數學問題:三角數值問題

考慮 1 的 2n+1 個 2n+1 次方根 : x²ⁿ⁺¹ = 1
由棣美弗定理 , x = cos 2kπ/(2n+1) + i sin 2kπ/(2n+1) , (k = 0 , 1 , 2 , ... , 2n)
故 x²ⁿ⁺¹ - 1 的複分解=
( x - cos 2(0)π/(2n+1) - i sin 2(0)π/(2n+1) )
*
( x - cos 2(1)π/(2n+1) - i sin 2(1)π/(2n+1) )
( x - cos 2(2n)π/(2n+1) - i sin 2(2n)π/(2n+1) )
*
( x - cos 2(2)π/(2n+1) - i sin 2(2)π/(2n+1) )
( x - cos 2(2n-1)π/(2n+1) - i sin 2(2n-1)π/(2n+1) )
*
( x - cos 2(3)π/(2n+1) - i sin 2(3)π/(2n+1) )
( x - cos 2(2n-2)π/(2n+1) - i sin 2(2n-2)π/(2n+1) )
*
......................
*
( x - cos 2(n)π/(2n+1) - i sin 2(n)π/(2n+1) )
( x - cos 2(n+1)π/(2n+1) - i sin 2(n+1)π/(2n+1) )
=

(x - 1)
*
( x - cos 2(1)π/(2n+1) - i sin 2(1)π/(2n+1) )
( x - cos 2(1)π/(2n+1) + i sin 2(1)π/(2n+1) )
*
( x - cos 2(2)π/(2n+1) - i sin 2(2)π/(2n+1) )
( x - cos 2(2)π/(2n+1) + i sin 2(2)π/(2n+1) )
*
( x - cos 2(3)π/(2n+1) - i sin 2(3)π/(2n+1) )
( x - cos 2(3)π/(2n+1) + i sin 2(3)π/(2n+1) )
*
......................
*
( x - cos 2(n)π/(2n+1) - i sin 2(n)π/(2n+1) )
( x - cos 2(n)π/(2n+1) + i sin 2(n)π/(2n+1) )
對以上每個配對利用平方差公式得

x²ⁿ⁺¹ - 1
=
(x - 1) (x² - 2x cos 2π/(2n+1) + 1) (x² - 2x cos 4π/(2n+1) + 1)
(x² - 2x cos 6π/(2n+1) + 1) ... (x² - 2x cos 2nπ/(2n+1) + 1)
即
(x²ⁿ⁺¹ - 1) / (x - 1)
=
x²ⁿ + x²ⁿ⁻¹ + ... + x + 1
=
(x² - 2x cos 2π/(2n+1) + 1) (x² - 2x cos 4π/(2n+1) + 1)
(x² - 2x cos 6π/(2n+1) + 1) ... (x² - 2x cos 2nπ/(2n+1) + 1) ... ☆
令 x = 1 , 由 ☆ 即得
2n + 1
= 2ⁿ (1 - cos 2π/(2n+1)) (1 - cos 4π/(2n+1)) (1 - cos 6π/(2n+1)) ...
(1 - cos 2nπ/(2n+1))
利用公式 2sin² θ = 1 - cos2θ 得
2n + 1 = 2²ⁿ sin² π/(2n+1) sin² 2π/(2n+1) sin² 3π/(2n+1) ... sin² nπ/(2n+1)
開平方得
sin π/(2n+1) sin 2π/(2n+1) sin 3π/(2n+1) ... sin nπ/(2n+1) = √(2n+1) / 2ⁿ ...(1)
再令 x = - 1 , 由 ☆ 即得
1
= 2ⁿ (1 + cos 2π/(2n+1)) (1 + cos 4π/(2n+1)) (1 + cos 6π/(2n+1)) ...
(1 + cos 2nπ/(2n+1))
利用公式 2cos² θ = 1 + cos2θ 得
1 = 2²ⁿ cos² π/(2n+1) cos² 2π/(2n+1) cos² 3π/(2n+1) ... cos² nπ/(2n+1)
開平方得
cos π/(2n+1) cos 2π/(2n+1) cos 3π/(2n+1) ... cos nπ/(2n+1) = 1 / 2ⁿ ...(2)
(1) / (2) :tan π/(2n+1) tan 2π/(2n+1) tan 3π/(2n+1) ... tan nπ/(2n+1) = √(2n+1)

得所欲證 , 故原式成立。