Proving $\,\sin^272^\circ - \sin^260^\circ = \frac18\left(\sqrt5 - 1\right)$ I'm fairly new into proving trigonometric equations, but I believe I'm getting better at it day after day. Yet, I have stumbled upon one such which left me at an impasse. The problem is presented as follows:
$$\sin^272^\circ - \sin^260^\circ = \dfrac{\sqrt5 - 1}8$$
The goal is to prove the truth of the so-presented equation. I first complicated the right-hand side of the equation to $\dfrac{\sqrt5-1}4\cdot\dfrac12$ to then simplify it into $\sin18^\circ\cdot\sin30^\circ$ (knowing, of course, that $\dfrac{\sqrt5-1}4$ is equal to $\sin18^\circ$, and that $\dfrac12$ is equal to $\sin30^\circ$).
I then proceeded to work the proof from the left-hand side, now that both sides of the equation are termed in trigonometric ratios. The left-hand side is easily noticed to be a difference of squares, hence being apt for complication as $(\sin72^\circ - \sin60^\circ)(\sin72^\circ + \sin60^\circ)$. It is only a coincidence to notice that the expressions can be further simplified knowing the sum/difference to product rule (i.e. that $\sin A + \sin B = 2\sin\dfrac{A + B}2\cos\dfrac{A - B}2$, and also that  $\sin A - \sin B =2\cos\dfrac{A + B}2\sin\dfrac{A - B}2$).
Executing these simplifications, we get the next result: $2\sin66^\circ\cdot\cos66^\circ\cdot2\sin6^\circ\cdot\cos6^\circ$. It is evident to see how the double angle rules could be of use here (that $2\sin\theta\cos\theta =\sin2\theta$). The penultimate result with this rule applied is thus, so far: $\sin132^\circ\cdot\sin12^\circ$. I can only go as far as to simply right the ratios in terms of angles less that $45$ degrees, which is: $\cos42^\circ\cdot\sin12^\circ$.
For the most of it, I am stuck; and any rules I apply to simplify either expression gets me to a result previously acquired in the run. How am I to finish proving that $\cos42^\circ\cdot\sin12^\circ$ is equal to $\sin18^\circ\cdot\sin30^\circ$, or equivalently $\sin18^\circ\cdot\cos60^\circ$. This problem can be found in S.L Loney's book on plane trigonometry, a really helpful book for getting a good picture on trigonometry. Thank you in advance.

Now
$$\sin^2 72°=\cos^2(90°-72°)=\cos^2(18°)=\frac{1+\cos36°}{2}=\frac{1+\frac{\sqrt 5+1}{4}}{2}$$
$$=\frac{5+\sqrt 5}{8}$$
Thus
$$\sin^2 60°=\frac 34$$
Hence
$$\color{orange}{\sin^272^\circ - \sin^260^\circ} = \frac{5+\sqrt 5}{8}-\frac 34=\frac{5+\sqrt 5-6}{8}=\color{red}{\frac18\left(\sqrt5 - 1\right)}$$
