Let us consider the LHS

sin 5x

Then,

sin 5x = sin(3x + 2x)

But as we know,

Sin(x + y) = sin x cos y + cos x sin y…..(i)

Therefore,
sin 5x = sin 3x cos 2x + cos 3x sin 2x

= sin(2x + x) cos 2x + cos (2x + x) sin 2x……..(ii)

And

cos(x + y) = cos x cos y – sin x sin y……(iii)

Then, substituting equation (i) and (iii) in equation (ii), we get

sin 5x = (sin 2x cos x + cos 2x sin x) cos 2x + (cos 2x cos x – sin 2x sin x) sin 2x

= sin 2x cos 2x cos x + cos² 2x sin x + (sin 2x cos 2x cos x – sin² 2x sin x)

= 2sin 2x cos 2x cos x + cos² 2x sin x – sin² 2x sin x …….(iv)

Then sin 2x = 2sin x cos x………(v)

And cos 2x = cos²x – sin²x………(vi)

By substituting equation (v) and (vi) in equation (iv), we get

sin 5x = 2(2sin x cos x) (cos²x – sin²x) cos x + (cos²x – sin²x)² sin x – (2sin x cos x)² sin x

= 4(sin x cos² x) ([1 – sin²x] – sin²x) + ([1 – sin²x] – sin²x)² sin x – (4sin² x cos² x)sin x

(as cos²x + sin²x = 1 ⇒ cos²x = 1 – sin²x)

sin 5x = 4(sin x [1 – sin²x]) (1 – 2sin²x) + (1 – 2sin²x)² sin x – 4sin³ x [1 – sin²x]

= 4sin x (1 – sin²x) (1 – 2sin² x) + (1 – 4sin²x + 4sin⁴x) sin x – 4sin³ x + 4sin⁵x

= (4sin x – 4sin³x) (1 – 2sin²x) + sin x – 4sin³x + 4sin⁵x – 4sin³ x + 4sin⁵x

= 4sin x – 8sin³x – 4sin³x + 8sin⁵x + sin x – 8sin³x + 8sin⁵x

= 5sin x – 20sin³x + 16sin⁵x

= RHS

Thus proved.