Prove that sin A sin (60 – A) sin (60 + A) = \(1\over 4\) sin 3A.

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

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