Correct Answer - Option 1 : 1/16
We know that, the trigonometry identity is:
\(sin\left( {60^\circ + A} \right)\cdot sin\left( {60^\circ - A} \right)sinA = \frac{1}{4}sin3A\)
Now, applying it in the given equation,
\(\Rightarrow sin\;70^\circ {\rm{\;}}sin\;50^\circ {\rm{\;}}sin\;10^\circ = \frac{1}{4}\sin \left( {3 \times 10^\circ } \right)\)
\(\Rightarrow sin\;70^\circ {\rm{\;}}sin\;50^\circ {\rm{\;}}sin\;10^\circ = \frac{1}{4}sin\;30^\circ\)
Now,
⇒ (sin 70° sin 50° sin 10°)(sin30°)
\(= \frac{1}{4}\left( {sin\;30^\circ } \right)\left( {sin\;30^\circ } \right)\)
\(\Rightarrow sin\;70^\circ {\rm{\;}}sin\;50^\circ {\rm{\;}}sin\;10^\circ {\rm{\;}}sin\;30^\circ = \frac{1}{4}si{n^2}{\rm{\;}}30^\circ\)
∵ \(\sin 30^\circ = \frac{1}{2}\)
\(\Rightarrow sin\;70^\circ {\rm{\;}}sin\;50^\circ {\rm{\;}}sin\;10^\circ {\rm{\;}}sin\;30^\circ = \frac{1}{4} \times {\left( {\frac{1}{2}} \right)^2}\)
\(\Rightarrow sin\;70^\circ {\rm{\;}}sin\;50^\circ {\rm{\;}}sin\;10^\circ {\rm{\;}}sin\;30^\circ = \frac{1}{4} \times \frac{1}{4}\)
\(\therefore sin\;70^\circ {\rm{\;}}sin\;50^\circ {\rm{\;}}sin\;10^\circ {\rm{\;}}sin\;30^\circ = \frac{1}{{16}}\)
