(tanx=2,sin²x sinxcosx-2cos²=?

来源:学生作业帮助网 编辑:作业帮 时间:2024/07/13 21:31:12
(1+tanx)/(1-tanx)=3+2根号2,求(sin x+cosx)^2-(cos^3x)/sinx

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/

已知sin(x-45°)=四分之根号2,求sinxcosx,tanx+1/tanx

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

求证(1-2sinxcosx)/(cos^2x-sin^2x)=(1-tanx)/(1+tanx)

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

求证1+2sinxcosx/cos^2x-sin^2x=1+tanx/1-tanx

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

1-2×sinx×cosx/cos∧2 x -sin∧2 x=1-tanx/1+tanx

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=[(

证明:(1+2sinXcosX)/(sin^2X-cos^2X)=(tanX+1)/(tanX-1)

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

求证(1-2sinXcosX)/(cosX^2-sin^2X)=(1-tanX)/(1+tanX)

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

若f(x)=2tanx-2sin

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

已知sin(x-45°)=√2/4.求(1)sinxcosx值,(2)求tanx+(1/tanx)值

1、sin(x-45)=sinxcos45-cosxsin45=√2/2*(sinx-cosx)=√2/4sinx-cosx=1/2平方sin²x+cos²x-2sinxcosx=

sin^2x*tanx+cos^2x*cotx+2sinx*cosx=tanx+cotx

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

t->0,lim[tan(sinx)-sin(tanx)]/(tanx-sinx)=?

原式=lim{x->0}{tan(sinx)-tan(tanx)[1+cos(tanx)-1]}/(tanx-sinx)=lim{x->0}{tan(sinx-tanx)[1+tan(sinx)tan

提问数学难题求证:sin^2x*tanx+cos^2x/tanx+2sinx*cosx=tanx+1/tanx

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

已知函数fx=(1+1/tanx)sin^x-2sin(x+π/4)sin(x-π/4)

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

(cos^2x-sin^2x)/(1-2sinxcosx)=(1+tanx)/(1-tanx)

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

1-2sinx cosx /COS^2X-SIN^2X =1-tanx/1+tanx 求证

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+

matlab求解导数y=ln((2tanx+1)/(tanx+2)),y=sin(e^(x^2+3x-2))

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)

求证(tanxtan2x/tan2x-tanx)/(tan2x-tanx)+√3(sin^2x-cos^2x)=2sin

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

(1-2sinxcosx)/[(cos)^2]a-(sin^2)x=(1-tanx)/(1+tanx)证明

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