# Proof expression:$\cos(80°)*\sin(70°)*\cos(60°)*\sin(50°)=\frac{1}{16}$ [closed]

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?

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Did, José Carlos Santos, Arnaud D., darij grinberg
If this question can be reworded to fit the rules in the help center, please edit the question.

$$\cos80°\sin70°\cos60°\sin50°=\frac{8\sin20^{\circ}\cos20^{\circ}\cos40^{\circ}\cos80^{\circ}}{16\sin20^{\circ}}=$$ $$=\frac{\sin160^{\circ}}{16\sin20^{\circ}}=\frac{1}{16}$$
• I used $2\sin\alpha\cos\alpha=\sin2\alpha$ three times. – Michael Rozenberg Dec 16 '17 at 15:05
• How did you simplify in the first step? That you got: $\frac{8\sin(20)\cos(20)\cos(40)\cos(80)}{16\sin20}$ – Bili Debili Dec 16 '17 at 15:09
• $\sin70^{\circ}=\cos20^{\circ}$, $\sin50^{\circ}=\cos40^{\circ}$ and $\cos60^{\circ}=\frac{1}{2}$. – Michael Rozenberg Dec 16 '17 at 15:15
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.