# Cos 3A Formula – Proof and Examples

Here you will learn what is the formula of cos 3A with proof and examples based on it.

Let’s begin –

## Cos 3A Formula

The formula of cos 3A is $$4 cos^3 A – 3 cos A$$.

Proof :

We have,

cos (A + B) = cos A cos B – sin A sin B

Replacing B by 2A,

$$\implies$$ cos 3A = cos A cos 2A – sin A sin 2A

$$\implies$$ cos 3A = cos A ($$2 cos^2 A – 1$$) + sin A (2 sin A cos A)

[ $$\because$$   cos 2A = $$2 cos^2 A – 1$$ & sin 2A = 2 sin A cos A ]

$$\implies$$  cos 3A = $$2 cos^3 A$$ – cos A + 2 cos A ($$sin^2 A$$)

$$\implies$$  cos 3A = $$2 cos^3 A$$ – cos A + 2 cos A ($$1 – cos^2 A$$)

Hence, cos 3A = $$4 cos^3 A$$ – 3 cos A

We can also write above relation of angle A in terms of angle A/3, just replace A by A/3, we get

cos A = $$4 cos^3 {A\over 3}$$ – $$3 cos {A\over 3}$$

Example : Prove that : $$8 cos^3 {\pi\over 3}$$ – $$6 sin {\pi\over 9}$$ = 1.

Solution : We have,

L.H.S = 2($$8 cos^3 {\pi\over 3}$$ – $$6 sin {\pi\over 9}$$) = $$2 cos (3 \times {\pi\over 9})$$

L.H.S = $$2 cos {\pi\over 3}$$ = 1 = R.H.S

Example : 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 ($${-3\over 4} + cos^2 A$$)

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