Htykuut

How to solve this one?
$$sum_{k=1}^n q^k sin(kx) = S(q,n)$$ where $n$ is natural number and $q$ is a real number

I already got the formula $$S(q,n) = Imleft(frac{qexp(ix)(1-q^nexp(nix))}{1-qexp(ix)}right)$$
but I don't know how to simplify the $1-qexp(ix)$ using the formulas
$$ sin(φ)=frac{exp(iφ)-exp(-iφ)}{2i}$$
$$ cos(φ)=frac{exp(iφ)+exp(-iφ)}{2},.$$

Let me recast your formula using $n-1$ instead of $n$, for convenience. First, we reduce the calculation to the sum of a geometric progression
$$
sum_{k=0}^{n-1} q^k sin(kx) = Im sum_{k=0}^{n-1}(q, e^{ix})^k
= Im frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}},,$$
where we have used
$$
sum_{k=0}^{n-1} s^k = frac{1-s^n}{1-s}
$$
(this is easily proved for $sneq 1$ multiplying both sides by $1-s$ and noting the telescopic sum on the left-hand side; for $sto1$, the right-hand side reduces to $n$).
Now,
$$
frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}} =frac{1-q^ncos(nx)-iq^nsin(nx)}{1-qcos x -iq sin x}=(1-qcos x+i q sin x)frac{1-q^ncos(nx)-iq^nsin(nx)}{(1-qcos x)^2 + (q sin x)^2},,
$$
whose imaginary part is
$$
frac{qsin x(1-q^ncos(nx))-q^n sin(nx) (1-qcos x)}{(1-qcos x)^2+(q sin x)^2},.
$$
Simplifying slightly, by means of trigonometric formulas,
$$
sum_{k=0}^{n-1} q^k sin(kx) = frac{qsin x - q^n sin(nx)+q^{n+1}sin((n-1)x)}{1-2qcos x+q^2},.
$$

Well, I have this solution for the problem:
$$
sum_{k=1}^{n} q^k sin(kx) = Im sum_{k1}^{n}(q, e^{ix})^k
= Im frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}}q, e^{ix} ,,
$$
as a geometric series (I don't know why $|q, e^{ix}|<1$ as you said).

Then,
$$
frac{1-(q, e^{ix})^{n}}{1-q, e^{ix}}q, e^{ix} =frac{1-q^ncos(nx)-iq^nsin(nx)}{1-qcos x -iq sin x}q, e^{ix}=(1-qcos x+i q sin x)frac{1-q^ncos(nx)-iq^nsin(nx)}{(1-qcos x)^2 + (q sin x)^2}q, (cosx+isinx),,
$$
and
then simplifying the last expressions just by opening the brackets and then using the formula for the $sin$ of the sum we get
$$
sum_{k=1}^{n} q^k sin(kx) = qfrac{q^{n+1}sin(nx)-q^{n}sin((n+1)x)+sin x}{q^2-2qcos x+1}, .
$$
Sorry, I firstly haven't caught that in the sum k is going not from the 0 but from 1, that is my mistake.