close
標題:
數學問題集(56)---數學雜題(三)
發問:
最佳解答:
1)cos x = tan ycos2 x = tan2 y1 - sin2 x = (sin2 y) / (1 - sin2 y)同理 1 - sin2 y = (sin2 z) / (1 - sin2 z)1 - sin2 z = (sin2 x) / (1 - sin2 x)令 a = sin2 x , b = sin2 y , c = sin2 z , 則1 - a = b / (1 - b) ......(1)1 - b = c / (1 - c) ......(2)1 - c = a / (1 - a) ==> 2 - c = 1 / (1 - a) ==> (1 - a) = 1 / (2 - c) ......(3)(3) 代入 (1) :1 / (2 - c) = b / (1 - b) 1 + 1 / (2 - c) = 1 + b / (1 - b)(3 - c) / (2 - c) = 1 / (1 - b)(1 - b) = (2 - c) / (3 - c) , 代入 (2) :(2 - c) / (3 - c) = c / (1 - c) c(3 - c) = (c - 1)(c - 2)3c - c2 = c2 - 3c + 2c2 - 3c + 1 = 0c = sin2 z = (3 ± √5)/2 因 0 ≤ sin2 z ≤ 1 , sin2 z = (3 - √5)/2 因 sin36° = cos54° 2 sin18° cos18° = 4cos3 18° - 3cos18°2 sin18° = 4 cos2 18° - 32 sin18° = 4(1 - sin2 18°) - 34 sin2 18° + 2 sin18° - 1 = 0sin18° = (√5 - 1) / 4 故4 sin2 18° = 4 [(√5 - 1) / 4]2 = (3 - √5)/2 = sin2 z同理 (3 - √5)/2 = sin2 y (3 - √5)/2 = sin2 x即 sin2 x = sin2 y = sin2 z = 4 sin2 18°。 2) a, b互為相反數 , a + b = 0 c, d互為倒數 , cd = 1 |x| = 2 , x^2 = 4ax^2 + bx^2 - 2acd - 2b + 2cd + 2cdx^2= (a + b)x^2 - 2a - 2b + 2 + 2x^2= 0 - 2(a + b) + 2 + 2(4)= 0 - 0 + 2 + 8= 10 3)(2x2 - x - 1)3 = (a_6)x^6 + (a_5)x^5 + (a_4)x^4 + (a_3)x^3 + (a_2)x^2 + (a_1)x + (a_0)令x = 1 ,(2 - 1 - 1)3 = 0 = (a_6) + (a_5) + (a_4) + (a_3) + (a_2) + (a_1) + (a_0) .....(1)令x = - 1 ,(2 + 1 - 1)3 = 8 = (a_6) - (a_5) + (a_4) - (a_3) + (a_2) - (a_1) + (a_0) .....(2)(2) - (1) :2 (a_1 + a_3 + a_5) = 8(a_1 + a_3 + a_5) = 4 2010-10-12 14:41:51 補充： 修正 3) (1) - (2) : 2 (a_1 + a_3 + a_5) = 0 - 8 (a_1 + a_3 + a_5) = - 4
其他解答:
數學問題集(56)---數學雜題(三)
發問:
此文章來自奇摩知識+如有不便請留言告知
1. 證明：若 cos x = tan y, cos y = tan z, cos z = tan x，則 sin^2 x = sin^2 y = sin^2 z = 4 sin^2 18°。 2. 已知a, b互為相反數，c, d互為倒數，x的絕對值為2. 求 ax^2 + bx^2 - 2acd - 2b + 2cd + 2cdx^2 的值。 3. 若 (2x^2 - x - 1)^3 = (a_6)(x^6) + (a_5)(x^5) + ... + (a_1)x + a_0，求a_1 + a_3 + a_5 的值。最佳解答:
1)cos x = tan ycos2 x = tan2 y1 - sin2 x = (sin2 y) / (1 - sin2 y)同理 1 - sin2 y = (sin2 z) / (1 - sin2 z)1 - sin2 z = (sin2 x) / (1 - sin2 x)令 a = sin2 x , b = sin2 y , c = sin2 z , 則1 - a = b / (1 - b) ......(1)1 - b = c / (1 - c) ......(2)1 - c = a / (1 - a) ==> 2 - c = 1 / (1 - a) ==> (1 - a) = 1 / (2 - c) ......(3)(3) 代入 (1) :1 / (2 - c) = b / (1 - b) 1 + 1 / (2 - c) = 1 + b / (1 - b)(3 - c) / (2 - c) = 1 / (1 - b)(1 - b) = (2 - c) / (3 - c) , 代入 (2) :(2 - c) / (3 - c) = c / (1 - c) c(3 - c) = (c - 1)(c - 2)3c - c2 = c2 - 3c + 2c2 - 3c + 1 = 0c = sin2 z = (3 ± √5)/2 因 0 ≤ sin2 z ≤ 1 , sin2 z = (3 - √5)/2 因 sin36° = cos54° 2 sin18° cos18° = 4cos3 18° - 3cos18°2 sin18° = 4 cos2 18° - 32 sin18° = 4(1 - sin2 18°) - 34 sin2 18° + 2 sin18° - 1 = 0sin18° = (√5 - 1) / 4 故4 sin2 18° = 4 [(√5 - 1) / 4]2 = (3 - √5)/2 = sin2 z同理 (3 - √5)/2 = sin2 y (3 - √5)/2 = sin2 x即 sin2 x = sin2 y = sin2 z = 4 sin2 18°。 2) a, b互為相反數 , a + b = 0 c, d互為倒數 , cd = 1 |x| = 2 , x^2 = 4ax^2 + bx^2 - 2acd - 2b + 2cd + 2cdx^2= (a + b)x^2 - 2a - 2b + 2 + 2x^2= 0 - 2(a + b) + 2 + 2(4)= 0 - 0 + 2 + 8= 10 3)(2x2 - x - 1)3 = (a_6)x^6 + (a_5)x^5 + (a_4)x^4 + (a_3)x^3 + (a_2)x^2 + (a_1)x + (a_0)令x = 1 ,(2 - 1 - 1)3 = 0 = (a_6) + (a_5) + (a_4) + (a_3) + (a_2) + (a_1) + (a_0) .....(1)令x = - 1 ,(2 + 1 - 1)3 = 8 = (a_6) - (a_5) + (a_4) - (a_3) + (a_2) - (a_1) + (a_0) .....(2)(2) - (1) :2 (a_1 + a_3 + a_5) = 8(a_1 + a_3 + a_5) = 4 2010-10-12 14:41:51 補充： 修正 3) (1) - (2) : 2 (a_1 + a_3 + a_5) = 0 - 8 (a_1 + a_3 + a_5) = - 4
其他解答:
文章標籤
全站熱搜
留言列表
