prove that :
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LHS = cos 570^o sin 510^o + sin (-330^o) cos (-390^o)

We know that sin (-x) = -sin (x) and cos (-x) = +cos (x).

= cos 570^o sin 510^o + [-sin (330^o)] cos (390^o)

= cos 570^o sin 510^o - sin (330^o) cos (390^o)

= 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.