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