),解得:x=3k?12,k∈Z,则对称中心的横坐标为3k?12(k∈Z);(2)由已知b2=ac及余弦定理,得:cosx=a2+c2?b22ac=a2+c2?ac2a...

(1)化简可得f(x)=4sinx?1?cos(π2+x)2+cos2x=2sinx(1+sinx)+1-2sin2x=2sinx+1,∵x∈R,∴sinx∈[-1,1],∴f(x)的值域是[-1,3](2)当x∈[π6,2π3]时,sinx∈[12,1],∴f(x)∈[2,3]由|f(x)-m|<2可得-2<f(x)-m<2,∴f(x)-2<m<f(x...

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