若f(x)=开方3cos(3x-A)-sin(3x-A)为奇函数,则tanA=?
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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)=[sin(x+a/2)+根号3cos(x+a/2)]*cos(x+a/2)=sin(x+a/2)cos(x+a/2)+根号3cos(x+a/2)cos(x+a/2)=1/2sin(2x+a)
f(x)=√2sin(2 x-π/4)+2f(x)max=√2+2;此时X=3π/8令(2 x-π/4)∈【-π/2,π/2】,解得x∈【-π/8,3π/8】,因为x∈(0&nbs
f(sinx+2)=(cosx)^2+3=1-(sinx)^2+3=-(sinx+2)^2+4(sinx+2)f(x)=-x^2+4x
f(sin(pai/2-x))=cos[3(pai/2-x)]f(cosx)=cos(3pai/2-3x)f(cospai/9)=cos(3pai/2-pai/3)=-sinpai/3=-根号3/2f
两倍角公式2cos²x=cos2x+12sinxcosx=sin2xf(x)=-4cos²x+4√(3)sinxcosx+5=-2(cos2x+1)+2√3sin2x+5=2√3s
y=sinx^2+根3sinxcosx+2cosx^2=-1/2(1-2sinx^2)+1/2根3*2sinxcosx+2cosx^2-1+3/2=-1/2cos2x+二分之根3倍sin2x+cos2
f(x)=cos(3x)*cos(2x)+sin(3x)*sin(2x)=cos(3x-2x)=cosxf'(x)=-sinx
因为cos2x=1-2sin^2xf(sinx)=3-cos2x=3-1+2sin^2x=2+2sin^2x所以f(cosx)=2+2cos^2x=2+cos2x+1=3+cos2x选C
化简,f(x)=cosx(1)不用证了吧(2)1/2再问:过程再答:先将所有的带π的全部化成标准的“-sin^2(x+π)”化成“-sin^2(x)”“-2cos(-x-π)”化成“2cosx”“2c
cos^2(x+∏/3)+cos^2(x-∏/3)=(cosx/2-根号3*sinx/2)^2+(cosx/2+根号3*sinx/2)^2=(cosx)^2/2+3(sinx)^2/2=1/2+(si
f(x)=sin(2x+a)+2√3*[cos(2x+a)+1]/2-√3=sin(2x+a)+√3cos(2x+a)=2sin(2x+a+z)其中tanz=√3/1=tan(π/3)所以f(x)=2
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,(
是若f(sinx)=3-cos2x,则f(x)为吗,若是,则因为cos2x=cosx平方-sinx平方,又sinx平方+cosx平方=1,所以f(sinx)=1-2sinx平方,换元的结果再问:然后呢
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+
f(x)=cosx+cosxcosπ/3-sinxsinπ/3=cosx+cosx*1/2-sinx*√3/2=cosx*3/2-sinx*√3/2=√3(cosx*√3/2-sinx*1/2)=√3
f(x)=cos(2x-π\3)+sin²x-cos²x=1/2cos2x+√3/2sin2x-cos2x=√3/2sin2x-1/2cos2x=-cos(2x+π\3)-1
已知f(cosx)=cos3x,则f(sinx)=要有过程sinx=cos(π/2-x)f(sinx)=f(cos(π/2-x))=cos3(π/2-x)=sin3x
f(sinx)=3-cos2x=3-(1-2sin2x)=3-1+2sin2x=2sin2x+2,则f(cosx)=2cos2x+2=2cos2x-1+3=3+cos2x,故选:C.
0再问:确定对不?要是对就直接满意答案~~再答:200%肯定