Trigonometric series

Evaluation of sin(pi/7)+sin(2pi/7)+sin(4pi/7)=

sin(A) +sin(B) +sin(C) = 2sin[(A +B) /2] *cos[(A -B) /2] +sin(C) = 2cos(C/2) *cos[(A-B)/2] + sin(C) where A +B +C = π. Then sin(A) +sin(B) +sin(C) = 2cos(C/2) *(cos[(A -B)/2] +sin(C/2)) = 2cos(C/2) *(sin[(π-A+B)/2] + sin(C/2)) = 4cos(C/2) *sin[(π-A+B+C)/4] * cos[(π -A +B-C)/4] = 4cos(C/2) *cos(A/2) *cos(B/2)
Thus x = sin(pi/7)+sin(2pi/7)+sin(4pi/7)= 4 cos(π/14) *cos(π/7) *cos(2π/7) = 4sin(4π/7) *cos(π/7) *cos(2π/7)..?

Bonjour;

Tout d'abord , on a :

1 + e^(ix) + e^(i2x) + e^(i3x) = (e^(i4x) - 1)/(e(x) - 1)

= e^(i2x)/e^(ix/2) (e^(i2x) - e^(- i2x))/(e^(ix/2) - e^(- ix/2))

= e^(i 3/2 x) sin(2x)/sin(x/2) ;

donc :

sin(x) + sin(2x) + sin(3x) = sin(3/2x) sin(2x)/sin(x/2)

= 1/2 (cos(x/2) - cos(7/2 x))/sin(x/2) .


Conclusion :

si x = pi/7 donc : sin(pi/7) + sin(2pi/7) + sin(3pi/7) = 1/2 (cos(pi/14) - cos(pi/2))/sin(pi/14) ;

donc : sin(pi/7) + sin(2pi/7) + sin(4pi/7) = 1/2 cotang(pi/14) .

otherwise, sin(2π/7) = 2sin(π/7)cos(π/7) and sin(4π/7) = 2sin(2π/7)cos(2π/7)= 2sin(2π/7).[cos²(π/7) -sin²(π/7)], Therefore sin(π/7) +sin(2π/7) +sin(4π/7) = sin(π/7).[1 +2cos(π/7).(1+2[cos²(π/7) -sin²(π/7)])] = sin(π/7).(1 +2cos(π/7).[4cos²(π/7) -1]) =sin(π/7).[1 -2cos(π/7) +8cos³(π/7)]