lim(2sinx cosx)∧(1 x)

f(x)=sin²x+sinxcosx=[1-cos(2x)]/2+sin(2x)/2=sin(2x)/2-cos(2x)/2+1/2=(√2/2)sin(2x-π/4)+1/2最小正周期T

f(x)=a(cos²x+sinxcosx)+b=a(cos²x-1/2+sinxcosx+1/2)+b=a(cos2x/2+sin2x/2)+b=a根号下2sin(2x+π/4)

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/

sin²x-sinxcosx+2cos²x=1-sinxcosx+cos²x=1-½sin2x+(1+cos2x)/2=3/2+½cos2x-

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

y=2(cosx)^2+2√3sinxcosx=cos2x+1+2√3sinxcosx=cos2x+√3sin2x+1=2[1/2cos2x+√3/2sin2x)+1=2sin(2x+π/6)+1

趣味数学sin2X/2ncosX=2sinxcosx/2ncosX=six6

sinxcosx=(1/2)*sin（2x）=(1/2)*{2tanx/[1+(tanx)^2]}=2/5

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

1.lim(1+1/x)=1+0=1x→∞2.lim(1-3x)∧1/(2x)x→0=lim(1-3x)^[(-1/3x)*(-3/2)]x→0=lim[(1-3x)^(-1/3x)]^(-3/2)x

解题思路：利用三角函数正弦的和公式sin（x+x）可得结果解题过程：解：因为sin2x=sin（x+x）=sinxcosx+cosxsinx=2sinxcosx,所以y=2sinxcosx=sin2x

分两部分求2sin2x=4sinxcosx注：sin2x=2sinxcosx=4sinxcosx/{(cosx)^2+(sinx)^2}注：{(cosx)^2+(sinx)^2=1=4tanx/{1+

(1-2sinxcosx)(1+2sinxcosx)=(sin²x+cos²x-2sinxcosx)(sin²x+cos²x+2sinxcosx)=(sinx-

y=sinx+cosx+2sinxcosx+2=sinx+cosx+2sinxcosx+1+(sinx)^2+(cosx)^2=(sinx+cosx)^2+sinx+cosx+1=(sinx+cosx

n趋向什么呢?假设是无限吧lim[n→∞](n+1)/(n+2)=lim[n→∞](1+1/n)/(1+2/n)=(1+0)/(1+0)=1lim[n→∞](n²-1)/(2n²+

原式等于（3sinXcosX+cos²x-sin²x）/（sin²x+cos²x）再同时除以cos²x就行了

第一题用重要极限进行换元求解.第二题用罗比塔法则,或用等价无穷小代换

⑴接着你做的,sin(π/2)+cos(π/2)=1+0=1⑵f(α/2)=sinα+cosα=√2/2两边平方得1+2sinαcosα=1/2sin2α=-1/22α=150°,α=75°sinα=

(sinx)^2-sinxcosx-2(cosx)^2=0(sinx-2cosx)(sinx+cosx)=0sinx=2cosxtanx=2x=atan(2)+k*pi,k为整数或sinx=-cosx