prove that :
cos 570o sin 510o + sin (-330o) cos (-390o) = 0
LHS = cos 570o sin 510o + sin (-330o) cos (-390o)
We know that sin (-x) = -sin (x) and cos (-x) = +cos (x).
= cos 570o sin 510o + [-sin (330o)] cos (390o)
= cos 570o sin 510o - sin (330o) cos (390o)
= cos (90° × 6 + 30°) sin (90° × 5 + 60°) – sin (90° × 3 + 60°) cos (90° × 4 + 30°)
We know that cos is negative at 90° + θ i.e. in Q2 and when n is odd, sin → cos and cos → sin.
= -cos 30° cos 60° - [-cos 60°] cos 30°
= -cos 30° cos 60° + cos 60° cos 30°
= 0
= RHS
Hence proved.