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Proof expression: $$\cos(80^\circ)\sin(70^\circ)\cos(60^\circ)\sin(50^\circ)=\frac{1}{16}$$

I tried in different ways, but I always get $\sin(20^\circ)$ in expression, which I can't simplify. What is the fastest way to prove this?

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closed as off-topic by TheSimpliFire, Did, José Carlos Santos, Arnaud D., darij grinberg Jan 15 at 20:55

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$$\cos80°\sin70°\cos60°\sin50°=\frac{8\sin20^{\circ}\cos20^{\circ}\cos40^{\circ}\cos80^{\circ}}{16\sin20^{\circ}}=$$ $$=\frac{\sin160^{\circ}}{16\sin20^{\circ}}=\frac{1}{16}$$

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  • $\begingroup$ How did you get to the second step? $\endgroup$ – Bili Debili Dec 16 '17 at 15:04
  • $\begingroup$ I used $2\sin\alpha\cos\alpha=\sin2\alpha$ three times. $\endgroup$ – Michael Rozenberg Dec 16 '17 at 15:05
  • $\begingroup$ How did you simplify in the first step? That you got: $\frac{8\sin(20)\cos(20)\cos(40)\cos(80)}{16\sin20}$ $\endgroup$ – Bili Debili Dec 16 '17 at 15:09
  • $\begingroup$ $\sin70^{\circ}=\cos20^{\circ}$, $\sin50^{\circ}=\cos40^{\circ}$ and $\cos60^{\circ}=\frac{1}{2}$. $\endgroup$ – Michael Rozenberg Dec 16 '17 at 15:15
  • $\begingroup$ How did you get sin(20°) in denominator? $\endgroup$ – Bili Debili Dec 16 '17 at 15:17
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As @labbhattacharjee suggested, use the link with $x=20^{\circ}$ so that $$\cos(20^{\circ})\cos(-40^{\circ})\cos(80^{\circ})=\sin(70^{\circ})\sin(50^{\circ})\cos(80^{\circ})=\frac14 \cos(3\cdot 20^{\circ})=\frac18$$ Hence $$\cos(80^\circ)\sin(70^\circ)\cos(60^\circ)\sin(50^\circ)=\frac12 \cdot \frac18 =\frac{1}{16}$$ as required.

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