f(x)=cos²x sinxcosx化简

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

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

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

f(sinx)=cos(x),则f(cosx)令t=sinx(-1≤t≤1)cosx=±√1-sinx²=±√1-t²f(t)=±√1-t²∴f(cosx)=±√1-co

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

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

f'(x)=(-xsinx-cosx)/x^2=-(xsinx+cosx)/x^2

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

f'(x)=-sin(sinx)*(sinx)'=-sin(sinx)*cosx

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,（

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

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)=cos(2x-π\3)+sin²x-cos²x=1/2cos2x+√3/2sin2x-cos2x=√3/2sin2x-1/2cos2x=-cos(2x+π\3)-1

首先把这一题翻译一下就是问：f(x)=log3xx>0和f(x)=cosπxx>0的图像上有几个交点.你可以这样分析：f(x)=cosπx∈（-1,1）,周期T=2.当x>3时,两图像不会会有交点.因

你确定第一个符号是加号不是乘号?

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

（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

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

f(sin15)=f(cos75)=cos150=-cos30=-√3/2

1.cos(x+60)=cos60cosx-sin60sinx=0.5cosx-(√3/2)sinxcos(60-x)=cos60cosx+sin60sinx=0.5cosx+(√3/2)sinx∴f