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

Solution :

We have,

L.H.S = cos A cos (60 – A) cos (60 + A)

\(\implies\) L.H.S = cos A (\(cos^2 60 – sin^2 A\))

[ By using this formula, cos (A + B) cos (A – B) = \(cos^2 A\) – \(sin^2 B\)  above ]

\(\implies\) L.H.S = cos A (\(1\over 4\) – \(sin^2 A\)) = cos A \(({1\over 4} – (1 – cos^2 A))\)

\(\implies\) L.H.S = cos A (\({-3\over 4} + cos^2 A\))

\(\implies\) L.H.S = \(1\over 4\) cos A (\(-3 + 4 cos^2 A\)) = \(1\over 4\)(\(4 cos^3 A\) – 3 cos A)

Since \(4 cos^3 A\) – 3 cos A = cos 3A,

\(\implies\) L.H.S = \(1\over 4\) cos 3A = R.H.S

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