tanx-sinx x^2sinx求极限

tanx=根号2/2,所以sinx=根号3/3,cosx=根号6/3.因为tanx=sinx/cosx,所以原试等于（根号3/3+根号6/3)^2-cos^2x/tanx,代入得1+根号2/3-(2/

第一题

有sin(x-45°)=√2／4=sinxcos45°-cosxsin45°,得sinx-cosx=0.5,两边平方得1-2sinxcosx=0.25.sinxcosx=3／8.tanx+1／tanx

(1-2sinxcosx)/(cos²x-sin²x)=(sin²x+cos²x-2sinxcosx)/(cos²x-sin²x)=(cos

证明左=(sin²x+cos²x+2sinxcosx)/[(cosx+sinx)(cosx-sinx)]=(cosx+sinx)²/[(cosx+sinx)(cosx-s

lim(x→0)[(tanx-sinx)/(sin^22x)]=lim(x→0)[tanx(1-cosx)/(2x)^2]=lim(x→0)[x*x^2/2]/(2x)^2=0

证明：1-2×sinx×cosx/cos∧2x-sin∧2x=[(sinx)^2-2sinx*cosx+(cosx)^2]/[(cosx)^2-(sinx)^2]分子分母同时除以(cosx)^2=[(

左边=(sin²x+cos²x+2sinxcosx)/(sinx+cosx)(sinx-cosx)=(sinx+cosx)²/(sinx+cosx)(sinx-cosx)

(1-2sinxcosx)/(cos²x-sin²x)=(sin²x+cos²x-2sinxcosx)/(cos²x-sin²x)=(cos

原式=1/(sinxcosx)=(sin²x+cos²x)/sinxcosx此时分子分母同除以cos²x原式=(tan²x+1)/tanx是不是哪里有问题?

(sinx+tanx)/(cos^2x+sin^2x+cosx)=(sinx+sinx/cosx)/(1+cosx)=sinx(cosx+1)/[cosx(1+cosx)]=sinx/cosx=tan

没做完不好意思,要睡觉了

∵f（x）=2tanx-2sin2x2−1sinx2cosx2=2 (sinxcosx+cosxsinx)=2sinxcosx＝4sin2x∴f(π12) ＝4sinπ6＝8故答案

(sinx)^2tanx=[1-(cosx)^2]tanx=tanx-(cosx)^2tanx=tanx-(cosx)^2*sinx/cosx=tanx-sinxcosx(cosx)^2cotx=[1

(sinx)^2tanx=[1-(cosx)^2]tanx=tanx-(cosx)^2tanx=tanx-(cosx)^2*sinx/cosx=tanx-sinxcosx(cosx)^2cotx=[1

f(x)=(1+1/tanx)*(sinx)^2-2sin(x+π/2)sin(x-π/4)=(1+cosx/sinx)*(sinx)^2+2sin(x+π/4)cos[(x-π/4)+π/2]=(s

(cos²x-sin²x)/(1-2sinxcosx)1=cos²x+sin²x=(cos²x-sin²x)/(cos²x+sin

这里用到:(sin)^2+(cosx)^2=1,原式=(cosx-sinx)^2/(cosx+sinx)(cosx-sinx)=(cosx-sinx)/(cosx+sinx)=(1-tanx)/(1+

y1='log((2*tan(x)+1)/(tan(x)+2))'%log在matlab中求自然对数y11=diff(y1)%求导simple(y11)%化简y2='sin(e^(x^2+3*x-2)

tanxtan2x/(tan2x-tanx)=sinxsin2x/(sin2xcosx-sinxcos2x)=sinxsin2x/sin(2x-x)=sin2x(tanxtan2x/(tan2x-ta