Lời giải

Ta có: sin 3x = sin (2x + x) = sin 2x . cos x + cos 2x . sin x

= (2sin x. cos x) . cos x + (1 – 2sin²x) . sin x

= 2sin x. cos² x + sin x – 2 sin³x

= 2sin x . (1 – sin²x) + sin x – 2 sin³x

= 2sin x – 2 sin³x + sin x – 2 sin³x

= 3sin x – 4sin³x

Vậy sin 3x = 3sin x – 4sin³x

Ta có: cos 3x = cos (2x + x) = cos 2x . cos x – sin 2x . sin x

= (–1 + 2cos²x) . cos x – 2cos x . sin x . sin x

= – cos x + 2cos³ x – 2cos x . sin² x

= – cos x + 2cos³ x – 2cos x . (1 – cos² x)

= – cos x + 2cos³ x – 2cos x + 2cos³ x

= 4cos³x – 3cos x

Vậy cos 3x = 4cos³x – 3cos x .