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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

f(sinx)=cos2x+1 = 1-2sin²x+1 = 2 - 2sin²x 所以,f(x)=2-2x² 所以,f(cosx)=2-2cos²x=2sin²x

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(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

(1)由题设f(x)=-sin2x+1+cos2x+1=2cos(2x+π4)+2.∵f(x)-1=0,∴2cos(2x+π4)+2=1,∴cos(2x+π4)=?22,则2x+π4=2kπ+34π或2x+π4=2kπ+54π,k∈Z,得x=kπ+π4或x=kπ+π2,k∈Z,∵x∈(0,π),∴x1=π4,x2=π2,∴x1+x2=34π;(2)由函数y=f(x...

由 f(x)=sinx+cosx知 f'(x)=cosx-sinx(Ⅰ)由f(x)=2f'(x)得3sinx=cosx,有1+sin2xcos2x?sinxcosx=cos2x+2sin2xcos2x?sinxcosx=9sin2x+2sin2x9sin2x?3sinxsinx=116.(Ⅱ)由x∈[0,2π],g(x)=f(x)?f′(x)4+f(x)+f′(x)=2sinx4+2cosx=s...

y=-cos2x+cosx+m =-(2cos²x-1)+cosx+m =-2cos²x+cosx+m+1 设cosx=t,则 y=-2t²+t+m+1 (-1≤t≤1) 这是对称轴为 t=1/4,开口向下的抛物线的一部分 当 t=1/4 时,y最大值=m+9/8 当t=-1时,y最小值=m-2 由于 1≤y≤5 恒成立, 所以 m-2≥...

f'(x)=1/2cos2x*(2x)]+cosx =cos2x+cosx 显然这是偶函数 cos2x+cosx =2cos²x-1+cosx =2(cosx+1/4)²-9/8 -1

f(sinx)=1+cos2x = 1+ (1-2(sinx)^2) = 2 - 2(sinx)^2 f(x) =2-2x^2 lim(x-> ∞ )(3x^3-7x+4)/(7x^3-8x+9) =lim(x-> ∞ )(3-7/x+4/x^3)/(7-8/x^2+9/x^3) =3/7

f(x)=(sinx+cosx)²+cos2x =1+sin2x+cos2x =√2sin(2x+π/4)+1 ∴(1)T=2π/2=π; (2)x∈[0,π/2]时, 2x+π/4∈[π/4,5π/4] sin(2x+π/4)∈[-√2/2,1] ∴f(x)max=√2+1(x=3/8时取到) f(x)min=0(x=π/2时取到)

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