✔ 最佳答案
3)
cos π/11 + cos 3π/11 + cos 5π/11 + cos 7π/11 + cos 9π/11
= 2sin π/11 cos π/11 + 2sin π/11 cos 3π/11 + 2sin π/11 cos 5π/11
+ 2sin π/11cos 7π/11 + 2sin π/11 cos 9π/11
──────────────────────────────────────────────
............................................2sin π/11
=sin 2π/11 - sin 0 + sin 4π/11 - sin 2π/11 + sin 6π/11 - sin 4π/11
+ sin 8π/11 - sin 6π/11 + sin 10π/11 - sin 8π/11
──────────────────────────────────────────────
..............................................2sin π/11
=sin 10π/11
─────── = 1/2
2 sin π/11
4)
If x + y + z = 0 , then
(y + z)³ = - x³
y³ + 3y²z + 3yz² + z³ = - x³
x³ + y³ + z³ = - 3yz (y + z)
x³ + y³ + z³ = - 3yz (- x)
x³ + y³ + z³ = 3xyz .................. (1)
Let x = cis a , y = cis b , z = cis c , then
x + y + z
= (cos a + cos b + cos c) + i (sin a + sin b + sin c)
= 0 + i (0)
= 0
(cos 3a + cos 3b + cos 3c) + i (sin 3a + sin 3b + sin 3c)
= cis 3a + cis 3b + cis 3c
= (cis a)³ + (cis b)³ + (cis c)³
= 3 (cis a) (cis b) (cis c) ............... By (1)
= 3 cis (a + b + c)
= 3 cos (a + b + c) + 3 i sin (a + b + c)∴ cos 3a + cos 3b + cos 3c = 3 cos (a + b + c) .... real parts are equal.
sin 3a + sin 3b + sin 3c = 3 sin (a + b + c) .... imaginary parts are equal.
i.e. cos (a + b + c) = 1/3 (cos 3a + cos 3b + cos 3c )
and
sin(a + b + c) = 1/3 (sin 3a + sin 3b + sin 3c )
2012-11-08 23:03:30 補充：
Complex number method for Q1) :
Let
x = cos π/11 + cos 3π/11 + cos 5π/11 + cos 7π/11 + cos 9π/11
y = sin π/11 + sin 3π/11 + sin 5π/11 + sin 7π/11 + sin 9π/11
2012-11-08 23:03:48 補充：
then
x + i y
= cis π/11 + cis 3π/11 + cis 5π/11 + cis 7π/11 + cis 9π/11
= cis π/11 + (cis π/11)³ + (cis π/11)⁵ + (cis π/11)⁷ + (cis π/11)⁹
= cis π/11 (1 - (cis π/11)¹º) / (1 - (cis π/11)²)
= cis π/11 (1 - cos 10π/11 - i sin 10π/11) / (1 - cos 2π/11 - i sin 2π/11)
2012-11-08 23:04:12 補充：
=
cis π/11 (2 sin² 5π/11 - i 2sin 5π/11 cos 5π/11)
──────────────────────────────
2 sin² π/11 - i 2sin π/11 cos π/11
=
sin 5π/11 cis π/11 ( sin 5π/11 - i cos 5π/11 )
────────────────────────────
sin π/11 ( sin π/11 - i cos π/11 )
2012-11-08 23:04:58 補充：
=
- i sin 5π/11 cis π/11 ( i sin 5π/11 + cos 5π/11 )
──────────────────────────────
- i sin π/11 ( i sin π/11 + cos π/11 )
=
sin 5π/11 cis π/11 ( i sin 4π/11 + cos 4π/11 )
────────────────────────────
sin π/11
2012-11-08 23:05:04 補充：
=
sin 5π/11 ( i sin 5π/11 + cos 5π/11 )
───────────────────────
sin π/11
The real part = x
=
sin 5π/11 cos 5π/11
─────────────
sin π/11
=
(1/2) sin 10π/11
────────── = 1/2
sin π/11
2012-11-09 18:11:01 補充：
Alternatively :
3)
Consider z¹¹ + 1 = 0
z¹¹ = - 1
z¹¹ = cis π
z = cis (2kπ + π)/11 , (k = 0 , 1 , 2 , ... 10)
i.e.
z =
cos π/11 + i sin π/11 ,
cos 3π/11 + i sin 3π/11 ,
cos 5π/11 + i sin 5π/11 ,
cos 7π/11 + i sin 7π/11 ,
cos 9π/11 + i sin 9π/11 ,
cos 11π/11 + i sin 11π/11 = - 1 ,
2012-11-09 18:16:51 補充：
cos 13π/11 + i sin 13π/11 = cos 9π/11 - i sin 9π/11 ,
cos 15π/11 + i sin 15π/11 = cos 7π/11 - i sin 7π/11 ,
cos 17π/11 + i sin 17π/11 = cos 5π/11 - i sin 5π/11 ,
cos 19π/11 + i sin 19π/11 = cos 3π/11 - i sin 3π/11 ,
cos 21π/11 + i sin 21π/11 = cos π/11 - i sin π/11 ,
2012-11-09 18:16:57 補充：
Sum of roots of z¹¹ + 1 = 0 is 0.
i.e.
2 (cos π/11 + cos 3π/11 + cos 5π/11 + cos 7π/11 + cos 9π/11) - 1 = 0
cos π/11 + cos 3π/11 + cos 5π/11 + cos 7π/11 + cos 9π/11 = 1/2
2012-11-09 18:24:55 補充：
Please see 意見 008 and 009.
