Question

Find the value of sin 3x, only in form of sin.

Answer 1

To find sin 3x :

= sin(2x+x)

By using identity of sin(A+B),

= sin2x*cosx + sinx*cos2x

= 2 sinx. cosx. cosx + sinx. (1-2sin²x)

= 2sinx. cos²x + sinx - 2sin³x

= 2sinx(1-sin²x) + sinx - 2sin³x

= 2sinx - 2sin³x + sinx - 2sin³x

= 3sinx - 4sin³x.

Answer 2

Answer:

We know that :

sin ( 3 x ) can be written as sin ( 2 x + x )

We also know that :

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

⇒ sin ( 2 x + x ) = sin 2 x cos x + cos 2 x sin x

We know that cos 2 x can be written as 1 - 2 sin²x :

⇒ sin ( 3 x ) = sin 2 x cos x + ( 1 - 2 sin²x ) sin x

⇒ sin ( 3 x ) = sin 2 x cos x + sin x - 2 sin³x

We know that sin 2 x can be written as 2 sin x cos x :

⇒ sin ( 3 x ) = 2 sin x cos x . cos x + sin x - 2  sin³x

⇒ sin ( 3 x ) = 2 sin x cos²x + sin x - 2 sin³x

⇒ sin ( 3 x ) = 2 sin x ( 1 - sin²x ) + sin x - 2 sin³x

⇒ sin ( 3 x ) = 2 sin x - 2 sin³x + sin x - 2 sin³x

⇒ sin ( 3 x ) = 3 sin x - 4 sin³x

Another notable formula used is :

cos²x = 1 - sin²x