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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

换元法

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...

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=cosx/√(1-sin2x)+sinx/√(1-cos2x)+tanx/√tan2x 定义域tan2x>0 →kπ/2

解答:解:(1)函数f(x)=3?cos2x2-4t?sinx2cosx2+2t2-6t=sin2x-2tsinx+2t2-6t+1 =(sinx-t)2+t2-6t+1.当t<-1时,g(t)=(-1-t)2+t2-6t+1=2t2-4t+2.当-1≤t≤1时,g(t)=t2-6t+1.当t>1时,g(t)=(1-t)2+t2-6t+1=2t2-8t+2.综上可得...

f(x)=sinx3cosx3+3cos2x3=12sin2x3+32(1+cos2x3)=12sin2x3+32cos2x3+32=sin(2x3+π3)+

f(x)=(1+cos2x+8(sinx)^2)/(sin2x) =(1+(1-2(sinx)^2)+8(sinx)^2)/(2sinx*cosx) =(1+3(sinx)^2)/(sinx*cosx) =(4(sinx)^2+(cosx)^2)/(sinx*cosx) =4sinx/cosx + cosx/sinx 设t=sinx/cosx --> cosx/sinx=1/t f(x)=4t+1/t 当00 -->t=sinx/cosx>0 -...

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