# 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