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?
 A: 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
$$
X=\cos t,\quad Y=\sin t,\quad T=U=e^{it},\quad T^{-1}=V=e^{-it}.
$$
A: 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.
A: 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.
A: de Moivre's formula says
$$
\begin{align}
\cos(3x)+i\sin(3x)
&=(\cos(x)+i\sin(x))^3\\
&=\left(\cos^3(x)-3\cos(x)\sin^2(x)\right)+i\left(3\cos^2(x)\sin(x)-\sin^3(x)\right)\\
&=\left(4\cos^3(x)-3\cos(x)\right)+i\left(3\sin(x)-4\sin^3(x)\right)\tag{1}
\end{align}
$$
Therefore,
$$
\begin{align}
\cos(3x)&=4\cos^3(x)-3\cos(x)\tag{2}\\
\sin(3x)&=3\sin(x)-4\sin^3(x)\tag{3}
\end{align}
$$
Solving $(3)$ for $\sin^3(x)$ yields
$$
\sin^3(x)=\frac34\sin(x)-\frac14\sin(3x)\tag{4}
$$
A: \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}
A: \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:
$$ \sin(3x)=3\cdot \sin x -4\cdot \sin^3 x $$
A: 
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 $n$,
$$(2\mathrm i\sin(x))^{2n+1}=(\mathrm e^{\mathrm ix}-\mathrm e^{-\mathrm ix})^{2n+1}=\sum_{k=0}^{2n+1}(-1)^k{2n+1\choose k}\mathrm e^{(2n+1-2k)\mathrm ix}$$
$$
(-1)^n\cdot4^n\cdot2\mathrm i\cdot \sin^{2n+1}(x)=2\mathrm i\sum_{k=0}^n (-1)^k{2n+1\choose k}\sin((2n+1-2k)x)
$$
$$
\sin^{2n+1}(x)=\frac1{4^n}\sum_{k=0}^n (-1)^k{2n+1\choose n+k}\sin((2k+1)x)
$$
A: If you're ever unsure of a trig identity, just convert it to complex exponentials and expand. It's especially nice if you have a CAS that can do the expansion for you (or if you're just handy with such things). 
$\sin^3(x) = \left (\frac{e^{i x} - e^{-i x}}{2i}\right )^3 = \frac{e^{3ix}-3e^{ix}+3e^{-ix}-e^{-3ix}}{-8i}$ and $\frac{3}{4}\sin(x) - \frac{1}{4}\sin(3x) = \frac{3}{4}\left (\frac{e^{i x} - e^{-i x}}{2i}\right ) - \frac{1}{4}\left (\frac{e^{3i x} - e^{-3i x}}{2i}\right ) = \frac{e^{3ix}-3e^{ix}+3e^{-ix}-e^{-3ix}}{-8i}$.   
Alternately, we could derive the right hand side by converting the exponentials back to sines and cosines using $e^{ix} = \cos(x) + i\sin(x)$ and canceling.
