香港介紹ppt

標題:

有a.maths 唔識做

發問:

find the distance between (a cosθ, b sinθ) , (b cosθ , a sin θ)

最佳解答:

• What is the definition of a clause-
• 1)If a= 5, b= - 4 and c= - 3,
• 中國北方d女人靚d or 香港d女人靚d-

 

此文章來自奇摩知識+如有不便請留言告知

Distance between them = root [ (acosθ- bcosθ)2 (bsinθ- asinθ)2 ] = root { [(a-b) cos θ]2 [(b-a) sinθ]2 } = root [ (a-b)2 cos2 θ (b-a)2 sin2θ ] = root [ (a-b)2 cos2 θ (a-b)2 sin2θ ] = root [ (a-b)2 (cos2 θ sin2θ) ] = root [ (a-b)2 ( 1 ) ] = root (a-b)2 =│a-b│ Notes : (1) (a-b)2 = a2 - 2ab b2 = b 2 - 2ba a2 = (b-a)2 (2) root ( a2 ) = ( root a )2 =│a│,where denotes absolute value of a │a│= a ,if a≥0 and -a , if a〈 0 in simple, │a│must be positive with magnitude of a it is also important to state that the answer must be expressed as │a-b│ , otherwise marks will be deducted. the answer such as root(a-b)2 is not acceptable for final answer

其他解答:

我睇完waiwaikeith個答案學倒d野^^ thanks|||||by using distance formula, distance between 2 points (x1, y1) and (x2, y2) = sqrt[(x1-x2)2+(y1-y2)2] thus, the distance between (acosθ, bsinθ) and (bcosθ, asinθ) = sqrt[(acosθ-bcosθ)2+(bsinθ-asinθ)2] = sqrt[cos2θ(a-b)2+sin2θ(b-a)2] = sqrt[cos2θ(a-b)2+sin2θ(a-b)2] ((a-b)2 = (b-a)2) = sqrt[(cos2θ+sin2θ)(a-b)2] = sqrt(a-b)2 = a-b or b-a// as distance is always larger than 0, when a is larger than b, then the distance = a-b// when b is larger than a, then the distance = b-a//|||||Use distance formula The distance between=sqrt[(a2-b2)cos2θ+(b2-a2)sin2θ] =sqrt[a2-b2+(2b2-2a2)sin2θ] =sqrt[a2-b2-2sin2θ(a2-b2)] =sqrt[(a2-b2)(1-2sin2θ)] =sqrt[(a2-b2)cos2θ]