Solution :
We have,
L.H.S = sin A sin (60 – A) sin (60 + A)
\(\implies\) L.H.S = sin A (\(sin^2 60 – sin^2 A\))
[ By using this formula, sin (A + B) sin (A – B) = \(sin^2 A\) – \(sin^2 B\) above ]
\(\implies\) L.H.S = sin A (\(3\over 4\) – \(sin^2 A\)) = \(1\over 4\) sin A \((3 – 4 sin^2 A)\)
\(\implies\) L.H.S = \(1\over 4\)(3 sin A – \(4 sin^3 A\))
Since 3 sin A – \(4 sin^3 A\) = sin 3A,
\(\implies\) L.H.S = \(1\over 4\) sin 3A = R.H.S
