Is $\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$? $$\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$$
Is there any formula that tells this or why is it like that?

\begin{equation}
\text{You can use De Moivre's identity:}
\end{equation}
\begin{equation} 
\text{Let's Call:}\\\\
\end{equation}
\begin{equation}
\mathrm{z=\cos x+i \sin x}\\
\mathrm{\frac{1}{z}=\cos x-i \sin x}\\
\end{equation}
\begin{equation}
\text{Now subtracting both equations together, we get:}\\
\end{equation}
\begin{equation}
\mathrm{2i\sin x=z-\frac{1}{z}}\\
\text{And we know that:}\\
\end{equation}
\begin{equation}
\mathrm{z^n=(cis x)^n=cis~nx}\\
\end{equation}
\begin{equation}
\text{So:}\\
\mathrm{2i\sin x=z-\frac{1}{z}}\Rightarrow\\
\mathrm{-8i\sin^3 x=\left (z-\frac{1}{z}  \right )^{3}}\\
\end{equation}
\begin{equation}
\text{Expanding the RHS:}\\
\end{equation}
\begin{equation}
\mathrm{-8i\sin^3 x =z^3-\frac{1}{z^3}-3\left (z-\frac{1}{z}\right)}\\ 
\end{equation}
\begin{equation}
\mathrm{-8i\sin^3 x=2i\sin 3x-6i\sin x}\\
\end{equation}
\begin{equation}
\boxed{\boxed{\mathrm{\therefore\sin^{3} x=\frac{
3}{4}}\sin\mathrm{x}-\frac{1}{4}\sin 3x}}
\end{equation}
