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
