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?


Motto: The multiplicative structure of the $\color{red}{\text{complex exponential}}$ make it easier to use than the $\color{red}{\text{sine and cosine}}$ functions. Thus, to prove properties of the latter, it is often fruitful to go back to the former.

To wit, one has
$$\sin(3x)=\Im(\mathrm e^{3\mathrm ix})=\Im\left((\cos(x)+\mathrm i\sin(x))^3\right)$$
and $$(\cos(x)+\mathrm i\sin(x))^3=\cos^3(x)+3\mathrm i\ \cos^2(x)\sin(x)-3\cos(x)\sin^2(x)-\mathrm i\ \sin^3(x)$$
hence $$\sin(3x)=3\ \cos^2(x)\sin(x)-\sin^3(x)=3\sin(x)-4\sin^3(x)$$
and $$4\ \sin^3(x)=3\sin(x)-\sin(3x)$$
Exercise: Show that, likewise,
$$16\ \sin^5(x)=\sin(5x)-5\sin(3x)+10\sin(x)$$
Added later on: One can also go the other way round, which provides an easy generalization and explains the appearance of the binomial coefficients $(1,3)$ for $\sin^3$ and $(1,5,10)$ for $\sin^5$. To wit,
$$(2\mathrm i\sin(x))^3=(\mathrm e^{\mathrm ix}-\mathrm e^{-\mathrm ix})^3=\mathrm e^{3\mathrm ix}-3\mathrm e^{\mathrm ix}+3\mathrm e^{-\mathrm ix}-\mathrm e^{-3\mathrm ix}$$
$$
(-8\mathrm i)\ \sin^3(x)=(2\mathrm i)\ (\sin(3x)-3\sin(x))
$$
$$
-4\ \sin^3(x)=\sin(3x)-3\sin(x)
$$
Likewise, for every nonnegative integer $n$,
$$(2\mathrm i\sin(x))^{2n+1}=(\mathrm e^{\mathrm ix}-\mathrm e^{-\mathrm ix})^{2n+1}=\sum_{k=0}^{2n+1}(-1)^k{2n+1\choose k}\mathrm e^{(2n+1-2k)\mathrm ix}$$
$$
(-1)^n\cdot4^n\cdot2\mathrm i\cdot \sin^{2n+1}(x)=2\mathrm i\sum_{k=0}^n (-1)^k{2n+1\choose k}\sin((2n+1-2k)x)
$$
$$
\sin^{2n+1}(x)=\frac1{4^n}\sum_{k=0}^n (-1)^k{2n+1\choose n+k}\sin((2k+1)x)
$$
