sin pi/n + sin 2pi/n + .... + sin (n-1)pi/n = ctg pi/2n
dar ctg pi/2n = cos pi/2n /sin pi/2n
eliminam numitorul
sin pi/2n(sin pi/n + sin 2pi/n + .... + sin (n-1)pi/n ) = cos pi/2n
sin pi/n sin pi/2n + sin 2pi/n sin pi/2n+......+sin(n-1)pi/n sin pi/2n = cos pi/2n
1/2(cos pi/2n -cos 3pi/n+cos 3pi/n - cos 5pi/n+.....+ cos(2n-3)pi/2n - cos(2n-1)pi/2n = cos pi/2n
=1/2(cos pi/2n -cos (2n-1)pi/2n) = cos pi/2n
Notam pi/n= x
sin x + sin 2x + sin 3x + ... + sin (n-1)x = ctg x/2
sin x + sin 2x + sin 3x + ... + sin (n-1)x = cos x/2 / sin x/2
sin x/2 * sin x + sin x/2 * sin 2x + sin x/2 * sin 3x + ... + sin x/2 * sin (n-1)x =cos x/2
Aplicam formula: sin x * sin y = [cos (x-y) - cos (x+y)]/2 si obtinem:
cos x/2 - cos3x/2+
cos 3x/2 - cos 5x/2+
cos 5x/2 - cos 7x/2+
.
.
.
cos [x/2 - (n-1)x] - cos [x/2 + (n-1)x] = 2cos x/2 si le adunam
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se observa ca se reduc doi cate doi termeni si ramane
cos x/2 - cos [x/2 + (n-1)x] = 2cos x/2
Aplicam formula: cos x - cos y = -2 sin[(x+y)/2] * sin[(x-y)/2]
-2sin[x+(n-1)x] * sin[-(n-1)x] = 2cos x/2
2sin nx * sin[(n-1)x] = 2cos x/2 l : 2 si ne intoarcem la substitutia facuta, inlocuid pe x=pi/n
sin n*pi/n * sin (n-1)*pi/n = cos pi/2n
sin pi * sin[(n-1)*pi/n] = cos pi/2n
0 = cos pi/2n
pi/2n = pi/2 +- 2kpi l : pi/2
n = 1+- 4k, cu k∈N ⇔
n ∈{ -∞, ... , -11, -7, -3, 1, 5, 9, 13, ... , +∞}
Notam pi/n= x
sin x + sin 2x + sin 3x + ... + sin (n-1)x = ctg x/2
sin x + sin 2x + sin 3x + ... + sin (n-1)x = cos x/2 / sin x/2
sin x/2 * sin x + sin x/2 * sin 2x + sin x/2 * sin 3x + ... + sin x/2 * sin (n-1)x =cos x/2
Aplicam formula: sin x * sin y = [cos (x-y) - cos (x+y)]/2 si obtinem:
cos x/2 - cos3x/2+
cos 3x/2 - cos 5x/2+
cos 5x/2 - cos 7x/2+
.
.
.
cos [x/2 - (n-1)x] - cos [x/2 + (n-1)x] = 2cos x/2 si le adunam
-------------------------------------------------------------
se observa ca se reduc doi cate doi termeni si ramane
cos x/2 - cos [x/2 + (n-1)x] = 2cos x/2
Aplicam formula: cos x - cos y = -2 sin[(x+y)/2] * sin[(x-y)/2]
-2sin[x+(n-1)x] * sin[-(n-1)x] = 2cos x/2
2sin nx * sin[(n-1)x] = 2cos x/2 l : 2 si ne intoarcem la substitutia facuta, inlocuid pe x=pi/n
sin n*pi/n * sin (n-1)*pi/n = cos pi/2n
sin pi * sin[(n-1)*pi/n] = cos pi/2n
0 = cos pi/2n
pi/2n = pi/2 +- 2kpi l : pi/2
n = 1+- 4k, cu k∈N ⇔
n ∈{ -∞, ... , -11, -7, -3, 1, 5, 9, 13, ... , +∞}
Notam pi/n= x
sin x + sin 2x + sin 3x + ... + sin (n-1)x = ctg x/2
sin x + sin 2x + sin 3x + ... + sin (n-1)x = cos x/2 / sin x/2
sin x/2 * sin x + sin x/2 * sin 2x + sin x/2 * sin 3x + ... + sin x/2 * sin (n-1)x =cos x/2
Aplicam formula: sin x * sin y = [cos (x-y) - cos (x+y)]/2 si obtinem:
cos x/2 - cos3x/2+
cos 3x/2 - cos 5x/2+
cos 5x/2 - cos 7x/2+
.
.
.
cos [x/2 - (n-1)x] - cos [x/2 + (n-1)x] = 2cos x/2 si le adunam
-------------------------------------------------------------
se observa ca se reduc doi cate doi termeni si ramane
cos x/2 - cos [x/2 + (n-1)x] = 2cos x/2
Aplicam formula: cos x - cos y = -2 sin[(x+y)/2] * sin[(x-y)/2]
-2sin[x+(n-1)x] * sin[-(n-1)x] = 2cos x/2
2sin nx * sin[(n-1)x] = 2cos x/2 l : 2 si ne intoarcem la substitutia facuta, inlocuid pe x=pi/n
sin n*pi/n * sin (n-1)*pi/n = cos pi/2n
sin pi * sin[(n-1)*pi/n] = cos pi/2n
0 = cos pi/2n
pi/2n = pi/2 +- 2kpi l : pi/2
n = 1+- 4k, cu k∈N ⇔
n ∈{ -∞, ... , -11, -7, -3, 1, 5, 9, 13, ... , +∞}