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