Proving $\sin(x) + \sin(2x) + \sin(3x)+ \cdots +\sin(nx) = \frac{cos(1/2) - cos(n-1/2)}{2 \sin(1/2)2}$ via Ceasro Summation?

Question

I'm having trouble proving the following conjecture in $(1.)$, via Fourier methods, my intial attack can be seen in $(2)$

$(1.)$

$$\sin(x) + \sin(2x) + \sin(3x)+ \cdots +\sin(nx) = \frac{cos(1/2) - cos(n-1/2)}{2 \sin(1/2)2}$$  

$(2)$

On the RHS, side of our conjecture of $(1.)$, the following observations can be made due to applying the technique of Ceasro Summation, one can derive the Cesaro means from the RHS of $(2.1)$.

$$[S_{k} = \sin(x) + \sin(2x) + \sin(3x)+ \cdots + \sin (nx) = \sum_{n} \sin(n(x))\tag{2.1}]$$

$$[\sigma_{n} = \frac{\sin(x)) + \sin(2(x)) + \sin(3(x)) + \cdot \cdot \cdot + sin(n(x))_{n-1}}{N} \tag{2.2}]$$

$$[ \sin(x) + (1 - 1/k) \sin(2x) + \cdot \cdot \cdot + (1 - n/k) \sin(n(x)) + \cdot \cdot \cdot \frac{1}{k}\sin(n(x))_{k-1} \tag{2.3}]$$

One, can formally verify the observations made in $(2.2)-(2.3)$, by taking the $\lim_{n \rightarrow \infty}\sigma_{N}$, as follows in $(2.4)$

$$\lim_{n \rightarrow\infty} (\sin(x) + (1 - 1/k) \sin(2x) + \cdot \cdot \cdot + (1 - n/k) \sin(n(x)) + \cdot \cdot \cdot \frac{1}{k}\sin(n(x))_{k-1}) = S$$

Lemma:

The specifics of my questions, lie on applying Theorem $(3.23)$ to the RHS side, of $(2.3)$ for the application of the theorem to be successful it would following from the RHS of $(1.)$ to be written as Fourier polynomials as follows:

$$F_{n}(x) = a_{o}+(a_{1}\cos(x) +b_{1}\sin(x))) + \cdot \cdot \cdot + (a_{n} \cos(n(x) + b_{n}\sin(n(x))$$

Is my approach on the right track any hints would be helpful.

Answer 1

$$\sum_{k=1}^n\sin{k}=\frac{\sum\limits_{k=1}^n2\sin\frac{1}{2}\sin{k}}{2\sin\frac{1}{2}}=\frac{\sum\limits_{k=1}^n\left(\cos\left(k-\frac{1}{2}\right)-\cos\left(k+\frac{1}{2}\right)\right)}{2\sin\frac{1}{2}}=\frac{\cos\frac{1}{2}-\cos\left(n+\frac{1}{2}\right)}{2\sin\frac{1}{2}}.$$

Answer 2

You can also write $$series = Im\bigg( \sum_{k=1}^n e^{ikx}\bigg) =Im\bigg( e^{ix} \frac{1- e^{inx}}{1-e^{ix}} \bigg)= Im \bigg( e^{ix/2} \frac{1- e^{inx}}{e^{-ix/2} - e^{ix/2}} \bigg)\\ = Im \bigg(i \frac{e^{ix/2}- e^{i(n+0.5)x}}{2 \sin(x/2)} \bigg)= Re \bigg( \frac{e^{ix/2}- e^{i(n+0.5)x}}{2 \sin(x/2)} \bigg)= \frac{\cos(x/2) - \cos[(n+0.5)x]}{2 \sin(x/2)} $$

where Im and Re stand for imaginary and real parts of exponential function.