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

This one comes from the classic S.L. Loney's Plane Trigonometry. We have the identity $$1 + 2x\cos t + 2x^{2}\cos 2t + 2x^{3}\cos 3t + \cdots = \frac{1 - x^{2}}{1 - 2x\cos t + x^{2}}\tag{1}$$ This is easily proved by letting $C$ denote the LHS of $(1)$ and $S$ denote $$S = 2x\sin t + 2x^{2}\sin 2t + \cdots$$ so that $$C + iS = 1 + 2xe^{it} + 2x^{2}e^{2it} + \cdots = \frac{1 + xe^{it}}{1 - xe^{it}} = \frac{(1 + xe^{it})(1 - xe^{-it})}{(1 - xe^{it})(1 - xe^{-it})}$$ and thus $$C + iS = \frac{1 + 2ix\sin t - x^{2}}{1 - 2x\cos t + x^{2}}$$ Equating the real part we get the value of sum $C$ as RHS of $(1)$.

It follows that $2\cos nt$ is the coefficient of $x^{n}$ in $\dfrac{1 - x^{2}}{1 - 2x\cos t + x^{2}}$ and thus we have \begin{align} 2\cos nt &= \text{coeff. of }x^{n}\text{ in }\dfrac{1 - x^{2}}{1 - 2x\cos t + x^{2}}\notag\\ &= \text{coeff. of }x^{n}\text{ in }(1 - 2x\cos t + x^{2})^{-1} - \text{coeff. of }x^{n - 2}\text{ in }(1 - 2x\cos t + x^{2})^{-1}\notag\\ &= \text{coeff. of }x^{n}\text{ in }\{1 + x(x - 2\cos t)\}^{-1} - \text{coeff. of }x^{n - 2}\text{ in }\{1 + x(x - 2\cos t)\}^{-1}\notag \end{align} Note that each term in $\{1 + x(x - 2\cos t)\}^{-1}$ is of the form $(-1)^{r}x^{r}(x - 2\cos t)^{r}$ and we need to evaluate the coefficient of $x^{n}$ and $x^{n - 2}$ in this expansion and subtract the two and by the above argument this coefficient will finally be $2\cos nt$.

Let us assume $n$ to be odd positive integer. In this case the first contribution in the desired coefficient comes from the term corresponding to $r = (n - 1)/2$ and in this case we get the coefficient of $x^{n - 2}$ as $$(-1)^{(n - 1)/2}\cdot\frac{n - 1}{2}\cdot(-2\cos t)$$ and in the final calculation this needs to be subtracted and hence the contribution for $r = (n - 1)/2$ is $$-(-1)^{(n - 1)/2}\cdot\frac{n - 1}{2}\cdot(-2\cos t)$$ Similarly we try to find the contributions for $r = (n + 1)/2, (n + 3)/2, \dots$ and finally get \begin{align} 2\cos nt &= (-1)^{(n - 1)/2}\left[-\frac{n - 1}{2}(-2\cos t)\right]\notag\\ &\,\,\,\, + (-1)^{(n + 1)/2}\left[\frac{n + 1}{2}(-2\cos t) - \dfrac{\dfrac{n + 1}{2}\cdot\dfrac{n - 1}{2}\cdot\dfrac{n - 3}{2}}{3!}(-2\cos t)^{3}\right]\notag\\ &\,\,\,\, + (-1)^{(n + 3)/2}\left[\binom{(n + 3)/2}{3}(-2\cos t)^{3} - \binom{(n + 3)/2}{5}(-2\cos t)^{5}\right]\notag\\ &\,\,\,\, + \dots\notag\\ &\,\,\,\, + (2\cos t)^{n}\notag \end{align} Hence by multiplying with $(-1)^{(n - 1)/2}$ we get \begin{align} &(-1)^{(n - 1)/2}(2\cos nt)\notag\\ &\,\,\,\,\,\,\,\,= \cos t\{(n - 1) + (n + 1)\} - \frac{(n + 1)(n - 1)}{3!}\cos^{3}t \{(n - 3) + (n + 3)\} + \cdots\notag \end{align} Hence by dividing by $2$ we get $$(-1)^{(n - 1)/2}\cos nt = n\cos t - \frac{n(n^{2} - 1^{2})}{3!}\cos^{3}t + \frac{n(n^{2} - 1^{2})(n^{2} - 3^{2})}{5!}\cos^{5}t - \cdots$$ Replacing $t$ by $(\pi/2) - \theta$ we get $$\sin n\theta = n\sin \theta - \frac{n(n^{2} - 1^{2})}{3!}\sin^{3}\theta + \frac{n(n^{2} - 1^{2})(n^{2} - 3^{2})}{5!}\sin^{5}\theta - \cdots\tag{2}$$ Note the similarity with the series $$\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots\tag{3}$$ Clearly we can derive $(3)$ from $(2)$ in an intuitive (but non-rigorous) way by putting $x = n\theta$ and keeping $x$ fixed and letting $n \to \infty$ so that $\theta \to 0$. In that case $n\sin \theta = n\sin(x/n) \to x$ as $n \to \infty$ and the formula $(2)$ is transformed into the series in equation $(3)$.

Note: This and many similar expansions of $\sin n\theta, \cos n\theta$ in powers of $\sin \theta, \cos \theta$ are provided in Loney's book and I have not found such elementary proofs anywhere else. I don't know if this classic textbook is studied anywhere or not. Also note that the formula $(2)$ is proved under the assumption that $n$ is an odd positive integer. By changing sign of $n$ it is easily seen that the formula holds for negative odd integers also. The same formula holds for all values of $n$ (the RHS is an infinite series if $n$ is not an integer and the identity should be treated as such in case $n$ is not an integer), but the proof of the identity for general $n$ can not be given via the approach mentioned in the above solution.

Update: From OP's comment it appears that he is interested in a solution which does not use complex numbers at all. Note that the solution presented above uses complex numbers to establish the identity $(1)$. Apart from this no use is made of complex numbers anywhere else.

I provide the following proof of $(1)$ without any use of complex numbers. Consider the expression $$f(n) = x^{n + 1}\cos (n - 1)t - x^{n}\cos nt$$ and then we have \begin{align} f(n) - f(n - 1) &= x^{n + 1}\cos (n - 1)t - x^{n}\cos nt - x^{n}\cos (n - 2)t + x^{n - 1}\cos (n - 1)t\notag\\ &= x^{n - 1}(1 + x^{2})\cos (n - 1)t - 2x^{n}\cos (n - 1)t\cos t\notag\\ &= (1 - 2x\cos t + x^{2})x^{n - 1}\cos (n - 1)t\tag{4} \end{align} Putting $n = 1, 2, \ldots, n$ in $(4)$ and adding the resulting equations we get $$f(n) - f(0) = (1 - 2x\cos t + x^{2})\sum_{k = 0}^{n - 1}x^{k}\cos kt$$ Let us now assume that $|x| < 1$ and then let $n \to \infty$. It is then obvious that $x^{n} \to 0$ and hence $f(n) \to 0$ and thus we have $$\sum_{k = 0}^{\infty}x^{k}\cos kt = -\frac{f(0)}{1 - 2x\cos t + x^{2}} = \frac{1 - x\cos t}{1 - 2x\cos t + x^{2}}$$ The sum in LHS of $(1)$ is given by $$-1 + 2\sum_{k = 0}^{\infty}x^{k}\cos kt = -1 + \frac{2 - 2x\cos t}{1 - 2x\cos t + x^{2}} = \frac{1 - x^{2}}{1 - 2x\cos t + x^{2}}$$ and thus the identity $(1)$ is established for all values of $x, t$ with the constraint $|x| < 1$.

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 $\cos(x)=\sqrt{1-\sin^2(x)}$