Is $\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$?

\begin{align} \sin(3x) &= \sin(x+2x) \tag{1} \\ \sin(\alpha+\beta) &= \sin \alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta \tag{2} \\ \sin 2\alpha &= 2\cdot \sin \alpha \cdot \cos \alpha \tag{3} \\ \cos 2\alpha &= \cos^2 \alpha - \sin^2 \alpha \tag{4} \\ 1 &= \sin^2 \alpha + \cos^2 \alpha \tag{5} %% \end{align}

If you apply all these formulas you should get:

$$ \sin(3x)=3\cdot \sin x -4\cdot \sin^3 x $$

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 $n$,

$$(2\mathrm i\sin(x))^{2n+1}=(\mathrm e^{\mathrm ix}-\mathrm e^{-\mathrm ix})^{2n+1}=\sum_{k=0}^{2n+1}(-1)^k{2n+1\choose k}\mathrm e^{(2n+1-2k)\mathrm ix}$$ $$ (-1)^n\cdot4^n\cdot2\mathrm i\cdot \sin^{2n+1}(x)=2\mathrm i\sum_{k=0}^n (-1)^k{2n+1\choose k}\sin((2n+1-2k)x) $$ $$ \sin^{2n+1}(x)=\frac1{4^n}\sum_{k=0}^n (-1)^k{2n+1\choose n+k}\sin((2k+1)x) $$

Here is a mild generalization.

Let $X,Y\ $ be indeterminates, and let $f(X;Y)\in\mathbb C[X,Y\ ]$ be a polynomial.

The following observations give a purely algebraic method to decide if we have $$f(\cos t\,;\sin t)=0$$ for all $t$ in $\mathbb R$.

If $U,V,T$ are other indeterminates, then the following conditions are equivalent:

(a) $\ f(\cos t\,;\sin t)=0$ for all $t$ in $\mathbb R$,

(b) $\ X^2+Y^2-1$ divides $f(X;Y)$,

(c) $\ UV-1$ divides $$g(U,V):=f\left(\frac{U+V}{2}\ ;\ \frac{U-V}{2i}\right),$$

(d) $\ g(T,T^{-1})=0$.

The interpretation of the indeterminates $X,Y,U,V,T$ can be informally expressed by the equalities $$ X=\cos t,\quad Y=\sin t,\quad T=U=e^{it},\quad T^{-1}=V=e^{-it}. $$