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{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 $$ X=\cos t,\quad Y=\sin t,\quad T=U=e^{it},\quad T^{-1}=V=e^{-it}. $$

Continuing from my comment to bgins's answer, one can start from the (defining) Chebyshev polynomial identity

$$T_{2n-1}(\cos\,x)=\cos((2n-1)x)$$

and make the substitution $x=\frac{\pi}{2}-z$ to yield the identity

$$T_{2n-1}(\sin\,z)=(-1)^{n+1}\sin((2n-1)z)$$

(use the addition formula and mind the values of cosine and sine at integer multiples of $\pi$.)

Now, there is the identity

$$T_{2n-1}(u)=\sum_{m=0}^{n-1} (-1)^{n+m+1}\frac{2n-1}{2m+1}\binom{n+m-1}{2m} 4^m u^{2m+1}$$

which can be used to derive the matrix-vector identity

$$\begin{pmatrix}\sin\,x\\\sin\,3x\\\vdots\\\sin((2n-1)x)\end{pmatrix}=\mathbf L\mathbf D\begin{pmatrix}\sin\,x\\\sin^3 x\\\vdots\\\sin^{2n-1}x\end{pmatrix}$$

where the diagonal matrix $\mathbf D$ has the diagonal entries $d_{k,k}=(-4)^{k-1}$ and the unit lower triangular matrix $\mathbf L$ has the entries $\ell_{j,k}=\dfrac{2j-1}{2k-1}\dbinom{j+k-2}{2k-2}$. Inverting this relation gives

$$\begin{pmatrix}\sin\,x\\\sin^3 x\\\vdots\\\sin^{2n-1}x\end{pmatrix}=\mathbf D^{-1}\mathbf W\begin{pmatrix}\sin\,x\\\sin\,3x\\\vdots\\\sin((2n-1)x)\end{pmatrix}$$

where $\mathbf W=\mathbf L^{-1}$ is also a unit lower triangular matrix with entries $w_{j,k}=(-1)^{j+k} \dbinom{2j-1}{j+k-1}$. (The proof that $\mathbf L\mathbf W=\mathbf I$ is not too hard, and is left as an exercise.) In particular, the second row yields the OP's desired identity.

You might want to look at Chebyshev polynomials of the first and second kind, denoted $T_n$ and $U_n$ respectively, which are defined recursively by $$ \begin{matrix} T_0 = 1,& T_1(x) = x,& T_{n+1} = 2x T_n(x) - T_{n-1}(x) \\ U_0 = 1,& U_1(x) = 2x,& U_{n+1} = 2x U_n(x) - U_{n-1}(x) \end{matrix} $$ (hence each of degree $n$) and give the formulas $$ \cos(n\theta) = T_n(\cos\theta), \qquad \frac{\sin\left((n+1)\theta\right)}{\sin\theta} = U_n(\cos\theta), $$ and have various closed form expressions, for example involving binomial coefficients.