Mathematics F.3-F.4 (10)

1)x^4 - 5x^2 - 36= (x^2 - 9)(x^2 + 4)= (x - 3)(x + 3)(x^2 + 4)
2)h^6 - 1= (h^3)^2 - 1= (h^3 - 1)(h^3 + 1)= (h - 1)(h^2 + h + 1) (h + 1)(h^2 - h + 1)= (h - 1)(h + 1)(h^2 - h + 1)(h^2 + h + 1)
3)x^3 - (x - y)^3= (x - (x-y)) (x^2 + x(x-y) + (x-y)^2)= y(x^2 + x^2 - xy + x^2 - 2xy + y^2)= y(3x^2 - 3xy + y^2)
4)x^8 - 1= (x^4)^2 - 1= (x^4 - 1)(x^4 + 1)= (x^2 - 1)(x^2 + 1) (x^4 + 1)= (x - 1)(x + 1)(x^2 + 1)(x^4 + 1)
5)cos^2 1°＋cos^2 2°＋cos^2 3°＋...＋cos^2 89°
= (cos^2 1° + cos^2 89°) + (cos^2 2° + cos^2 88°) + (cos^2 3° + cos^2 87°)+ (cos^2 44° + cos^2 46°) + (cos^2 45° + cos^2 45°)
= (cos^2 1° + sin^2 1°) + (cos^2 2° + sin^2 2°) + (cos^2 3° + sin^2 3°) + ...+ (cos^2 44° + sin^2 44°) + (cos^2 45° + sin^2 45°)= 1 + 1 + 1 + ... + 1 + 1= 45

2010-08-01 17:32:20 補充：
4)

χ^1024 - 1

= (x^128)^8 - 1

= (x^128 - 1) (x^128 + 1)(x^256 + 1)(x^512 + 1)

= [(x^16)^8 - 1] (x^128 + 1)(x^256 + 1)(x^512 + 1)

= (x^16 - 1)(x^16 + 1)(x^32 + 1)(x^64 + 1) (x^128 + 1)(x^256 + 1)(x^512 + 1)

= [(x^8)^2 - 1] (x^16 + 1)(x^32 + 1)(x^64 + 1) (x^128 + 1)(x^256 + 1)(x^512 + 1)

2010-08-01 17:32:26 補充：
= (x^8 - 1) (x^8 + 1) (x^16 + 1)(x^32 + 1)(x^64 + 1) (x^128 + 1)(x^256 + 1)(x^512 + 1)

= (x - 1)(x + 1)(x^2 + 1)(x^4 + 1) (x^8 + 1) (x^16 + 1)(x^32 + 1)(x^64 + 1) (x^128 + 1)(x^256 + 1)(x^512 + 1)

2010-08-01 18:14:26 補充：
Corrections :

5)

cos^2 1°＋cos^2 2°＋cos^2 3°＋...＋cos^2 89°

= (cos^2 1° + cos^2 89°) + (cos^2 2° + cos^2 88°) + (cos^2 3° + cos^2 87°)
+ (cos^2 44° + cos^2 46°) + cos^2 45°

= (cos^2 1° + sin^2 1°) + (cos^2 2° + sin^2 2°) + (cos^2 3° + sin^2 3°) + ...
+ (cos^2 44° + sin^2 44°) + cos^2 45°

2010-08-01 18:14:30 補充：
= (1 + 1 + 1 + ... + 1) + cos^2 45°

= 44 + 1/2

= 44.5