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

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