LHS is

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

But we know,

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

⇒ sin 5x = sin 3x cos 2x+cos 3x sin 2x

⇒ sin 5x = sin (2x+x) cos 2x+cos (2x+x) sin 2x……..(ii)

And

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

Now 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 5x = sin 2x cos 2x cos x+cos² 2x sin x+(sin 2x cos 2x cos x – sin² 2x sin x)

⇒ sin 5x = 2sin 2x cos 2x cos x+cos² 2x sin x– sin² 2x sin x …….(iv)

Now sin 2x = 2sin x cos x………(v)

And cos 2x = cos²x –sin²x………(vi)

Substituting equation (v) and (vi) in equation (iv), we get

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

⇒ sin 5x = 4(sin x cos² x)([1–sin²x] –sin²x)+([1–sin²x]–sin²x)²sin 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)²sin x–4sin³ x [1–sin²x]

⇒ sin 5x = 4sin x(1–sin²x)(1–2sin²x)+(1–4sin²x+4sin⁴x)sin x–4sin³ x +4sin⁵x

⇒ sin 5x = (4sin x–4sin³x)( 1–2sin²x) +sin x–4sin³x+4sin⁵x–4sin³ x +4sin⁵x

⇒ sin 5x = 4sin x–8sin³x–4sin³x+8sin⁵x+sin x–8sin³x+8sin⁵x

⇒ sin 5x = 5sin x–20sin³x+16sin⁵x

Hence LHS = RHS

[Hence proved]