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
