Sin 3A Formula – Proof and Examples

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

Let’s begin –

Sin 3A Formula

The formula of sin 3A is \(3 sin A – 4 sin^3 A\).

Proof :

We have,

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

Replacing B by 2A,

\(\implies\) sin 3A = sin A cos 2A + cos A sin 2A

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

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

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

\(\implies\)  sin 3A = 3 sin A – \(4 sin^3 A\)

Hence, sin 3A = 3 sin A – \(4 sin^3 A\)

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

sin A = \(3 sin {A\over 3}\) – \(4 sin^3 {A\over 3}\)

Example : Prove that : \(6 sin {\pi\over 9}\) – \(8 sin^3 {\pi\over 3}\) = \(\sqrt{3}\).

Solution : We have,

L.H.S = 2(\(3 sin {\pi\over 9}\) – \(4 sin^3 {\pi\over 3}\)) = \(2 sin (3 \times {\pi\over 9})\)

L.H.S = \(2 sin {\pi\over 3}\) = \(\sqrt{3}\) = R.H.S

Example : Prove that sin A sin (60 – A) sin (60 + A) = \(1\over 4\) sin 3A.

Solution : We have,

L.H.S = sin A sin (60 – A) sin (60 + A)

\(\implies\) L.H.S = sin A (\(sin^2 60 – sin^2 A\))

[ By using this formula, sin (A + B) sin (A – B) = \(sin^2 A\) – \(sin^2 B\)  above ]

\(\implies\) L.H.S = sin A (\(3\over 4\) – \(sin^2 A\)) = \(1\over 4\) sin A \((3 – 4 sin^2 A)\)

\(\implies\) L.H.S = \(1\over 4\)(3 sin A – \(4 sin^3 A\))

Since 3 sin A – \(4 sin^3 A\) = sin 3A,

\(\implies\) L.H.S = \(1\over 4\) sin 3A = R.H.S

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