數學問題集(56)---數學雜題(三)

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