Prove that : $\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = \dfrac{3}{{16}}$

Hint: Use trigonometry identity $2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right)$
and $2\cos A\sin B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right)$

Here we have to prove Left hand side(LHS) equal to Right hand side(RHS).
Let’s take a left hand side of the question.
  $LHS = \sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ $
As we know the value of $\sin 60^\circ = \dfrac{{\sqrt 3 }}{2}$
$ \Rightarrow \dfrac{{\sqrt 3 }}{2}\left( {\sin 20^\circ \sin 40^\circ \sin 80^\circ } \right)$
Now, multiply by 2 in numerator and denominator
$
   \Rightarrow \dfrac{{\sqrt 3 }}{4}\left( {2\sin 20^\circ \sin 40^\circ \sin 80^\circ } \right) \\
   \Rightarrow \dfrac{{\sqrt 3 }}{4}\left( {\left( {2\sin 20^\circ \sin 40^\circ } \right)\sin 80^\circ } \right) \\
 $
Use identity $2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right)$
\[
   \Rightarrow \dfrac{{\sqrt 3 }}{4}\left( {\left( {\cos \left( {40^\circ - 20^\circ } \right) - \cos \left( {40^\circ + 20^\circ } \right)} \right)\sin 80^\circ } \right) \\
   \Rightarrow \dfrac{{\sqrt 3 }}{4}\left( {\left( {\cos \left( {20^\circ } \right) - \cos \left( {60^\circ } \right)} \right)\sin 80^\circ } \right) \\
  \therefore \cos 60^\circ = \dfrac{1}{2} \\
   \Rightarrow \dfrac{{\sqrt 3 }}{4}\left( {\cos 20^\circ \sin 80^\circ } \right) - \dfrac{{\sqrt 3 }}{8}\sin 80^\circ \\
   \Rightarrow \dfrac{{\sqrt 3 }}{8}\left( {2\cos 20^\circ \sin 80^\circ } \right) - \dfrac{{\sqrt 3 }}{8}\sin 80^\circ \\
 \]
Use identity $2\cos A\sin B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right)$
 \[
   \Rightarrow \dfrac{{\sqrt 3 }}{8}\left( {\sin 100^\circ + \sin 60^\circ } \right) - \dfrac{{\sqrt 3 }}{8}\sin 80^\circ \\
   \Rightarrow \dfrac{{\sqrt 3 }}{8}\sin 60^\circ + \dfrac{{\sqrt 3 }}{8}\sin 100^\circ - \dfrac{{\sqrt 3 }}{8}\sin 80^\circ \\
 \]
As we know $\sin \left( {180^\circ - A} \right) = \sin \left( A \right)$
\[
   \Rightarrow \dfrac{{\sqrt 3 }}{8}\sin 60^\circ + \dfrac{{\sqrt 3 }}{8}\sin \left( {180^\circ - 80^\circ } \right) - \dfrac{{\sqrt 3 }}{8}\sin 80^\circ \\
   \Rightarrow \dfrac{{\sqrt 3 }}{8}\sin 60^\circ + \dfrac{{\sqrt 3 }}{8}\sin \left( {80^\circ } \right) - \dfrac{{\sqrt 3 }}{8}\sin 80^\circ \\
   \Rightarrow \dfrac{{\sqrt 3 }}{8}\sin 60^\circ \\
   \Rightarrow \dfrac{{\sqrt 3 }}{8} \times \dfrac{{\sqrt 3 }}{2} \\
   \Rightarrow LHS = \dfrac{3}{{16}} \\
 \]
So, $LHS = \dfrac{3}{{16}} = RHS$
Hence proved, $\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = \dfrac{3}{{16}}$

Note:Whenever we come across these types of problems first substitute the values of known trigonometric angles and collect the rest of trigonometric terms then use the product to sum formulas to convert unknown trigonometric angles to known trigonometric angles .