在不可使用計算機的情況下,證明sin(pi/7)sin(2pi/7)sin(4pi/7) = ((7)^1/2) /8 (平方根7除以8), pi=圓周率
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2011-05-13 10:49 pm
回答 (2)
2011-05-13 11:42 pm
✔ 最佳答案
首先, sin (4π/7) = sin (π - 4π/7) = sin (3π/7)所以:
sin (π/7) sin (2π/7) sin (4π/7) = sin (π/7) sin (2π/7) sin (3π/7)
考慮:
[sin(π/7) sin(2π/7) sin(3π/7)]²
= sin(π/7) sin(2π/7) sin(3π/7) * sin(π/7) sin(2π/7) sin(3π/7)
= sin(π/7) sin(2π/7) sin(3π/7) * sin(6π/7) sin(5π/7) sin(4π/7)
= [sin(π/7) sin(6π/7)] * [sin(2π/7) sin(5π/7)] * [sin(3π/7) sin(4π/7)]
= ½[cos(5π/7) - cos(π)] * ½[cos(3π/7) - cos(π)] * ½[cos(π/7) - cos(π)] ﹝積化和差﹞
= (1/8) [cos(5π/7) + 1] [cos(3π/7) + 1] [cos(π/7) + 1]
= (1/8) [ cos(5π/7)cos(3π/7)cos(π/7)
+ cos(5π/7)cos(3π/7) + cos(5π/7)cos(π/7) + cos(3π/7)cos(π/7)
+ cos(5π/7) + cos(3π/7) + cos(π/7)
+ 1 ]
= (1/8) { ½[cos(2π/7) + cos(8π/7)]cos(π/7)
+ ½[cos(2π/7) + cos(8π/7)] + ½[cos(4π/7) + cos(6π/7)]
+ ½[cos(2π/7) + cos(4π/7)]
+ cos(5π/7) + cos(3π/7) + cos(π/7)
+ 1 } ﹝積化和差﹞
= (1/8) { ½[cos(2π/7) - cos(π/7)]cos(π/7)
+ ½[cos(2π/7) - cos(π/7)] + ½[-cos(3π/7) - cos(π/7)]
+ ½[cos(2π/7) - cos(3π/7)]
- cos(2π/7) + cos(3π/7) + cos(π/7)
+ 1 }
= (1/16) { cos(2π/7)cos(π/7) - cos(π/7)cos(π/7)
+ cos(2π/7) - cos(π/7) - cos(3π/7) - cos(π/7) + cos(2π/7) - cos(3π/7)
- 2cos(2π/7) + 2cos(3π/7) + 2cos(π/7)
+ 2 }
= (1/16) { cos(2π/7)cos(π/7) - cos(π/7)cos(π/7) + 2 }
= (1/16) { ½[cos(π/7) + cos(3π/7)] - ½[1 + cos(2π/7)] + 2 } ﹝積化和差﹞
= (1/32) { cos(π/7) + cos(3π/7) - cos(2π/7) + 3 }
cos(π/7) + cos(3π/7) - cos(2π/7)
= sin(3π/7)[cos(π/7) + cos(3π/7) - cos(2π/7)] / sin(3π/7)
= [sin(3π/7)cos(π/7) + sin(3π/7)cos(3π/7)
- sin(3π/7)cos(2π/7)] / sin(3π/7)
= { ½[sin(2π/7) + sin(4π/7)] + ½[sin(0π/7) + sin(6π/7)]
- ½[sin(π/7) + sin(5π/7)] } / sin(3π/7)
= ½ { [sin(2π/7) + sin(3π/7)] + [0 + sin(π/7)]
- [sin(π/7) + sin(2π/7)] } / sin(3π/7)
= ½ { sin(3π/7) } / sin(3π/7)
= ½
所以
[sin(π/7) sin(2π/7) sin(3π/7)]²
= (1/32) { 1/2 + 3 }
= (1/64) { 1 + 6 }
= 7/64
sin(π/7) sin(2π/7) sin(3π/7) = √7 / 64
2011-05-13 21:43:11 補充：
最後一行應為:
sin(π/7) sin(2π/7) sin(3π/7) = √7 / 8, 不是 64
Thanks to sioieng
參考: 原創答案
2011-05-14 11:08 pm
Suggested solution:
z^n - 1 = (z - 1)[z^(n-1) + z^(n-2) + ... + 1]
The roots of z^(n-1) + z^(n-2) + ... + 1 are w, w^2,... w^(n-1)
Then z^(n-1) + z^(n-2) + ... + 1 = (z - w)(z - w^2)...(z - w^(n-1))
Set z = 1
Π |1 - w^k| = n (k from 1 to n - 1)
Since
1 - w^k = 1 - exp(i2kπ/n) = -2isin(kπ/n)exp(ikπ/n)
So, Π |-2isin(kπ/n)exp(ikπ/n)| = n (k from 1 to n - 1)
If n is odd. Let n = 2t + 1
2^(2t) Π sin(kπ/2t + 1) = 2t + 1
Π sin(kπ/2t + 1) = (2t + 1)/2^(2t)
As sin(kπ/n) = sin[(n - k)π/n]
Π sin(kπ/2t + 1) = √(2t + 1)/2^t (k from 1 to t)
Change back t to n
Π sin(kπ/2n + 1) = √(2n + 1)/2^n (k from 1 to n)
Finally, sub. n = 3
sin(π/7)sin(2π/7)sin(3π/7) = √7/8 or
sin(π/7)sin(2π/7)sin(4π/7) = √7/8
z^n - 1 = (z - 1)[z^(n-1) + z^(n-2) + ... + 1]
The roots of z^(n-1) + z^(n-2) + ... + 1 are w, w^2,... w^(n-1)
Then z^(n-1) + z^(n-2) + ... + 1 = (z - w)(z - w^2)...(z - w^(n-1))
Set z = 1
Π |1 - w^k| = n (k from 1 to n - 1)
Since
1 - w^k = 1 - exp(i2kπ/n) = -2isin(kπ/n)exp(ikπ/n)
So, Π |-2isin(kπ/n)exp(ikπ/n)| = n (k from 1 to n - 1)
If n is odd. Let n = 2t + 1
2^(2t) Π sin(kπ/2t + 1) = 2t + 1
Π sin(kπ/2t + 1) = (2t + 1)/2^(2t)
As sin(kπ/n) = sin[(n - k)π/n]
Π sin(kπ/2t + 1) = √(2t + 1)/2^t (k from 1 to t)
Change back t to n
Π sin(kπ/2n + 1) = √(2n + 1)/2^n (k from 1 to n)
Finally, sub. n = 3
sin(π/7)sin(2π/7)sin(3π/7) = √7/8 or
sin(π/7)sin(2π/7)sin(4π/7) = √7/8
收錄日期: 2021-04-23 22:28:06
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110513000051KK00438