Prove that $\sin n\theta=n\sin \theta-\frac{n(n^2-1)}{3!}\sin^3\theta+\frac{n(n^2-1)(n^2-3^2)}{5!}\sin^5\theta+\cdots$ 
Prove that 
  $$\sin n\theta=n\sin \theta-\frac{n(n^2-1)}{3!}\sin^3\theta+\frac{n(n^2-1)(n^2-3^2)}{5!}\sin^5\theta+-\cdots$$

If I am not mistaken, this identity was either proven by Newton or known to him, so if possible I would really like to see the way he approached it, though any solution will suffice. 
My brief efforts involved induction on $n$ which failed since I ended up with having to manipulate $\sin( n+1)\theta=\sin( n\theta +\theta)=\sin n\theta \cos\theta+\cos n\theta\sin\theta $, which involves $\cos n\theta$. 
I tried the "familiar" method of expansion of $$\sin n\theta=n\theta-\frac{(n\theta)^3}{3!}+\frac{(n\theta)^5}{5!}-+\cdots$$
but this only made it more complicated$$\sin n\theta=n\theta-\frac{(n\theta)^3}{3!}+\frac{(n\theta)^5}{5!}-+\cdots=\\n\Big(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-+\cdots\Big)-\frac{n(n^2-1)}{3!}\Big(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-+\cdots\Big)^3+-\cdots$$
Another attempt would be to use the identity $$\sin n\theta=\frac{1}{2i}(e^{in\theta}-e^{-in\theta})$$
though this was obviously not known to Newton. 
In any case, any thougths, ideas are welcome..
EDITA thorough answer has been provided below, but since it  involves the use of complex numbers, I deem that the search for another answer, one based solely on the mathematical knowledge up to Newton's time is open.
After a bit more research, it appears that Newton came up with the formula after reading a book by Vieta, but I have been unable to gather further info on whether the formula was known to Vieta as well.
 A: The series can be retrieved by expressing the Maclaurin expansion of $\sin(nx)$ considered as a function of $\sin x$, avoiding then the use of complex numbers. Moreover, it can be extended to the case of non-integer $n$. 
Let $u$ a real number, we introduce the notations  $y=\sin x$ et $f(y)=\sin ux$. Successive derivations of $f(y)$ with respect to $x$ give
\begin{align*}
\cos xf^{\prime}(y)&=u\cos ux\\
-\sin xf^{\prime}(y)+(1-\sin^{2}x)f^{\prime\prime}(y)&=-u^{2}\sin ux    %
\end{align*}
The latter expression can be written as
\begin{equation}
 (1-y^{2})f^{\prime\prime}(y)=yf^{\prime}(y)-u^{2}f
\end{equation} 
Now, successive derivation with respect to $y$ are
\begin{align*}
(1-y^{2})f^{(3)}(y)  &  =3yf^{\prime\prime}(y)+(1-u^{2})f^{\prime}(y)\\
(1-y^{2})f^{(4)}(y)  &  =5yf^{(3)}(y)+(4-u^{2})f^{\prime\prime}(y)
\end{align*}
This form suggest the recurrence relation:
\begin{equation}
 (1-y^{2})f^{(n+2)}(y)=(2n+1)yf^{(n+1)}(y)+(n^{2}-u^{2})f^{(n)}(y)
\end{equation} 
which is easily established.
As $f(0)=0$ and $f^{\prime}(0)=u$, we deduce for $p\geq1$
\begin{align}
f^{(2p)}(0)&=0\\
f^{(2p+1)}(0)&=u(1-u^{2})(3^{2}-u^{2})\dots\left[  (2p-1)^{2}-u^{2}\right] 
\end{align}
 The Maclaurin expansion of $f(y)$ gives:
\begin{equation}
\sin ux=u\sum\limits_{p=0}^{\infty}(1^{2}-u^{2})(3^{2}-u^{2})\dots\left[
(2p-1)^{2}-u^{2}\right]  \frac{\sin^{2p+1}x}{(2p+1)!}
\end{equation}
(For $p=0$, the product of the factors in the summation is taken to be equal to 1 by definition).

Additional expansions can be obtained. By taking $y=\sin x$ and $g(y)=\cos ux$, and using the same method, we obtain the same recurrence expression. Now, $g(0)=1$ and $g^{\prime}(0)=0.$ Then
\begin{align}
g^{(2p+1)}(0)&=0\\
g^{(2p)}(0)&=u^{2}(2^{2}-u^{2})(4^{2}-u^{2})\dots\left[  (2p-2)^{2}-u^{2}\right]
\end{align}
Thus,
\begin{equation}
\cos ux=1+\sum\limits_{p=0}^{\infty}(0-u^{2})(2^{2}-u^{2})(4^{2}%
-u^{2})\dots\left[  (2p)^{2}-u^{2}\right]  \frac{\sin^{2p+2}x}{(2p+2)!}
\label{cosux}%
\end{equation}
By changing $u\to2u$ in the previous expression, we obtain also
\begin{equation*}
 \sin^2ux=\sum\limits_{p=0}^{\infty}(u^{2}-0)(u^{2}-1^{2})(%
u^{2}-2^{2})\dots\left(  u^{2}-p^{2}\right)  \frac{\left( -1 \right)^p2^{2p+1}\sin^{2p+2}x}{(2p+2)!}
\end{equation*}
From derivation of the above expressions, we can also find the following expansions
\begin{align}
\frac{\cos ux}{\cos x}&=\sum\limits_{p=0}^{\infty}(1^{2}-u^{2})(3^{2}%
-u^{2})\dots\left[  (2p-1)^{2}-u^{2}\right]  \frac{\sin^{2p}x}{(2p)!}\\
\frac{\sin ux}{\cos x}&=\frac{1}{u}\sum\limits_{p=0}^{\infty}(0-u^{2}%
)(2^{2}-u^{2})(4^{2}-u^{2})\dots\left[  (2p)^{2}-u^{2}\right]  \frac{\sin
^{2p+1}x}{(2p+1)!} \label{Dercosux}%
\end{align}
A: hint $$\sin(nx)=\frac{e^{inx}-e^{-inx}}{2i}$$ then you can use $e^{ix}=\cos(x)+i\sin(x)$ and then use binomial for $(a+b)^n-(a-b)^n$ by using the relation that $\cos(x)=\sqrt{1-\sin^2(x)}$
A: I suppose you could also make use of the formula $\left(\begin{matrix} \cos n\theta & \sin n \theta \\ -\sin n \theta & \cos n \theta \end{matrix} \right)=\left(\begin{matrix} \cos \theta & \sin \theta\\ -\sin \theta & \cos \theta \end{matrix} \right)^n$ and substitute $s = \sin \theta, c = \cos \theta = \sqrt{1-s^2}$? Ultimately, though, this is equivalent to the other answers.
