(2sinx cosx)^1 x的极限 用洛必达法则

y=√2/2*sin2x+(1+cos2x)/2-1/2=√2/2*sin2x+1/2*cos2x=√3/2*sin(2x+z)其中tanz=(1/2)/(√2/2)=√2/2所以T=2π/2=π

y=sinxcosx-cos^2x=1/2sin2x-1/2(1+cos2x)=1/2(sin2x-cos2x-1)=1/2[√2*sin(2x-派/4)-1]=√2/2*sin(2x-派/4)-1/

y=√3sin2x-cos2x=2sin(2x-30°)ymin=-2

f(x)=(1-cos2x)/2+(√3/2)sin2x+1/2=(√3/2)sin2x-(1/2)cos2x+1=sin(2x-π/6)+1（1）最小正周期T=2π/2=π（2）f(x)max=2,

f(x)=√3sinxcosx+cos?x=√3/2sin2x+(cos2x+1)/2=√3/2sin2x+1/2cos2x+1/2=sin(2x+π/6)+1/2f(x)的最小正周期为2π/2=π∵

sin^2xtanx+cos^2x/tanx+2sinxcosx-(1+cosx/sinxcosx)=sin^3x/cosx+cos^3x/sinx+2sinxcosx-(1+cosx/sinxcos

f(x)=√3/2sin2x-1/2(1-cos2x)+1/2=√3/2sin2x-1/2+1/2cos2x+1/2=√3/2sin2x+1/2cos2x=sin(2x+π/3)

x=0代入f(0)=cos0-sin0+2(3sin0cos0+1)=1-0+2(0+1)=3

2*(sinxcosx-cos平方x)+1=2sinxcosx-2cos²x+1=2sinxcosx-cos2x=sin2x-cos2x=√2sin(2x-45°)

y=sin^x+2sinxcosx=1/2-cos2x/2+sin2x=根号下(5/4)*[2sin2x/根号5-cos2x/根号5]+1/2设cosa=2/根号5,sina=-1/根号5上式=根号下

y=1/2sin(2X+TT/3)-sinXcosX=cos(2x+TT/6)sin(TT/6)=1/2cos(2x+TT/6)单调递减区间是2kπ≤2x+π/6≤2kπ+πkπ-π/12≤x≤kπ+

y=1/2sin(2x+π/3)-sinxcosx=1/2(1/2sin2x+√3/2cos2x)-1/2sin2x=1/2(-1/2sin2x+√3/2cos2x)=1/2(cosπ/6cos2x-

g

sin2x=(2tanx)/(1+tan^2x)=4/5(1)sinxcosx=1/2sin2x=2/5(2)(2sin^2x-3sinxcosx-4cos^2x)/(sinxcosx)=2tanx-

y=sin²x+√3sinxcosx-1=[1-cos(2x)]/2+(√3/2)sin(2x)-1=(√3/2)sin(2x)-(1/2)cos(2x)-1/2=sin(2x-π/6)-1

f(x)=2sinxcosx-(2cos²x-1)=sin2x-cos2x=√2sin(2x-π/4)所以值域是[-√2,√2]

首先把分母化为1=(sinx)^2+(cosx)^2所以原式就为[(sinx)^2+(cosx)^2]/[2sinxcosx+(cosx)^2],然后分子分母同除以一个(cosx)^2式子就可以化为[

=sinxcosx/1+sinxcosx/sin^2x+sinxcosx/cos^2x=sinxcosx/1+sinxcosx/1=sinxcosx/2=2sinxcosx/4=sin2x/4sin2

sinx+cosx/sinx-cosx=2化简得sinx=3cosxsin²=9cos²x解出sin²=9/10sinxcosx=3/10sin^2x+2sinxcosx