f(x)=(2cos^2x-1)sin2x 1 2cos4x

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已知f(x)=cos^2x/1+sin^2x求f'(π/4)

f(x)=[1+cos2x)/2]/[1+(1-cos2x)/2]=(1+cos2x)/(3-cos2x)=-1+4/(3-cos2x)f'(x)=-4/(2-cos2x)^2*(2-cos2x)'=

已知函数f(x)=6cos^4x-5cos^2x+1/2cos^2x-1,求f(x)的定义域值域

f(x)=(6cos^4x-5cos^2x+1)/(2cos^2x-1)=3cos^2x-1,2cos^2x-1≠0cos2x≠02x≠kπ+π/2,k∈Z解出x得定义域f(x)=3cos^2x-1=

f(x)=(1+cos2x)/[4sin(pai/2+x)]-asin(x/2)cos(pai-x/

诱导公式f(x)=(1+2cos²x-1)/(4cosx)+asin(x/2)cos(x/2)=(cosx)/2+a/2*sinx=(a/2)sinx+(1/2)cosx=√[(a/2)&s

设2f(x)cos x=d/dx [f(x)]²,f(0)=1,则f(x)=

因为2f(x)cosx=d/dx[f(x)]²=2f(x)f'(x),所以2f(x)[f'(x)-cosx]=0,有f'(x)=cosx得:f(x)=sinx+C因为f(0)=1,所以f(x

已知函数f(x)=cos^2(x/2)-sin(x/2)*cos(x/2)-1/2

1f(x)=cos²(x/2)-sin(x/2)*cos(x/2)-1/2=1/2*[2cos²(x/2)-1]-1/2*sinx=1/2*cosx-1/2*sinx=√2/2*(

已知函数f(x)=sin(x/2)cos(x/2)+cos(x/2)的平方-1/2

f(x)=1/2sinx+1/2cosx(二倍角的正弦、余弦公式)=根号2/2(sinxcos45°+cosxsin45°)=根号2/2sin(x+45°)(1)f(a)=根号2/2sin(a+45°

f(x)为奇函数,x>0,f(x)=sin 2x+cos x,则x

设x0所以f(-x)=sin2(-x)+cos(-x)=-sin2x+cosx因为f(x)为奇函数,所以f(-x)=-f(x)得f(x)=-f(-x)=sin2x-cosx(x

函数f(x)=-√2(sin2x+π/4)+6 sin x cos x-2cos²x+1

f(x)=-√2sin(2x+π/4)+6sinxcosx-2cos²x+1=-√2(sin2xcosπ/4+cos2xsinπ/4)+3sin2x-2×(1+cos2x)/2+1=-√2(

求导f(x) = cos(3x) * cos(2x) + sin(3x) * sin(2x).

f(x)=cos(3x)*cos(2x)+sin(3x)*sin(2x)=cos(3x-2x)=cosxf'(x)=-sinx

f ' (sinx)=cos^2x,求f(x)

f'(sinx)=cos²x=1-sin²xf'(x)=1-x²f(x)=x-x^3/3

f(x)有定义,f(2x)=f(x)cos x,lim f(x)=f(0)=1(x趋于0时),求f(x)

对任意x均有f(x)=f(x/2)*cosx/2=f(x/4)*cosx/2*cosx/4=……=∏(i=∞)cos(x/2^i)*1f(x)=∏(i=∞)cos(x/2^i)

设f=[sin(2/x)]=1+cosx,求f(x),f[cos(2/x)].

cosx=1-2(sinx/2)^2f=[sin(2/x)]=1+cosx=2-2(sinx/2)^2f(x)=2-2x^2f[cos(2/x)]=2-2[cos(2/x)]^2

f(x)=cos(2x-派/3)-2sin x*cos (派/2+x)

令F’(x)=√3cos2x+sin2x=0,x1=kπ/2-π/6(k为偶数),x2=kπ/2-π/6(k为奇数)∴f(x)在x1极小,在x2处取极大值∴f(x)单调递减区间为[kπ/2-π/6,(

已知f(x)=cos(x/2)[sin(x/2)-cos(x/2)],其导为?

f'(x)=-1/2*sin(x/2)*[sin(x/2)-cos(x/2)]+cos(x/2)[1/2cos(x/2)+1/2sin(x/2)]=-1/2*sin²(x/2)+1/2sin

已知函数f(x)=cos(2x-π/3)+sin^2 x-cos^2 x

f(x)=cos(2x-π/3)-(cos^2x-sin^2x)=cos(2x-π/3)-cos2x=2sin(2x-π/6)sinπ/6=sin(2x-π/6)因为y=sinx的单减区间为[π/2+

已知f(x)=sin(x/2) + cos(x/2) +[cos(x/2)]^2-1/2

你确定第一个符号是加号不是乘号?

化简f(x)=2cos(x/2)·(sin(x/2)+cos(x/2))-1

(1)f(x)=2cos(x/2)·(sin(x/2)+cos(x/2))-1=2cos(x/2)·sin(x/2)+2cos^2(x/2)-1=sinx+cosx(倍角公式)=√2sin(x+π/4

若 f(sinx+1/sinx)=csc^2x-cos^2x,求f(x)

令sinx+1/sinx=t,则两边求平方得(sinx)的平方+2sinx(1/sinx)+1/(sinx的平方)=t的平方化简式子左边得到,(sinx)的平方+1/(sinx的平方)+2=t的平方即

已知函数f(x)=2Cos x(Sin x-Cos x)+1

f(x)=2cosx*sinx-2cosx^2+1f(x)=sin2x-cos2xf(x)=根号2*sin(2x-45)周期T=π

已知f(x)=-1/2+sin(π/6-2x)+cos(2x-π/3)+cos平方x.

解:⑴f(x)=-1/2+sin(π/6-2x)+cos(2x-π/3)+(cosx)^2=-1/2+sinπ/6cos2x-sin2xcosπ/6+cos2xcosπ/3+sin2xsinπ/3+(