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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 $\sin36^\circ$.

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2 Answers

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Use the following: $$\sin\theta \sin\left(\frac{\pi}{3}-\theta\right)\sin\left(\frac{\pi}{3}+\theta\right)=\frac{1}{4}\sin(3\theta)$$ Then you have: $$\sin(12^{\circ})\sin(60^{\circ}-12^{\circ})\sin(60^{\circ}+12^{\circ})=\frac{1}{4}\sin(36^{\circ})$$ $$\Rightarrow \sin(12^{\circ})\sin(48^{\circ})=\frac{1}{4}\frac{\sin(36^{\circ})}{\sin(72^{\circ})}$$ $$\Rightarrow \sin(12^{\circ})\sin(48^{\circ})\sin(54^{\circ})=\frac{1}{4}\frac{\sin(36^{\circ})}{\sin(72^{\circ})}\sin(54^{\circ})=\frac{1}{4}\frac{\sin(36^{\circ})}{\sin(72^{\circ})}\cos(36^{\circ})$$ $$\Rightarrow \sin(12^{\circ})\sin(48^{\circ})\sin(54^{\circ})=\boxed{\dfrac{1}{8}}$$

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$2\sin 12\cdot \sin 48 = \cos (48 - 12) - \cos (48 + 12) = \cos 36 - \dfrac{1}{2}$. 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.

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