化简整数f(x)=2cos²x 2-搜答案-神马作文网

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

f(x)=sin²x+2sin2x+3cos²x=1+2sin2x+2cos²x=1+2sin2x+cos2x+1=2sin2x+cos2x+2=√5sin(2x+fai

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(x)=sin²x+2√3sin(x+π/4)sin[π/2-(x-π/4)]-cos²x-√3=sin²x+2√3sin(x+π/4)sin(3π/4-x)-cos&

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

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

解法一：f(x)=sin(2X+π/4)+cos(2X+π/4)=√2[√2/2sin(2X+π/4)+√2/2cos(2X+π/4)]=√2sin(2x+π/4+π/4)=√2sin(2x+π/2)

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

=(1+cos2x)/2+sin2x/2=根号2/2sin(2x+pi/4)+1/2

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,（

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

你确定第一个符号是加号不是乘号?

（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)=cos^2x-2sinxcosx-sin^2x+1=(cos2x+1)/2-sin2x+1/2-sin^2x+1/2=(1/2)*cos2x+1/2-sin2x+(1/2)*cos2x+1/

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

周期循环的结果1/2再问：能写一下具体过程吗？