1. Dans tout ce qui suit on suppose $~(x\neq k\pi)~$: \begin{align*} \sin^3 x&= \sin x~\sin^2 x\\ \sin^3 x&= \sin x\dfrac{1-\cos 2x}{2}\\ \sin^3 x&=\dfrac{\sin x}{2}-\dfrac{1}{2}\sin x\cos (2x)\\ \sin^3 x&=\dfrac{\sin x}{2}-\dfrac{1}{4}(\sin(3x)-\sin x)\\ \sin^3 x&=\dfrac{3\sin x}{4}-\dfrac{\sin(3x)}{4} \end{align*}
  2. On calcule tout d'abord: $$~~\sum\limits_0^n{\sin(kx)}=\Im\left( \sum\limits_0^n{e^{ikx}}\right)$$ On a: \begin{align*} \sum\limits_0^n{e^{ikx}}&=\dfrac{e^{i(n+1)x}-1}{e^{ix}-1}\\ \sum\limits_0^n{e^{ikx}}&=\dfrac{e^{i\frac{(n+1)x}{2}}}{e^{i\frac{x}{2}}}~ \dfrac{2i\sin(\frac{(n+1)x}{2})}{2i\sin(\frac{x}{2})}\\ \sum\limits_0^n{e^{ikx}}&=e^{i(\frac{nx}{2})}~ \dfrac{\sin(\frac{(n+1)x}{2}) } {\sin(\frac{x}{2})} \end{align*} On en déduit: $$~~\sum\limits_0^n{\sin(kx)}=\dfrac{\sin(\frac{nx}{2})\sin(\frac{(n+1)x}{2})}{\sin(\frac{x}{2})}$$ D'autre part: \begin{align*} \sum\limits_0^n{\sin^3(kx)}&=\dfrac{1}{4}\sum\limits_0^n{(3\sin(kx)-\sin(3kx))}\\ \sum\limits_0^n{\sin^3(kx)}&=\dfrac{1}{4}\sum\limits_0^n{(3\sin(kx)-\sin(ky))}\qquad \text{avec:}\qquad (y=3x) \\ \sum\limits_0^n{\sin^3(kx)}&=\dfrac{3}{4}\sum\limits_0^n{\sin(kx)}-\dfrac{1}{4}\sum\limits_0^n{\sin(ky)}\\ \sum\limits_0^n{\sin^3(kx)}&=\dfrac{3}{4}~\dfrac{ \sin(\frac{nx}{2}) \sin(\frac{(n+1)x}{2}) } {\sin(\frac{x}{2})}-\dfrac{1}{4}~\dfrac{ \sin(\frac{ny}{2}) \sin(\frac{(n+1)y}{2}) } {\sin(\frac{y}{2})} \end{align*} Enfin: $$\color{magenta}\boxed{\sum\limits_0^n{\sin^3(kx)}=\dfrac{3}{4}~\dfrac{ \sin(\frac{nx}{2}) \sin(\frac{(n+1)x}{2}) } {\sin(\frac{x}{2})}-\dfrac{1}{4}~\dfrac{ \sin(\frac{3nx}{2}) \sin(\frac{3(n+1)x}{2}) } {\sin(\frac{3x}{2})}}$$