Improving my way of showing $\sin^212^\circ+\sin^221^\circ+\sin^239^\circ+\sin^248^\circ=1+\sin^29^\circ+\sin^218^\circ$

Question

This problem is from 1904 and was given to students studying for the Cambridge and Oxford entry examinations. My solution is presented below, but I am of the opinion that it can be improved. All ideas welcome.

Show that $$\sin^{2}{12^{\circ}}+\sin^{2}{21^{\circ}}+\sin^{2}{39^{\circ}}+\sin^{2}{48^{\circ}}=1+\sin^{2}{9^{\circ}}+\sin^{2}{18^{\circ}}$$

A solution

$$\begin{align} \sin^{2}{12^{\circ}}=\sin^{2}{(30^{\circ}-18^{\circ})} &=(\sin{30^{\circ}}\cos{18^{\circ}}-\cos{30^{\circ}}\sin{18^{\circ}})^{2} \tag1\\ &=\left(\frac{1}{2}\cos{18^{\circ}}-\frac{\sqrt{3}}{2}\sin{18^{\circ}}\right)^{2} \tag2\\ &=\frac{1}{4}\cos^{2}{18^{\circ}}+\frac{3}{4}\sin^{2}{18^{\circ}}-\frac{\sqrt{3}}{2}\cos{18^{\circ}}\sin{18^{\circ}} \tag3 \\ \\ \\ \sin^{2}{48^{\circ}} &=\sin^{2}{(30^{\circ}+18^{\circ})} \tag4 \\ &= (\sin{30^{\circ}}\cos{18^{\circ}}+\cos{30^{\circ}}\sin{18^{\circ}})^{2} \tag5 \\ &=\left(\frac{1}{2}\cos{18^{\circ}}+\frac{\sqrt{3}}{2}\sin{18^{\circ}}\right)^{2} \tag6 \\ &=\frac{1}{4}\cos^{2}{18^{\circ}}+\frac{3}{4}\sin^{2}{18^{\circ}}+\frac{\sqrt{3}}{2}\cos{18^{\circ}}\sin{18^{\circ}} \tag7 \\ \\ \\ \sin^{2}{21^{\circ}} &=\sin^{2}{(30^{\circ}-9^{\circ})} \tag8\\ &=\left(\sin{30^{\circ}}\cos{9^{\circ}}-\cos{30^{\circ}}\sin{9^{\circ}}\right)^{2} \tag9\\ &=\left(\frac{1}{2}\cos{9^{\circ}}-\frac{\sqrt{3}}{2}\sin{9^{\circ}}\right)^{2} \tag{10}\\ &=\frac{1}{4}\cos^{2}{9^{\circ}}+\frac{3}{4}\sin^{2}{9^{\circ}}-\frac{\sqrt{3}}{2}\cos{9^{\circ}}\sin{9^{\circ}} \tag{11} \\ \\ \\ \sin^{2}{39^{\circ}}=\sin^{2}{(30^{\circ}+9^{\circ})} &=(\sin{30^{\circ}}\cos{9^{\circ}}+\cos{30^{\circ}}\sin{9^{\circ}})^{2} \tag{12} \\ &=(\frac{1}{2}\cos{9^{\circ}}+\frac{\sqrt{3}}{2}\sin{9^{\circ}})^{2} \tag{13}\\ &=\frac{1}{4}\cos^{2}{9^{\circ}}+\frac{3}{4}\sin^{2}{9^{\circ}}+\frac{\sqrt{3}}{2}\cos{9^{\circ}}\sin{9^{\circ}} \tag{14} \end{align}$$

So, $$\begin{align} \text{LHS} &=\frac{3}{2}\sin^{2}{18^{\circ}}+\frac{1}{2}\cos^{2}{^18{\circ}}+\frac{3}{2}\sin^{2}{9^{\circ}}+\frac{1}{2}\cos^{2}{9^{\circ}} \tag{15}\\ &=\sin^{2}{18^{\circ}}+sin^{2}{9^{\circ}}+\frac{1}{2}(sin^{2}{18^{\circ}}+\cos^{2}{18^{\circ}})+\frac{1}{2}(sin^{2}{9^{\circ}}+\cos^{2}{9^{\circ}}) \tag{16} \\ &=\sin^{2}{9^{\circ}}+\sin^{2}{18^{\circ}}+\frac{1}{2}+\frac{1}{2} \tag{17} \\ &=1+\sin^{2}{9^{\circ}}+\sin^{2}{18^{\circ}} \tag{18} \\ &=RHS \tag{19} \end{align}$$

Answer 1

Well, I think the most obvious improvement would be to establish a general identity and then use it in the specific cases, rather than to redo the same computation four times:

$$\begin{align} \sin^2 (x+y) + \sin^2 (x-y) &= (\sin x \cos y + \cos x \sin y)^2 + (\sin x \cos y - \cos x \sin y)^2 \\ &= 2 \left( \sin^2 x \cos^2 y + \cos^2 x \sin^2 y \right) \tag{1} \\ \end{align}$$ so that with $x = 30^\circ$, $$\sin^2 (30^\circ + y) + \sin^2 (30^\circ - y) = \frac{\cos^2 y + 3 \sin^2 y}{2} = \frac{1}{2} + \sin^2 y. \tag{2}$$ Now substituting $y = 9^\circ$ and $y = 18^\circ$, we immediately find the LHS of the claimed identity equals $$1 + \sin^2 9^\circ + \sin^2 18^\circ.$$

Answer 2

As an alternative, using that $\sin^2 \theta = \frac{1-\cos(2\theta)}{2}$ we obtain the equivalent

$$\cos 18^\circ-\cos 24^\circ+\cos 36^\circ-\cos 42^\circ-\sin 12^\circ+\sin 6^\circ=0$$

and by sum to product identities applied to $\cos 18^\circ-\cos 24^\circ$, $\cos 36^\circ-\cos 42^\circ$ and $-\sin 12^\circ+\sin 6^\circ$ we obtain

$$\require{cancel} \cancel 2\sin 21^\circ \cancel{\sin 3^\circ}+\cancel2\sin 39^\circ \cancel{\sin 3^\circ}-\cancel2\cos 9^\circ \cancel{\sin 3^\circ} =0$$

and again by sum to product identities

$$2 \sin 30^\circ {\cos 9^\circ}-{\cos 9^\circ}=0$$ $$\iff \sin 30^\circ=\frac12$$

which is true.