Prove that $\sin(12^\circ)\sin(48^\circ)\sin(54^\circ)=\frac18$ Prove that $$\sin(12^\circ)\sin(48^\circ)\sin(54^\circ)=\frac18.$$ Without using a calculator. I tried all identities I know but I have no idea how to proceed. I always get stuck on finding $\sin36^\circ$.

$2\sin 12\cdot \sin 48 = \cos (48 - 12) - \cos (48 + 12) = \cos 36 - \dfrac{1}{2}$. So the problem is to find $\cos 36$. Use $1 - 2x^2 = 3x - 4x^3$ to solve for $\sin 18$ ( not hard ), then find $\cos 36$, and $\sin 54$. In fact, the equation is: $4x^3 - 2x^2 - 3x + 1 = 0 \to (x-1)(4x^2 + 2x - 1) = 0$, so $\sin 18 = x = \dfrac{\sqrt{5} - 1}{4}$. You can take it from here.
