We have,
sin12θ+sin4θ
⇒sin2(6θ)+sin2(2θ)
⇒2sin6θcos6θ+2sin2θcos2θ
⇒2sin2(3θ)cos6θ+4sinθcosθcos2θ
⇒4sin3θcos3θcos6θ+4sinθcosθcos2θ
Hence, this is the answer.
Express each of the following as an algebraic sum of sines or cosines :
(i) 2 sin 3x cos 2x
(ii) 2 cos 4x sin 2x
(iii) 2 cos 6x cos 4x
(iv) 2 sin 3x sin 5x