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求证Cosθ=sinθ╱1-Cotθ

sina+cosa=(√3+1)/2 平方 1+2sinacosa=1+√3/2 m/2=sinacosa=√3/4 m=√3/2 sinθ/(1-cotθ)+cos/(1-tanθ) =sina/(1-cosa/sina)+cosa/(1-sina/cosa) =sin^2a/(sina-cosa)-cos^2a/(sina/cosa) =(sin^2a-cos^2a)/(sina-cosa) =sina+cosa =(√3+1)/2

(1+z)/(1-z)=z 解析: 1±z =(1+cosθ)±isinθ =2cos²(θ/2)±i2sin(θ/2)cos(θ/2) =2cos(θ/2)[cos(θ/2)±isin(θ/2)] ∴ (1+z)/(1-z) =[cos(θ/2)+isin(θ/2)]/[cos(θ/2)-isin(θ/2)] =e^(iθ/2)/[e^(-iθ/2)] =e^(iθ) =z

左边 =[1+(1/cos2θ)]/[sinθ/cos2θ] =[(1+cos2θ)/(cos2θ)]×[cos2θ/sin2θ] =cos2θ/sin2θ =cot2θ =右边 则: (1+sec2θ)/tan2θ=cot2θ

用极坐标来做,令x=rcosθ,y=rsinθx²+y²≤a²即r²≤a²,r的范围是0到a,而θ的范围是0到2π那么原积分=∫(0到2π)dθ*∫(0到a)r²*rdr=2π*1/4*a^4=a^4*π/2=8π于是a^4=16解得a=2

由题意,S100=(1α+1β)100?1(1α+1β)?1=0?(1α+1β)100=11α+1β≠1?1α+1β=?1?α+βαβ=?1,∵α+β=-sin2θ,αβ=-sinθcotθ=-cosθ,∴2sinθ=?1?sinθ=?12,∵0<θ<2π,∴θ=7π6或11π6.

1+cscb+cotb)/(1+cscb-cotb)=(sinb+1+cosb)/(sinb+1-cosb) cscb+cotb=(1+cosb)/sinb ∵(sinb+1+cosb)*sinb=sin^2b+sinb+sinbcosb (sinb+1-cosb)*(1+cosb)=sinb+sinbcosb+1-cos^2b=sin^2b+sinb+sinbcosb ∴(sinb+1+cosb)*sinb=(sinb+1-cosb)*(1+cos...

倒数关系: tanα ·cotα=1 sinα ·cscα=1 cosα ·secα=1 商的关系: sinα/cosα=tanα=secα/cscα cosα/sinα=cotα=cscα/secα 平方关系: sin^2(α)+cos^2(α)=1 1+tan^2(α)=sec^2(α) 1+cot^2(α)=csc^2(α) 平常针对不同条件的常用的两个公式 sin^2(α)+cos...

反三角函数公式: arcsin(-x)=-arcsinx arccos(-x)=∏-arccosx arctan(-x)=-arctanx arccot(-x)=∏-arccotx arcsinx+arccosx=∏/2=arctanx...

反三角函数公式: arcsin(-x)=-arcsinx arccos(-x)=∏-arccosx arctan(-x)=-arctanx arccot(-x)=∏-arccotx arcsinx+arccosx=∏/2=arctanx...

csc^2x-1 =1/sin^2x-1 =(1-sin^2x)/sin^2x =cos^2x/sin^2x =cot^2x 扩展资料: 二倍角公式 sin2α=2sinαcosα tan2α=2tanα/(1-tan^2(α)) cos2α=cos^2(α)-sin^2(α)=2cos^2(α)-1=1-2sin^2(α) 同角三角函数的基本关系式 倒数关系:tanα ·cotα=1、sinα...

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