根号(sinx-2)和sin根号2哪个是复合函数

cos2x/sin{x-π/4}=(cos²x-sin²x)/[根号2/2(sinx-cosx)]=-根号2(cosx+sinx)由已知得,-根号2(cosx+sinx)=-根号2

答案是选A的请看大图

sinx+cosx=√2/2(sinx+cosx)^2=1/2(sinx)^2+2sinxcosx+(cosx)^2=1/21+2sinxcosx=1/22sinxcosx=-1/2sinxcosx=

1+cos²x-sin²x>02cos²x>0cos²x>0即cos²x≠0x≠kπ+π/2,k∈Z定义域为{x∈R|x≠kπ+π/2,k∈Z}再问：

cos(x-π/4)=cosxcosπ/4+sinxsinπ/4=√2/2*(cosx+sinx)=√2/10sinx+cosx=1/5cosx=1/5-sinx平方cos²x=1-sin&

=√{[sin(x/2)]^2+[cos(x/2)]^2-2sin(x/2)cos(x/2)}+√{[sin(x/2)]^2+[cos(x/2)]^2+2sin(x/2)cos(x/2)}=√{[si

f(x)=2sin^2-2√3*sinX*cosX+1π=1-cos2X-√3*sin2X+1=2-2sin(2X+π/3)T=2π/2=π

√2sin(x-π/4)=√2(sinxcosπ/4-cosxsinπ/4)=√2[(1/√2)sinx-(1/√2)cosx]=sinx-cosx

答案是1/2cos2X可以分解成(cosX+sinX)(cosX-sinX),分母可以化成根2/2乘以(sinX-cosX)追问：分母可以化成根2/2乘以(sinX-cosX)?是如何化的?回答：si

再问：嗯再答：可以直接将x=0代入，因为分母不为0啊！求极限的结果=0

因为sin(π/6)=1/2；cos(π/6)=(√3)/2所以原式等于cos(π/6)sinx+sin(π/6)cosx利用两角和公式sin(A+B)=sinAcosB+cosAsinB得到sin（

(cos2x)/[sin(x-π/4)]=[(cosx)^2-(sinx)^2]/[((根号2)/2)sinx-((根号2)/2)cosx]=(cosx-sinx)(cosx+sinx)/[-((根号

配方,Y=-(sinx+√3/2)^2-1/2,显然当sinx=-√3/2,最大值-1/2,sinx=1,最小值-9/4-√3

[sinx+cos(π+x)]/[sinx+sin(π/2-x)]=(sinx-cosx)/(sinx+cosx)=(tanx-1)/(tanx+1)=1-[2/(tanx+1)]sinx=√3/3∴

y=2sin²x+2√3sinxcosx-2=1-cos(2x)+√3sin(2x)-2=2sin(2x-π/6)-1最小正周期=2π/2=π当sin(2x-π/6)=1时,y有最大值yma

f(x)=2根号3*sinx/3*cosx/3-2sin平方x/3=4sinx/3﹙√3cosx/3﹣sinx/3﹚/2=4sinx/3sin﹙π﹣x﹚/3=﹣2[cosπ/3－cos﹙2x-π﹚/3

sin(π/6)=1/2；cos(π/6)=√3/2∴√3/2*sinx+1/2*cosx=cos(π/6)sinx+sin(π/6)cosx=sin(x+π/6)

y=√（-2sin^2x+sinx）-2sin^2x+sinx≥0,0≤sinx≤1/2.2kπ≤x≤2kπ+π/6或2kπ+5π/6≤x≤2kπ+π,(k∈Z).所以函数定义域是[2kπ,2kπ+π

（1）化简可得f(x)=(sin(x/2))^2+((√3)/2)sinx-0.5f'(x)=sin(x/2)cos(x/2)+((√3)/2)cosx=sinx+√3cosx=0√3cosx=-si