三角恒等式 (1)

=
一些三角恒等式整理
2014 年 8 月 22 日
(1)
(2)
(3)
(4)
(5)
tan 210◦ + tan 250◦ + ta n 270◦ = 9
tan 410◦ + tan 450◦ + tan 470◦ = 59
ta n 610◦ + tan 650◦ + tan 670◦ = 433
tan 810◦ + tan 850◦ + tan 870◦ = 3251
cos 3
π
− cos 3 2π + cos 3 3π − cos 3 4π + cos 3 5π = 1
(6)
11 11 11 11 11 2
csc π − csc 2π + csc 3π
= 0
7 7 7
(7)
cos π − c os 2π + cos 3π = 1
7 7 7 2
(8)
cos 2 π + cos 2 2π + cos 2 3π = 5
(9)
7 7 7 4
π 2π 3π 1
(10)
cos 7 cos 7 cos 7 =
8
(11)
π sin 7
sin
π
2π sin
3π ¸
7 7 7 8 2π
3π ¸
tan 7 tan 7 tan 7
= 7
= =3
sin220◦+cos250◦+sin220◦cos250◦= (13)
3
cos220◦+sin250◦−cos220◦sin250◦= (14)
(15)
(16)
sin20◦sin40◦sin60◦sin80◦
3
16 cos20◦cos40◦cos60◦cos80◦
1
16
sin2
π
+ sin2
2π
+ sin2
3π
=
7
7 7 7 4
(17)
sin4π
+ sin4
2π
+ sin4
3π
=
21 7 7 7 16
(18)
sin6π
+ sin6
2π
+ sin6
3π
=
35 7 7 7 32
(19)
sin8π
+ sin8
2π
+ sin8
3π
=
245
(20)
7 7 7 256
cos4π
+ cos4
2π
+ cos4
3π
=
13 7 7 7 16
(21)
cos6π
+ cos6
2π
+ cos6
3π
=
19 7 7 7 32
(22)
cos8π
+ cos8
2π
+ cos8
3π
=
117
(23)
7 7 7 256
[
sin2k
π
+sin2k
2π
+sin2k
3π
]
−
[
cos2k
π
+cos2k
2π
+cos2k
3π
]
=
1
(k∈N)
(24)
7 7 7 7 7 7 2
tan2π
+ tan2
2π
+ tan2
3π
= 21 7 7 7
(25)
tan4π
+ tan4
2π
+ tan4
3π
= 371 7 7 7
4
4
4
4
2
∑ sin 2 i π =
11
∑ sin 4 i π =
33 ∑ sin 6 i π =
55 ∑ sin 8 i π =
385 i =1
sin 1 5 = tan 6 π + tan 6 2π + tan 6 3π
= 7077
7 7 7
(27)
tan 8 π + tan 8 2π + tan 8 3π
= 135779
7 7 7
(28)
(29)
i =1
11 4
(30)
i =1
11 16
(31)
i =1
11 32
(32)
i =1
11 256
(33)
(34)
(35)
sin 220◦ + sin 240◦ + sin 260◦ + sin 280◦ = 9
cos 220◦ + cos 240◦ + cos 260◦ + cos 280◦ =
7 tan 220◦ + tan 240◦ + tan 260◦ + tan 280◦ = 36 2n π 4n π
6n π 1
(36)
cos 7 + cos 7 + cos 7 = 2
(n ∈ N ∗)
n
∑
sin i ≤ 1
(37) 记 a n = cos 2n π + cos 4n π + cos 6n π , T n = cos n 2n π + cos n 4n π + cos n 6n π (n ∈ N )，则我们有
7
7
7
( 1 )n −1 ∑n
C k a 7
7
7
(
n 为奇数)
2 k =0 n
n −2k T n = ( 1 )n m ( 1 )n −1 m ∑−1 k
( )
(38)
3 · 2
C n + 2
k 0
C n a n −2k
n 为偶数
sin 47◦ + sin 61◦ − sin 11◦ − sin 25◦ = cos 7◦
5 5 5
= + = ∏ 4 ¸
∑ 1
= −
tan θ
∑
( )+
s i n
β cos α + β2
2
2
∑ cos i θ = 2 2
( ) (39)
(40)
(41)
(42)
(43)
1
tan 10◦
¸
3
sin 40◦
cos 7x + 7 cos 5x + 21 cos 3x + 35 cos x = 64cos 7x
¸
2 tan 18◦
+ tan 18◦
tan 12◦
+ ¸
3 tan 12◦ = 1
cos 6◦ cos 42◦ cos 66◦ cos 78◦
1
16
(44)
90◦
i =1◦
sin i =
1 45 × 6 10
(45)
(46)
n cot x cot 2n x i =1 sin 2i
x
tan θ tan 2θ + tan 2θ tan 3θ + · · · + tan (n − 1) θ tan n θ = tan n θ
− n
(47)
n 2i tan 2i θ = cot θ − 2n +1 cot 2n +1θ i =0
(48)
cos α + cos (α + β) + · · · + cos (α + n β) =
n 1 n
2
sin β n sin n θ cos n +1 θ
(49)
i =1 sin θ cos π + c os 3π + cos 5π = 1
9 7 7 2
(50)
cos π + c os 3π + c os 5π + cos 7π = 1
9 9 9 9 2
(51)
cos
π
cos 2π cos 3π cos 4π cos 5π = 1 11 11 11 11 11 32
4 ∑ ( )2 l n ∑
tan i π = ¸2n + 1 ∑
tan 2 i π = n (2n + 1)
∑
2 sin x = tan (2n + 1) x − tan x =
sin nx
∑
sin x =
sin nx
sin 2n = 2
n −1
2 (52)
(53)
cos 210◦ + cos 250◦ − sin 40◦ sin 80◦ = 3
(54)
cos θ cos 2θ cos 4θ · · · cos 2n sin 2n +1θ
θ =
2n +1 sin θ
(55)
i =1 n 2n + 1
(56)
i =1
2 s in x
2n + 1 cos x + cos (x + 2θ)
= tan (x + θ) − tan θ
(57)
n
(58)
k =1 cos x +
cos 2k x
n 2 2 cos n x + cos (n + 1) x
(59)
k =1 cos 2k x −
cos x
2 sin x
cos (n + 1) x − cos nx cos (x + 2θ) − cos x
= cot (x + θ) − cot θ
(60)
2 sin x
x + y x − y = tan − tan
(61)
cos x + cos y
n ∏
−1 2 2
k π n
(62)
k =1
2 sin
n ∑
−1 n = 2n −1
k π
n ln n
(63)
π n ln n < k =1
csc n ≤ 2 ln 2
(64)
n −1
k =1
csc k π n 2 2γ = π n ln n + π −
π
π n + O (1)
n ∏
−1
k π ¸
n
k =1
n
k=1
∑∏
sin
kπ
=
2n + 1
∑
csc2
kπ
=
2 (
n2−1
)
∑
sin2
kπ
= 2n + 1
(65)
n
¸(66)
k=1
n
2n + 1 2n
(67)
∑m(k=1
2n
i π
)m
3
i πx m m+1
(68) i =1
cos
m
n
cos
m
=
2m m
(69)
n k=1
2n + 1 4
2
∑
sin6k1π
sin4
k2π
sin2
k3π
= (2n + 1) (2n −3) (n −2) (6n −7) , n≥3
(70) k1=1
k1<k2<k3
2n + 12n + 12n + 112288
tan
2π
+ 4 sin
6π
=
√
13 + 2
¸
13
13 13
(71)
tan 4π
+ 4 sin
π=
√
13 − 2
¸
13 13 13
(72)
tan 3π
+ 4 sin
2π
=
¸
11
(73)
11
n∑−1
11
kππ
(74)
sin
k=1
n
= cot
2n
n∑−12k2π
¸
n (nπnπ
)
(75) k=1
sin
n
=
2 1 + c os 2
− sin
2
(76) 证明：n−1
k=1
cos
2k2π
¸
n
n
=
2 1 + c os
nπnπ
2
+ sin
2 2
(
cos
4π
+ cos
6π
+ cos
10π
)
19 19 19
是方程√
4 +
√
4 +
¸
4 −x = x 的一个根.
2
C
()
n
∏ ( )
∏
n 4 7
7
7
∑
1
= 2 (n + 3) + (n − 3) (77)
π
3π 9π 1 + ¸
13
cos + cos + cos =
(78)
13 13 13 4
cos 5π
+ cos 7π
+ cos 11π
= 1 − ¸
13
(79)
13 13 13 4
(80)
tan 610◦ + tan 650◦ + tan 670◦
√
3
cos
2π + √3
cos
4π + √3
cos
8π
(81) △ABC 三个内角 A , B ,C 此次成等差数列，设 f (x ) = x n ，则使得 f (a ) + f (c ) ≤ 2 f (b ) 对任何
一个这样的三角形都恒成立的最大整数 n （a , b , c 分别为 △A B C 三边），求：
tan n (5n )◦ − n ¸n − 1tan n −1 (5n )◦ + C 2 t an n −2 (5n )◦ + n ¸
n − 1 t an (5n )◦ . (82)
(83)
89 2 − sec 2n ◦ = 0 n =1
(84)
89
n =1,n ̸=45
n
(
2 − sec 2n ◦)
= 288
n
(85)
k =0 1 + 8sin 2 k π
3 (2n − 1) √
3 cos 2π
+ √
3 cos 4π
+ √
3 cos 8π
= √3 1
(3¸3 9 − 6)
9 9 9 2
(86)
√
3
cos 2π cos 4π + √
3 cos 4π cos 8π + √
3 cos 8π cos 2π
= −
√3 3 (
¸3
9 − 1)
(87)
9 9 9 9 9 9
4
2∑n −1
4 k π
(2n − 2) (2n − 1) (4n 2 + 6n − 13)
(88)
k =1
cot
2n =
45
2∑n −1
4 k π
(4n 2 − 1) (4n 2 + 11)
(89)
n
k =1
csc
2n =
45
∑ sin 2k x = 1 [(2n + 1) sin x − sin (2n + 1) x ] csc x n = 2
− cos (n + 1) x sin nx 2 sin x
k=1
n n n n n ∑ sin 4k x = 1
[3n − 4 cos (n + 1) x sin nx csc x + cos 2 (n + 1) x sin 2nx csc 2x ]
k =1
8
(90)
∑ cos 2k x = n − 1 + 1 cos nx sin (n + 1) x csc x
n cos (n + 1) x sin n x
(91)
k =1
2 2
= 2
+
2 sin x ∑ sin 3k x =
3 sin
n + 1
n
x − 1 sin
3 (n + 1) x sin 3nx csc 3x (92)
k =1
4
2 x sin 2 x csc 2 4 2 2 2
∑ cos 3k x = 3 cos
n + 1
n
x + 1 cos 3 (n + 1) x sin 3nx csc 3x (93)
k =1
4
2 x sin 2 x csc
2 4 2 2 2
k =1
8 (94)
∑
cos 4k x = 1
[3n + 4 cos (n + 1) x sin nx csc x + cos 2 (n + 1) x sin 2nx csc 2x ]
一类三角恒等式
天书
本文将要证明如下两组结论： π 2π n π
一、n 个角度为{
, ,..., }的平方三角函数满足如下方程：
2n +1 2n +1 2n +1
{tan 2
π , tan 2 2π ,..., tan 2 n π }满足： 2n + 1 2n + 1 2n + 1 t n - C 2 t n -1 + C 4 t n -2 - ...(-1)n C 2n
= 0
2n +1 2n +1 2n +1 {cos 2
π , cos 2 2π ,..., cos 2 n π }满足： 2n + 1 2n + 1 2n + 1 C 1 C 2 C n t n - 2n -1 t n -1 + 2n -2 t n -2
- ...(-1)n n = 0 41
42 4n {cot 2 π , cot 2 2π ,..., cot 2 n π }满足：
2n + 1 2n + 1 2n + 1 C 1 t n - C 3 t n -1 + C 5 t n -2 - ...(-1)n C 2n +1 = 0 2n +1 2n +1 2n +1 2n +1
{sec 2
π , sec 2 2π ,..., sec 2 n π }满足： 2n + 1 2n + 1 2n + 1 t n - 41 C 2 t n -1 + 42 C 4 t n -2 - ...(-1)n 4n
C 2n
= 0
n +1 n +2 2n {sin 2
π , sin 2 2π ,..., sin 2 n π }满足： 2n + 1 2n + 1 2n + 1
4n !! (2n + 1)!
t n - 4 ⨯ 6 ⨯ ...⨯ (4n - 2) .t n -1 + 6 ⨯ 8 ⨯ ...(4n - 4)
..
(2n - 1)! (2n - 3)! +(-1)n
(2n + 4)(2n + 2)(2n )(2n - 2) t 2 + (-1)n -1 (2n + 2)2n
t + (-1)n = 0
5! 3!
{csc 2 π , csc 2 2π ,..., csc 2 n π }满足：
2n + 1 2n + 1 2n + 1
t n - (2n + 2)2n t n -1 + (2n + 4)(2n + 2)(2n )(2n - 2) t n -2 + ... + 4n !! = 0
3! 5! (2n + 1)!
π 2π (n - 1)π
二、n-1 个角度为{ , ,..., } 的平方三角函数满足如下方程：
2n 2n 2n
n n n n ⎧ ⎪ {tan 2 π , tan 2 2π
2
(n - 1)π
2n
2n ,..., tan
}
2n
{cot
2
π , cot
2
2π
2 (n - 1)π
2n 2n ,..., cot }
2n C 1 t n -1
- C 3 t n -2 + ...(-1)n -1 C 2n -1
= 0 2n 2n 2n
{cos 2 π , cos 2
2π
2 (n - 1)π
2n
2n C 1 ,..., cos }
2n
C 2
C n -1
t n -1 - 2n -2
t n -2 + 2n -3 t n -3 - ...(-1)n -1 n = 0 4 42 {sec 2 π , sec 2 2π
2 (n - 1)π
4n -1
2n 2n ,..., s ec }
2n
C 1t n -1 - 4C 3 t n -2 + 42 C 5 t n -3 - ...(-1)n -1 4n -1 C 2n -1
= 0 n n +1 {csc 2 π , csc 2 2π
n +2 2n -1 2 (n - 1)π
2n 2n ,..., c sc }
2n
t n -1
- (4n 2 - 4) t n -2 + (4n 2 - 4)(4n 2 - 16) t
n -3 + ... + (-1) n -1 (4n 2 - 4)(4n 2 - 16)...[4n 2 - (2n - 2)2
] = 3! 5!
(2n - 1)! {sin 2 π , sin 2 2π
2 (n - 1)π
2n 2n ,..., s in }
2n
(4n 2 - 4)(4n 2 - 16)...[4n 2 - (2n - 2)2 ] t n -1 - (4n 2 - 4)(4n 2 - 16)...[4n 2 - (2n - 4)2 ] t n -2 + ...(-1) n -1
= 0
(2n - 1)! (2n - 3)!
根据棣莫佛定理,我们可以得到 n 倍角公式:
(cos a + i sin a )n = cos na + i sin na
= (cos n a - C 2 cos n -2 a sin 2 a + C 4 cos n -4 a sin 4 a - ...) + (C 1 sin a cos n -1 a - C 3 sin 3 a cos n -3
a + ...)i ⎧⎪cos na = C 0 cos n a - C 2 cos n -2 a sin 2 a + C 4 cos n -4 a sin 4
a - ...
so : ⎨ n n n (1) ⎪sin na = C 1 sin a cos n -1 a - C 3 sin 3 a cos n -3 a + ...
⎩ n n
事实上上面的基本 n 倍角展开式还可以由下面的式子直接得到:
(cos a + i sin a )n + (cos a - i sin a )n
cos na =
sin na =
2
(cos a + i sin a )n - (cos a - i sin a )n
2i
下面给出只含 cos 的 n 倍角公式及证明:
n
[ ] 2 ⎪cos n θ= ∑(C
k + C k -1 )(-1)k 2n -1-2k cos n -2k θ
⎪ k =0
⎨ [ n -1]
2 n -k n -1-k (2) ⎪sin n θ= ∑ C k (-1)k 2
n -1-2k
cos n -1-2k θsin θ ⎩⎪ k =0
n -1-k 证 用数学归纳法来证明这两个公式。

∑ C
0-0
sin n θ= ∑ C
2
2
- ∑ C 2
2
n -k
1 当 n = 1 时，
[ 1 ]
2 ∑(C
k
+ C
k -1 )(-1)k 21-1-2k cos 1-2k θ= (C 0 + C 0-1 )(-1)0 20-2⨯0 cos 1-0 θ = cos θ ，
k =0 [1-1
] 2 1-k
1-1-k
1-0
0-0
k 1-1-k (-1)k 21-1-2k cos 1-1-2k θsin θ = C 0 (-1)0 20-0 cos 0-0 θsin θ= sin θ ， k =0
公式显然成立。

2 设已知对某个给定的正整数 n ，公式成立，有
sin(2n +1)a = 0 cos n θ a k =
k π
2n +1
， , k ∈[]
[ n -1
] 2
k
n -1-k
(-1)k 2n -1-2k cos n -1-2k θsin θ 。

k =0
下面看 n + 1
时的情形：
cos(n + 1)θ= cos n θcos θ- sin n θsin θ
n
[ ] = ∑(C k
+ C k -1 [ n -1] )(-1)k 2n -1-2k cos n +1-2k θ - ∑ C k
- (-1)k 2n -1-2k cos n -1-2k θsin 2 θ
k =0
n [ ]
2 n -k n -1-k
k =0
n -1 k = ∑(C k + C k -1 )(-1)k 2n -1-2k cos n +1-2k θ
k =0 [ n -1
] 2 n -k n -1-k
k
n -1-k (-1)k 2n -1-2k (cos n -1-2k θ- cos n +1-2k θ)
k =0
n [ ] = ∑(C k
+ C k -1
+ C k
[ n -1]
- )(-1)k 2n -1-2k cos n +1-2k θ+ ∑ C k
- (-1)k +1 2n -1-2k cos n -1-2k θ
k =0 n [ ]
2 n -k n -1-k n -1 k [ n -1
]+1
2 k =0 n -1 k
= ∑(C k + C k )(-1)k 2n -1-2k cos n +1-2k θ+ ∑ C k -1 - (-1)k -1+1 2n -1-2( k -1) cos n -1-2( k -1) θ
k =0 n [ ]
2 n -k n -k [ n +1]
2
k -1=0
n -1-( k 1) k
n -k
k =0 [ n +1]
2 (-1)k 2n -2k cos n +1-2k θ+ ∑ C k -1 (-1)k 2n +1-2k cos n +1-2k θ k =1
= ∑ (C k + 2C k -1 )(-1)k 2n -2k cos n +1-2k θ
k =0
[ n +1]
2 n -k n -k
= ∑ (C k + C k -1 + C k -1 )(-1)k 2n -2k cos n +1-2k θ n -k n -k n -k
k =0
= ∑C
2
2
2 2
⎧
[ n +1] = ∑ (C k
+ C k -1
)(-1)k 2n -2k cos n +1-2k θ ； n +1-k
n -k
k =0
sin(n + 1)θ= sin n θcos θ+ cos n θsin θ
[ n -1] = ∑ C
k
n [ ] - (-1)k 2
n -1-2k
cos
n -2k
θsin θ +∑(C
k
+ C
k -1
)(-1)k 2n -1-2k cos n -2k θsin θ
k =0 n
[ ]
2 n -1 k
k =0
n -k
n -1-k
= ∑(C k + C k -1 + C k )(-1)k 2n -1-2k cos n -2k θsin θ
k =0
n [ ]
2 n -k n -1-k n -1-k
= ∑(C k + C k )(-1)k 2n -1-2k cos n -2k θsin θ k =0 n
[ ] = ∑C k n -k n -k
(-1)k 2n -2k cos n -2k θsin θ 。

n -k
k =0
可见，当 n + 1 时，公式也成立。

3 所以，对任何正整数 n ，公式都成立。

我们用一个简单的组合变换C k + C k -1 = n C k 来把(2)变形
[ n ]
n 2
C k
n -k n -1-k n - k
n -k
⎪cos n θ= ∑(-1)k n -k (2 cos θ)n -2k
⎪ ⎨ ⎪sin n θ 2 k =0
[ n -1]
2 n - k (3) ⎪ = ∑ C k (-1)k (2 cos θ)n -1-2k
⎪⎩ sin θ k =0
n -1-k
分 n 的奇偶情况,我们把第二个式子里面的 cos 换成 sin,就得到只含有 sin 的式子:
⎧ sin na = 1+ ⎪ n sin a
(1- n 2 ) 3! sin 2 a + (1- n 2 )(9 - n 2 ) 5!
sin 4
a + ...n ⇒ odd ⎨ sin na (n 2 - 4) (n 2 - 4)(n 2 -16)
(4) ⎪ = 1- ⎩ n cos a sin a
sin 2 a + 3! 5! sin 4 a + ...n ⇒ even
接下来我们证明篇首的恒等式
首先取: sin(2n + 1)a = 0 ,那么 a 的值为: 接下来我们把式子展开:
a = k π
, k ∈ [1, 2n ]
k
2n + 1
sin(2n + 1)a = C 1
sin a cos 2n a - C 3
sin 3 a cos 2n -2 a + ... = 0
2n +1
2n +1
为了换成 tan,两边除以cos
2n +1
a ,得到:
C 1
tan a - C 3
tan 3 a + C 5
tan 5 a ... + (-1)n C 2n +1 tan 2n +1 a = 0
2n +1
2n +1
2n +1
2n +1
∑ k k ∏ 我们除掉一个 tana,再用 tan 2
a = t 换元,得到方程:
t n - C 2 t n -1 + C 4 t n -2 - ...(-1)n C 2n = 0
2n +1 2n +1 2n +1
k π
他的 n 个根为: a k =
2n +1
, k ∈[1, 2n ]
{cot 2
π , cot 2 2π ,..., cot 2 n π }满足： 取倒数,就得到: 2n + 1 2n + 1 2n + 1 C 1 t n - C 3 t n -1 + C 5 t n -2
- ...(-1)n C 2n +1 = 0 2n +1 2n +1 2n +1 2n +1
[ n -1]
sin na
2
k k n -1-2k
接下来我们考虑恒等式(3):
sin a
= ∑ C n -1-k (-1) (2 cos a )
k =0
取 n ⇒ 2n + 1得到: sin(2n + 1)a = n sin a k =0
k 2n -k
(-1)k
(2
c os a )2n -2k
令sin(2n + 1)a = 0 ⇒ a = k π
2n + 1
, k ∈ [1, 2n ]再令cos 2 a = t 得到：
C 1 C 2 C n
t n -
2n -1
t n -1 + 2n -2 t n -2
- ...(-1)n n = 0 4 42 4n
再取倒数，就可以得到 sec 一组：
{sec 2
π
, sec 2 2π ,..., sec 2 n π
}满足： 2n + 1 2n + 1 2n + 1 t n - 41 C 2 t n -1 + 42 C 4 t n -2 - ...(-1)n 4n
C 2n
= 0
n +1 n +2 2n
用同样方法处理等式(4)，得到：
{sin 2
π
, s in 2 2π ,..., s in 2
n π } 2n +1 2n +1 2n +1
n
[(2n +1)2 - (2k -1)2 ] = a ;∏
a k =1
= 4n !! = 4n
(2n )!
(∏ a k n -1
( a k )(2n +1)! )t n
- k =1
t n -1 + .... + (-1)n (2n +1)! = 0 (2n -1)! k =1
{csc 2 π , csc 2 2π ,..., csc 2 n π }
2n +1 2n +1 2n +1
t n - (2n + 2)2n t n -1 + (2n + 4)(2n + 2)(2n )(2n - 2) t n -2 + ... + 4n !! = 0
3! 5! (2n +1)!
这样我们证明了四个三角函数，角度为固定值的时候所满足的方程，根据韦达定理，可以得 到
一系列恒等式，下面给出几个常见的 ：
n
C
tan
2
π
+ tan 2 2π + ... + tan 2
n π = n (2n +1) 2n +1 2n +1 2n +1 tan π tan 2π ... tan n π
=
2n +1 2n +1 2n +1 cos 2
π + cos 2 2π + ... + cos 2 n π = 2n -1 2n +1 2n +1 2n +1 4 cos
π cos 2π ...cos n π = 1 2n +1 2n +1 2n +1 2n
cot 2
π + cot 2 2π + ... + cot 2 n π = n (2n -1) 2n +1 2n +1 2n +1 3 cot
π cot 2π ...cot n π = 1 2n +1 2n +1 2n +1 sec 2
π + sec 2 2π + ... + sec 2
= 2n (n +1) 2n +1 2n +1 2n +1 sec
π sec 2π ...sec n π = 2n 2n +1 2n +1 2n +1 sin 2
π + sin 2 2π + ... + sin 2 n π = 2n +1 2n +1 2n +1 2n +1 4
sin
π sin 2π ...sin n π = 2n +1 2n +1 2n +1 2n
π 2π (n -1)π
接下来我们用同样的方法可以得到角度为{ , ,..., }的方程：
2n 2n 2n
{tan2 π
, tan2
2π
2
(n - 1)π
2n 2n
,..., tan }
2n {cot2
π
, cot2
2π
2
(n - 1)π
2n 2n
,..., cot }
2n
C1 t n-1 -C3 t n-2 + ...(-1)n-1 C 2n-1 = 0
2n 2n 2n {cos2
π
, cos2
2π
2
(n - 1)π
2n 2n
C1
,..., cos }
2n
C 2 C n-1 t n-1 -2n-2 t n-2 +2n-3 t n-3 - ...(-1)n-1n = 0
4 42
{sec2
π
, sec2
2π
2
(n - 1)π
4n-1
2n 2n
,..., s ec }
2n
C1t n-1 - 4C3 t n-2 + 42 C5 t n-3 - ...(-1)n-1 4n-1 C 2n-1 = 0 n n+1
{csc2
π
, csc2
2π
n+2 2n-1
2
(n - 1)π
2n 2n
,..., c sc }
2n
t n-1 -
(4n2 - 4)
t n-2 +
(4n2 - 4)(4n2 - 16)
t n-3 + ... + (-1) n-1
(4n2 - 4)(4n2 - 16)...[4n2 - (2n - 2)2 ]
=
3! 5! (2n -1)!
{sin2
π
, sin2
2π
2
(n - 1)π
2n 2n
,..., s in }
2n
(4n2 - 4)(4n2 - 16)...[4n2 - (2n - 2)2 ]
t n-1 -
(4n2 - 4)(4n2 - 16)...[4n2 - (2n - 4)2 ]
t n-2 +...(-1) n-1 = 0 (2n -1)! (2n -3)!
这样，利用韦达定理，又可以得出一系列恒等式，下面是一些常用的：
tan2
π
+ tan2
2π
+ ... + tan2
(n - 1)π
= cot2
π
+ cot2
2π
+ ... + cot2
(n - 1)π
=
(2n - 1)(n - 1)
2n 2n
tan
π
tan
2π
2n
(n - 1)π= 1
2n 2n 2n 3 2n 2n
... tan
2n
cos2
π
+ cos2
2π
+ ... + cos2
(n - 1)π
=
n - 1
2n 2n
cos
π
cos
2π
2n 2
(n - 1)π=
...cos
2n 2n 2n 2n-1
sec2
π
+ sec2
2π+ ... + sec2 (n - 1)π=2(n + 1)(n - 1) 2n 2n
π2π(n -1)π
2n 3
2n-1
sec sec ...sec =
2n
csc2
π
2n
2n
+csc 2
2π
2n
2n
+ ... + csc 2
(n - 1)π=
2n
2(n2 - 1)
3 π2π(n -1)π 2n-1
csc csc ...csc =
2n 2n 2n
sin2
π
+ sin2
2π
+ ... + sin2
(n - 1)π
=
n - 1
2n 2n
sin
π
sin
2π
2n 2
(n - 1)π=
(i)
2n 2n 2n 2n-1
n
n
n
n。