標題:
（急）f.4 A MATHS
發問:
第1條.) sin3x+cos3x=sin2x+cos2x sin3x-sin2x=cos2x-cos3x 2cos(3x+2x/2)sin(3x-2x/2)=-2sin(2x+3x/2)sin(2x-3x/2) cos(3x+2x/2)sin(3x-2x/2)= -sin(2x+3x/2)sin(2x-3x/2) cos(5x/2)sin(x/2)= -sin(5x/2)sin(-x/2) ... 顯示更多 第1條.) sin3x+cos3x=sin2x+cos2x sin3x-sin2x=cos2x-cos3x 2cos(3x+2x/2)sin(3x-2x/2)=-2sin(2x+3x/2)sin(2x-3x/2) cos(3x+2x/2)sin(3x-2x/2)= -sin(2x+3x/2)sin(2x-3x/2) cos(5x/2)sin(x/2)= -sin(5x/2)sin(-x/2) cos(5x/2)sin(x/2)=sin(5x/2)sin(x/2)<~E一part係朋友教我,我想問下呢點解個負可以唔見左．．．． cos(5x/2)sin(x/2)-sin(5x/2)sin(x/2)=0 我做到E度就唔識再點計落去＞3＜有冇人識計可以寫埋步驟出黎比我睇ｔｈｘ 第2條.)O and A are the points (0,0)and(6,0) respectively. P(x,y)is a variable point such that PO+PA =10 . Find the equation of the locus of P , giving the answer in the form ax^2+by^2+cx+d=0 . (√(x-0)^2+(y-0)^2) +( √(x-6)^2+(y-0)^2 )= 10 ((√x^2+y^2 )+(√(x-6)^2+y^2))^2=100 ((√x^2+y^2)^2+2(√x^2+y^2)(√(x-6)^2+y^2)+(√(x-6)^2+y^2)^2=100 我做到E度就唔識再點計落去＞3＜有冇人識計可以寫埋步驟出黎比我睇ｔｈｘ THX
cos(5x/2)sin(x/2)= -sin(5x/2)sin(-x/2) sin(-x/2)個負可以抽出黎,變成-sin(x/2) tan同埋sin都有e個特點 而cos呢,就可以將個負消除, 姐係cos(-x/2)會變成cos(x/2) cos(5x/2)sin(x/2)-sin(5x/2)sin(x/2)=0 sin(x/2)[cos(5x/2)-sin(5x/2)]=0 sin(x/2)=0 or [cos(5x/2)-sin(5x/2)=0]----(1) x=0,360度 or [1-tan(5x/2)=0]------ (1)成條式除cos(5x/2) or 5x/2=45,225 or x=18度,90度 所以x=0,18,90,360度 第二條 ((√x^2+y^2)^2+2(√x^2+y^2)(√(x-6)^2+y^2)+(√(x-6)^2+y^2)^2=100 x^2+y^2+x^2+12x+36+y^2+(2√(x^2+y^2)(x^2+12x+36+y^2)=100 2x^2+2y^2-12x+36+2(√x^4-12x^3+36x^2+x^2y^2+x^2y^2-12xy^2+36y^2 +y^4)=100 x^2+y^2-6x+18+(√x^4+2x^2y^2+y^4-12x^3-12xy^2+36x^2+36y^2)=50 (√x^4+2x^2y^2+y^4-12x^3-12xy^2+36x^2+36y^2)=50-(x^2+y^2-6x+18) x^4+2x^2y^2+y^4-12x^3-12xy^2+36x^2+36y^2=[50-(x^2+y^2-6x+18)]^2 x^4+2x^2y^2+y^4-12x^3-12xy^2+36x^2+36y^2=2500-2(50)(x^2+y^2-6x+18)+(x^2+y^2-6x+18)(x^2+y^2-6x+18) x^4+2x^2y^2+y^4-12x^3-12xy^2+36x^2+36y^2=2500-100x^2-100y^2+600x-1800+x^4+x^2y^2-6x^3+18x^2+x^2y^2+y^4-6xy^2+18y^2-6x^3-6xy^2+36x^2-108x+18x^2+18y^2-108x+324 x^4+2x^2y^2+y^4-12x^3-12xy^2+36x^2+36y^2=376-10x^2-74y^2+x^4+y^4-2x^2y^2-12x^3-12xy^2+384x 46x^2+110y^2+384x+376=0 23x^2+55y^2+192x+188=0------answer(i think) 建議你把一切寫在白紙上才研究 因為數式太複雜
其他解答:
cos(5x/2)sin(x/2)= -sin(5x/2)sin(-x/2) because sin(-A) = -sinA sin(-x/2) = -sin(x/2) -sin(5x/2)sin(-x/2) = [-sin(5x/2)] [-sin(x/2)] = sin(5x/2)sin(x/2) -------------------------------------------------------------------------- cos(5x/2)sin(x/2)-sin(5x/2)sin(x/2)=0 [cos(5x/2)-sin(5x/2)]sin(x/2)=0 cos(5x/2)-sin(5x/2) = 0 or sin(x/2)=0 tan(5x/2) = 0 or sin(x/2)=0 識得搞埋佢啦 -------------------------------------------------------------------------- (√(x-0)^2+(y-0)^2) +( √(x-6)^2+(y-0)^2 )= 10 (√x^2+y^2) + (√(x-6)^2+y^2 )= 10 (√(x-6)^2+y^2 )= 10 - (√x^2+y^2) (x-6)^2+y^2 = 100 - 20(√x^2+y^2) + x^2+y^2 -12x + 36 = 100 - 20(√x^2+y^2) 20(√x^2+y^2) = 12x + 64 5(√x^2+y^2) = 3x + 16 25(x^2+y^2) = 9x^2 + 256 + 96x 16x^2 + 25y^2 - 96x - 256 = 0FAD2A23AB937987B
（急）f.4 A MATHS
發問:
第1條.) sin3x+cos3x=sin2x+cos2x sin3x-sin2x=cos2x-cos3x 2cos(3x+2x/2)sin(3x-2x/2)=-2sin(2x+3x/2)sin(2x-3x/2) cos(3x+2x/2)sin(3x-2x/2)= -sin(2x+3x/2)sin(2x-3x/2) cos(5x/2)sin(x/2)= -sin(5x/2)sin(-x/2) ... 顯示更多 第1條.) sin3x+cos3x=sin2x+cos2x sin3x-sin2x=cos2x-cos3x 2cos(3x+2x/2)sin(3x-2x/2)=-2sin(2x+3x/2)sin(2x-3x/2) cos(3x+2x/2)sin(3x-2x/2)= -sin(2x+3x/2)sin(2x-3x/2) cos(5x/2)sin(x/2)= -sin(5x/2)sin(-x/2) cos(5x/2)sin(x/2)=sin(5x/2)sin(x/2)<~E一part係朋友教我,我想問下呢點解個負可以唔見左．．．． cos(5x/2)sin(x/2)-sin(5x/2)sin(x/2)=0 我做到E度就唔識再點計落去＞3＜有冇人識計可以寫埋步驟出黎比我睇ｔｈｘ 第2條.)O and A are the points (0,0)and(6,0) respectively. P(x,y)is a variable point such that PO+PA =10 . Find the equation of the locus of P , giving the answer in the form ax^2+by^2+cx+d=0 . (√(x-0)^2+(y-0)^2) +( √(x-6)^2+(y-0)^2 )= 10 ((√x^2+y^2 )+(√(x-6)^2+y^2))^2=100 ((√x^2+y^2)^2+2(√x^2+y^2)(√(x-6)^2+y^2)+(√(x-6)^2+y^2)^2=100 我做到E度就唔識再點計落去＞3＜有冇人識計可以寫埋步驟出黎比我睇ｔｈｘ THX
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最佳解答:cos(5x/2)sin(x/2)= -sin(5x/2)sin(-x/2) sin(-x/2)個負可以抽出黎,變成-sin(x/2) tan同埋sin都有e個特點 而cos呢,就可以將個負消除, 姐係cos(-x/2)會變成cos(x/2) cos(5x/2)sin(x/2)-sin(5x/2)sin(x/2)=0 sin(x/2)[cos(5x/2)-sin(5x/2)]=0 sin(x/2)=0 or [cos(5x/2)-sin(5x/2)=0]----(1) x=0,360度 or [1-tan(5x/2)=0]------ (1)成條式除cos(5x/2) or 5x/2=45,225 or x=18度,90度 所以x=0,18,90,360度 第二條 ((√x^2+y^2)^2+2(√x^2+y^2)(√(x-6)^2+y^2)+(√(x-6)^2+y^2)^2=100 x^2+y^2+x^2+12x+36+y^2+(2√(x^2+y^2)(x^2+12x+36+y^2)=100 2x^2+2y^2-12x+36+2(√x^4-12x^3+36x^2+x^2y^2+x^2y^2-12xy^2+36y^2 +y^4)=100 x^2+y^2-6x+18+(√x^4+2x^2y^2+y^4-12x^3-12xy^2+36x^2+36y^2)=50 (√x^4+2x^2y^2+y^4-12x^3-12xy^2+36x^2+36y^2)=50-(x^2+y^2-6x+18) x^4+2x^2y^2+y^4-12x^3-12xy^2+36x^2+36y^2=[50-(x^2+y^2-6x+18)]^2 x^4+2x^2y^2+y^4-12x^3-12xy^2+36x^2+36y^2=2500-2(50)(x^2+y^2-6x+18)+(x^2+y^2-6x+18)(x^2+y^2-6x+18) x^4+2x^2y^2+y^4-12x^3-12xy^2+36x^2+36y^2=2500-100x^2-100y^2+600x-1800+x^4+x^2y^2-6x^3+18x^2+x^2y^2+y^4-6xy^2+18y^2-6x^3-6xy^2+36x^2-108x+18x^2+18y^2-108x+324 x^4+2x^2y^2+y^4-12x^3-12xy^2+36x^2+36y^2=376-10x^2-74y^2+x^4+y^4-2x^2y^2-12x^3-12xy^2+384x 46x^2+110y^2+384x+376=0 23x^2+55y^2+192x+188=0------answer(i think) 建議你把一切寫在白紙上才研究 因為數式太複雜
其他解答:
cos(5x/2)sin(x/2)= -sin(5x/2)sin(-x/2) because sin(-A) = -sinA sin(-x/2) = -sin(x/2) -sin(5x/2)sin(-x/2) = [-sin(5x/2)] [-sin(x/2)] = sin(5x/2)sin(x/2) -------------------------------------------------------------------------- cos(5x/2)sin(x/2)-sin(5x/2)sin(x/2)=0 [cos(5x/2)-sin(5x/2)]sin(x/2)=0 cos(5x/2)-sin(5x/2) = 0 or sin(x/2)=0 tan(5x/2) = 0 or sin(x/2)=0 識得搞埋佢啦 -------------------------------------------------------------------------- (√(x-0)^2+(y-0)^2) +( √(x-6)^2+(y-0)^2 )= 10 (√x^2+y^2) + (√(x-6)^2+y^2 )= 10 (√(x-6)^2+y^2 )= 10 - (√x^2+y^2) (x-6)^2+y^2 = 100 - 20(√x^2+y^2) + x^2+y^2 -12x + 36 = 100 - 20(√x^2+y^2) 20(√x^2+y^2) = 12x + 64 5(√x^2+y^2) = 3x + 16 25(x^2+y^2) = 9x^2 + 256 + 96x 16x^2 + 25y^2 - 96x - 256 = 0FAD2A23AB937987B
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