设F(x)= x²=cos是f(x)的原函数,则f(x)
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∵f(x)=cos(2x-π/3)+(sinx)^2-(cosx)^2=cos(2x-π/3)-cos2x=2sin(2x-π/6)sin(π/6)=sin(2x-π/6).∴g(x)=[sin(2x
f(x)=(1/2^0)·sin(x/2)+(2^0)·cos(2x)f‘(x)=(1/2)·cos(x/2)+(-2)·sin(2x)=(1/2^1)·cos(x/2)+(-2^1)·sin(2x)
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
(1)f(x)=cos(x+2π/3)+2cos²(x/2)=-(cosx)/2-(√3sinx)/2+1+cosx=1-[(√3sinx)/2-(cosx)/2]=1-[sin(x-π/6
1)f(x)=1+cos(2x+π/3)-(1+cos2x)/2=1/2-sin2x根号3/2最小值1/2-根号3/2最小正周期π2)c带入得sinC=根号3/2C=π/3A=π-B-C=2π/3-a
帮楼上补充的f(x)为了好表示肯定可以用y等价表示咯,
f(x)=sin²x+sin2x+3cos²x=1+2cos²x+sin2x=sin2x+cos2x周期=π再问:要过程,还有第三题的图像再答:(1)f(x)=sin
y'=f'(sin²x)*(sin²x)'+f'(cos²x)*(cos²x)'=f'(sin²x)*(2sinxcos)+f'(cos²x
见图,复合函数求导.
f(x)=2-2x^2f(cosx/2)=1-cosx
两边对x求导:2cos(x^2+y)*(-sin(x^2+y))*(2x+y')=1所以y'=-1/sin(2x^2+2y)-2x再问:求f'(x)```再答:y'就是f'(x)啊。。。。。
f(f(f(x)))=f(f(arcsin(cos(x))))=f(arcsin(cos(arcsin(cos(x)))))=arcsin(cos(arcsin(cos(arcsin(cos(x)))
(1)解析:∵函数f(x)=cos(wx+f)(w>0,-π/2<f<0)的最小正周期为π∴w=2π/π=2,f(x)=cos(2x+f)∵f(π/4)=√3/2f(π/4)=cos
f(sinx)=cos2x+1=1-2sin^2x+1=2-2sin^2xf(cosx)=2-2cos^2x=2(1-cos^2x)=2sin^2x很高兴为您解答,【the1900】团队为您答题.请点
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
x∈(0,π),则f(x)=cos²x+sinx=1-sin²x+sinx=-(sinx-1/2)²+5/4当sinx=1/2时,取最大值=5/4此时x=π/6或5π/6
证明:f(x)=sinx-cosx+x+a求导:f'(x)=cosx+sinx+1=√2sin(x+π/4)+10
(1)f(x)=sinwxcoswx+coswxcoswx=1/2sin2wx+1/2cos2wx+1/2=√(根号)2/2sin(2wx+π/4)+1/2因为f(x)的周期为π,所以w=1f(x)=