The question asks to put $\sin^5 \theta$ in the form $$ A\sin 5\theta + B\sin 3\theta + C\sin \theta. $$
I have tried changing the $\sin$ into a $\sin^4$ and then putting it in terms of cos, and then using the reducing formula.
But I end up with a very long equation, as it is only a 4 mark question on the paper I am doing I assume I have made a mistake along the way. Does anyone have any alternate ideas as how best to solve it?
One way to solve the problem is to assume there exist constants $A$, $B$ and $C$ where
$$\sin^5\theta = A\sin 5\theta + B\sin 3\theta + C\sin \theta \tag{1}\label{eq1A}$$
is true for all angles $\theta$. Then let $\theta$ be several convenient values to get linear equations which can be solved for $A$, $B$ and $C$. First, with $\theta = 90^{\circ}$ we get
$$1 = A - B + C \tag{2}\label{eq2A}$$
Next, $\theta = 45^{\circ}$ gives
$$\frac{1}{4\sqrt{2}} = -\frac{A}{\sqrt{2}} + \frac{B}{\sqrt{2}} + \frac{C}{\sqrt{2}} \;\;\to\;\; 1 = -4A + 4B + 4C \tag{3}\label{eq3A}$$
Multiplying \eqref{eq2A} by $4$ and adding \eqref{eq3A} results in
$$5 = 8C \;\;\to\;\; C = \frac{5}{8} = \frac{10}{16} \tag{4}\label{eq4A}$$
Using this value of $C$, and $\theta = 30^{\circ}$, in \eqref{eq1A} gives
$$\frac{1}{32} = \frac{A}{2} + B + \frac{5}{16} \;\;\to\;\; 1 = 16A + 32B + 10 \;\;\to\;\; -9 = 16A + 32B \tag{5}\label{eq5A}$$
With \eqref{eq4A} in \eqref{eq2A}, we get
$$1 = A - B + \frac{10}{16} \;\;\to\;\; 16 = 16A - 16B + 10 \;\;\to\;\; 6 = 16A - 16B \tag{6}\label{eq6A}$$
Next, \eqref{eq5A} minus \eqref{eq6A} gives
$$-15 = 48B \;\;\to\;\; B = -\frac{5}{16} \tag{7}\label{eq7A}$$
This in \eqref{eq6A} shows that
$$6 = 16A + 5 \;\;\to\;\; A = \frac{1}{16} \tag{8}\label{eq8A}$$
Finally, substituting \eqref{eq4A}, \eqref{eq7A} and \eqref{eq8A} into \eqref{eq1A} gets what your image shows as being the answer, i.e.,
$$\sin^5\theta = \frac{1}{16}(\sin 5\theta - 5\sin 3\theta + 10\sin\theta) \tag{9}\label{eq9A}$$
Another way is to use the angle sum formulas to express $\sin 5\theta$ and $\sin 3\theta$ as polynomials in $\sin\theta$, then ensure all of the exponents are $0$ except for the fifth power term. First, we have
$$\sin 5\theta = \sin(3\theta + 2\theta) = \sin 3\theta\cos 2\theta + \cos 3\theta\sin 2\theta \tag{10}\label{eq10A}$$
Next, we get
$$\begin{equation}\begin{aligned} \sin 3\theta & = \sin(2\theta + \theta) \\ & = \sin 2\theta\cos\theta + \cos 2\theta\sin\theta \\ & = (2\sin\theta\cos\theta)\cos\theta + (1 - 2\sin^2\theta)\sin\theta \\ & = 2\sin\theta(1 - \sin^2\theta) + \sin\theta - 2\sin^3\theta \\ & = 3\sin\theta - 4\sin^3\theta \end{aligned}\end{equation}\tag{11}\label{eq11A}$$
and
$$\begin{equation}\begin{aligned} \cos 3\theta & = \cos(2\theta + \theta) \\ & = \cos 2\theta\cos\theta - \sin 2\theta\sin\theta \\ & = (1 - 2\sin^2\theta)\cos\theta - (2\sin\theta\cos\theta)\sin\theta \\ & = \cos\theta(1 - 4\sin^2\theta) \end{aligned}\end{equation}\tag{12}\label{eq12A}$$
Using \eqref{eq11A} and \eqref{eq12A} in \eqref{eq10A}, along with $\cos 2\theta = 1 - 2\sin^2\theta$ and $\sin 2\theta = 2\sin\theta\cos\theta$, results in
$$\begin{equation}\begin{aligned} \sin 5\theta & = (3\sin\theta - 4\sin^3\theta)(1 - 2\sin^2\theta) + \cos\theta(1 - 4\sin^2\theta)(2\sin\theta\cos\theta) \\ & = (3\sin\theta - 10\sin^3\theta + 8\sin^5\theta) + 2\sin\theta(1 - 4\sin^2\theta)(1 - \sin^2\theta) \\ & = (3\sin\theta - 10\sin^3\theta + 8\sin^5\theta) + 2\sin\theta(1 - 5\sin^2\theta + 4\sin^2\theta) \\ & = 5\sin\theta - 20\sin^3\theta + 16\sin^5\theta \end{aligned}\end{equation}\tag{13}\label{eq13A}$$
Using \eqref{eq11A} and \eqref{eq13A} in \eqref{eq1A}, then gathering terms according to powers of $\sin\theta$, we get the following for the coefficients of the $\sin\theta$, $\sin^3\theta$ and $\sin^5\theta$ terms, respectively,
$$5A + 3B + C = 0 \;\;\to\;\; C = -5A - 3B \tag{14}\label{eq14A}$$
$$-20A - 4B = 0 \;\;\to\;\; B = -5A \tag{15}\label{eq15A}$$
$$16A = 1 \;\;\to\;\; A = \frac{1}{16} \tag{16}\label{eq16A}$$
Using \eqref{eq16A} in \eqref{eq15A} gives $B = -\frac{5}{16}$. Using the determined values of $A$ and $B$ in \eqref{eq14A} shows that $C = -5\left(\frac{1}{16}\right) - 3\left(-\frac{5}{16}\right) = \frac{10}{16}$. This matches the results in \eqref{eq8A}, \eqref{eq7A} and \eqref{eq4A}, so we then also get \eqref{eq9A}.
Last, but not least, as suggested in the linked questions in Gerry Myerson's question comment, and by Jean Marie's comment, an easier and more direct method is to use $\sin \theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$ and Euler's formula to get, since the $\cos$ terms cancel, that
$$\begin{equation}\begin{aligned} \sin^5\theta & = \frac{1}{(2i)^{5}}(e^{i(5\theta)} - 5e^{i(3\theta)} + 10e^{i\theta} - 10e^{-i\theta} + 5e^{-i(3\theta)} - e^{-i(5\theta)}) \\ & = \frac{1}{32i}(i(2\sin 5\theta) + 5i(2\sin 3\theta) + 10i(2\sin\theta))) \\ & = \frac{1}{16}(\sin 5\theta + 5\sin 3\theta + 10\sin\theta) \end{aligned}\end{equation}\tag{17}\label{eq17A}$$
I didn't mention this initially since I didn't want to assume you have already studied or know about complex numbers.