Prove the given trigonometric expression:
\[\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{60}^{\circ }}\sin {{80}^{\circ }}=\dfrac{3}{16}\]

Hint: At first convert the value of \[\sin {{60}^{\circ }}\] as \[\dfrac{\sqrt{3}}{2}\] as it is a value of the standard angle. Then use the product sum rules which are \[2\sin A\sin B=\cos \left( B-A \right)-\cos \left( A+B \right)\]and \[2\cos A\sin B=\sin \left( A+B \right)-\sin \left( A-B \right)\] Then after all this, finally use the identity \[\sin \left( \pi -\theta \right)=\sin \theta \] to get the desired results.

Complete step by step answer:
We are given the expression \[\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{60}^{\circ }}\sin {{80}^{\circ }}\] and we have to prove its product is \[\dfrac{3}{16}\]. So at first, we will write the expression as it is
\[\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{60}^{\circ }}\sin {{80}^{\circ }}\]
We can further rearrange and write it as,
\[\sin {{60}^{\circ }}\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{20}^{\circ }}\]
Now, by the use of standard values of the standard angles, \[\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}\], we can write the expression as,
\[\dfrac{\sqrt{3}}{2}\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{80}^{\circ }}\]
Now we will multiply and divide 2 and write it as,
\[\dfrac{\sqrt{3}}{4}\left( 2\sin {{20}^{\circ }}\sin {{40}^{\circ }} \right)\sin {{80}^{\circ }}\]
Now, we will apply the identity, 2 sin A sin B = cos (B – A) – cos (B + A).
\[2\sin {{20}^{\circ }}\sin {{40}^{\circ }}=\cos \left( {{40}^{\circ }}-{{20}^{\circ }} \right)-\cos \left( {{40}^{\circ }}+{{20}^{\circ }} \right)\]
So, the expression holds as,
\[\dfrac{\sqrt{3}}{4}\left[ \cos {{20}^{\circ }}-\cos {{60}^{\circ }} \right]\sin {{80}^{\circ }}\]
We know that value of \[\cos {{60}^{\circ }}\text{ is }\dfrac{1}{2}\]. So, we can write it as,
\[\dfrac{\sqrt{3}}{4}\left( \cos {{20}^{\circ }}-\dfrac{1}{2} \right)\sin {{80}^{\circ }}\]
Now, let’s expand.
\[\dfrac{\sqrt{3}}{4}\cos {{20}^{\circ }}\sin {{80}^{\circ }}-\dfrac{\sqrt{3}}{8}\sin {{80}^{\circ }}\]
Now, let’s rewrite the expression as
\[\dfrac{\sqrt{3}}{8}\left( 2\cos {{20}^{\circ }}\sin {{80}^{\circ }} \right)-\dfrac{\sqrt{3}}{8}\sin {{80}^{\circ }}\]
Here, we will apply the identity, 2 cos A sin B = sin (A + B) – sin (A – B).
\[2\cos {{20}^{\circ }}\sin {{80}^{\circ }}=\sin \left( {{20}^{\circ }}+{{80}^{\circ }} \right)-\sin \left( {{20}^{\circ }}-{{80}^{\circ }} \right)\]
\[2\cos {{20}^{\circ }}\sin {{80}^{\circ }}=\sin \left( {{100}^{\circ }} \right)-\sin \left( -{{60}^{\circ }} \right)\]
We know that the value of \[\sin \left( -{{60}^{\circ }} \right)\text{ as }-\sin {{60}^{\circ }}\] which is equal to \[\dfrac{-\sqrt{3}}{2}\].
Hence, we get the expression as,
\[\dfrac{\sqrt{3}}{8}\left( \sin {{100}^{\circ }}-\dfrac{-\sqrt{3}}{2} \right)-\dfrac{\sqrt{3}}{8}\sin {{80}^{\circ }}\]
\[\dfrac{\sqrt{3}}{8}\sin {{100}^{\circ }}+\dfrac{\sqrt{3}}{8}\times \dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3}}{8}\sin {{80}^{\circ }}\]
We will apply \[\sin \left( \pi -\theta \right)=\sin \theta \]
\[\sin \left( {{180}^{\circ }}-80 \right)=\sin {{80}^{\circ }}\]
\[\sin {{100}^{\circ }}=\sin {{80}^{\circ }}\]
Here, we get the expression as,
\[\dfrac{\sqrt{3}}{8}\left( \sin {{100}^{\circ }}-\sin {{80}^{\circ }} \right)+\dfrac{3}{16}\]
\[=\dfrac{3}{16}\]

Note:
One can also do the same question by another method. Instead of using the identities one can check out trigonometric table to see the exact values of \[\sin {{20}^{o}},\sin {{40}^{o}},\sin {{60}^{o}},\sin {{80}^{o}}\] and then find their product to get the answer. The process would be very time consuming and will not give accurate answers.