已知f(x)=sinx,f[a(x)]=1-x2,求a(x)

f(x)=a*b=(cosx+sinx)*(cosx-sinx)+sinx*(-2cosx)=(cosx)^2-(sinx)^2-2sinx*cosx=cos2x-sin2x=根号2*(根号2*cos

1）f(x)=a·b=-2sin^x+2√3sinxcosx=cos2x-1+√3sin2x=2sin(2x+π/6)-1,它的增区间由（2k-1/2)π

f(x)=ab=2sin^2x+2sinxcosx=(1-cos2x)+sin2x=根号2*sin(2x-Pi/4)+1又-1

f(x)=2sinx(sinX+cosX)=2sinxsinx+2sinxcosx=1-cos2x+sin2x=√2sin(2x-π/4)+1所以f(x)的最小正周期=2π/2=π最大值=1+√2

(1)f(x)=a*b+1=2sin²x+2sinxcosx+1=1-cos2x+sin2x+1=√2sin(2x-π/4)+2所以函数f(x)的最小正周期是T=2π/2=π(2)x∈[0,

f(x)=a*b=2sinxcosx+(sinx+cosx)(cosx-sinx)=sin2x+cos2x=√2sin(2x+π/4)1）当x∈[0,π/2]时2x+π/4∈[π/4,π/4+π]当2

f(x)=sinx+5x,x∈(-1,1)∵f(-x)=sin(-x)-5x=-(sinx+5x)∴f(x)是奇函数f'(x)=cosx+5∴f'(x)恒>0∴f(x)是增函数f(1-a)+f(1-a

f(x)=ab=2sinx(1+sinx)+(cosx+sinx)(cosx-sinx)=2sinx+2sin^2x+cos^2x-sin^2x=2sinx+sin^2x+cos^2x=2sinx+1

函数f(x)=(sinx一cosx)sin2x/sinxsinx≠0,所以x≠kπ,k∈Z.函数定义域是{x|x≠kπ,k∈Z}.f(x)=(sinx一cosx)sin2x/sinx=(sinx一co

1、f(x)=2√3sinx+2cosx=4sin(x+π/6)f(x)的最大值为4,此时x∈{x|x=π/3+2kπ,k∈Z}.2、由f(x)=2bc-bc=bc所以bc

字数限制f(x)=cos2x+(1-cos2x)/2+sin2x/2=(cos2x+sin2x)/2+1/2=cos(2x+π/4)/根号2+1/2其最小正周期为π,最大值为：(1+根号2）/2x在[

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

ab=（cosx+sinx）^2-2sin^2x=sin2x+1-2sin^2x=sin2x+cos2x=√2sin（x+π/4）∴f（x）=√2sin（x+π/4）+1T=2π/1=2π（Ⅱ）∵x∈

f(x)=a·b=(cosx+sinx)²-2sin²x=cos²x+sin²x+2sinxcosx-2sin²x=1-2sin²x+2si

解析：∵a*b=(cosx+sinx,sinx)*(cosx-sinx,2cosx)=(cosx+sinx)(cosx-sinx)+2sinxcosx=[(cosx)^2-(sinx)^2]+2sin

f（x）=x*sinxf'(x)=sinx+xcosx,x∈(0,π/2)时,f'(x)>0,f(x)递增f(-x)=f(x),f(x)是偶函数∵A,B是锐角三角形两个内角∴cosA=sin(π/2-

f(x)=2sinx(sinx+cosx)   =2sin²x+2sinxcosx  =2sin²x-1+2sinxcosx+1&

（1）因为f（x）=0，即a＝sin2x−sinx＝(sinx−12)2−14，a的最大值等于(−1−12)2 −14=2，a的最小值等于-14，所以，a∈[−14，2]．（2）f（x）=-

f(x)=1-2x^2

（1）当sinX=-（2a/-2）=a时,f（x）最大,故g（a）=a^2+a-1