$\displaystyle \sin80^{\circ}=4\cdot\sin20^{\circ}\cdot\sin50^{\c irc}\cdot\sin70^{\circ} $

Of course it is correct , by writing $\displaystyle \sin{70^o} = \cos{20^o}~ ~,~ \sin{50^o} = \cos{40^o} $

$\displaystyle 4\cdot\sin20^{\circ}\cdot\sin50^{\circ}\cdot\sin70 ^{\circ} = (2\sin{20^o}\cos{20^o}) (2\cos{\color{red}50\color{black}^o}) $

$\displaystyle = 2\sin{40^o}\cos{40^o} = \sin{80^o} $

It is a nice trigonometric solution to the problem , mine is a pure geometrical solution , see if you can find any

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