## trigo

given that summation:
∑ z^k= (1-z^(n+1))/ (1- z) for k= 0,1,2,3......n

choose z= e^(ix),
solve ∑ sin (kx) from k=0,1,2,...n..

to solve this, i expressed
sin (kx)= 1/2( e^((ikx) - e^(-ikx))

thus ∑ sin(kx)= 1/2 [(1-(e^(ix)(n+1))/ (1- e(ix)) - ((1- e(ix))/(1-(e^(ix)(n+1)) ]

to get the summation of the part in green, i just took the reciprocal of the formula that is in red. can it be done that way?

becos i cant seem to simplify the answer after that