16sin (20 °)sin (40 °) (80 °) is equal to

We know that $sin\theta sin\left(60 - \theta\right)\sin \left(60 + \theta\right) = \dfrac{1}{4}\sin 3\theta$

Now given $16\sin 20 ° \sin 40 ° \sin 80 °$

Comparing with above formula $\theta = 20 °$

we get $16\sin 20 ° \sin 40 ° \sin 80 ° = 16 \times \dfrac{1}{4} \times \sin \left(3 \times 20 °\right)$

$= 16 \times \dfrac{1}{4} \times \sin 60 °$ $= 4 \times \dfrac{\sqrt{3}}{2} = 2\sqrt{3}$