Proving that $\sin(54°)\sin(66°) = \sin(48°)\sin(96°)$ I'm trying to prove that $\sin(54°)\sin(66°) = \sin(48°)\sin(96°)$ but I don't really have a way to approach it. Most of what I tried was replacing $\sin(2x)$ with $2\sin(x)\cos(x)$ or changing sines with cosines but none of that has really simplified it.
Would appreciate any solution. 

As $\sin54^\circ=\cos36^\circ$  and $$\sin48^\circ=2\cos24^\circ\sin24^\circ$$
the proposition reduces to $$\cos36^\circ=2\sin24^\circ\sin96^\circ=\cos(96-24)^\circ-\cos(96+24)^\circ=\cos72^\circ-\left(-\dfrac12\right)$$
Now use How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?.
